Linear Systems

Solving Linear Systems

Systems of linear equations are common in science and mathematics. These two examples from high school science [Onan] give a sense of how they arise.

The first example is from Statics. Suppose that we have three objects, we know that one has a mass of 2 kg, and we want to find the two unknown masses. Experimentation with a meter stick produces these two balances.

A balanced metre stick: masses h and c are 40 and 15 units to the left of the pivot; a mass of 2 is 50 units to the right. A balanced metre stick: mass c is 25 units to the left of the pivot; a mass of 2 is 25 units and mass h is 50 units to the right.

For the masses to balance we must have that the sum of moments on the left equals the sum of moments on the right, where the moment of an object is its mass times its distance from the balance point. That gives a system of two linear equations.

$\begin{aligned} 40 h + 15 c & = 100 \\ 25 c & = 50 + 50 h \end{aligned}$

The second example is from Chemistry. We can mix, under controlled conditions, toluene $C_{7} H_{8}$ and nitric acid $H N O_{3}$ to produce trinitrotoluene $C_{7} H_{5} O_{6} N_{3}$ along with the byproduct water (conditions have to be very well controlled—trinitrotoluene is better known as TNT). In what proportion should we mix them? The number of atoms of each element present before the reaction

$x C_{7} H_{8} + y H N O_{3} ⟶ z C_{7} H_{5} O_{6} N_{3} + w H_{2} O$

must equal the number present afterward. Applying that in turn to the elements C, H, N, and O gives this system.

$\begin{aligned} 7 x & = 7 z \\ 8 x + 1 y & = 5 z + 2 w \\ 1 y & = 3 z \\ 3 y & = 6 z + 1 w \end{aligned}$

Both examples come down to solving a system of equations. In each system, the equations involve only the first power of each variable. This chapter shows how to solve any such system of equations.

Gauss’s Method

Definition 1.1 A linear combination of $x_{1}$ , …, $x_{n}$ has the form

$a_{1} x_{1} + a_{2} x_{2} + a_{3} x_{3} + \dots + a_{n} x_{n}$

where the numbers $a_{1}, \dots, a_{n} \in ℝ$ are the combination’s coefficients. A linear equation in the variables $x_{1}$ , …, $x_{n}$ has the form $a_{1} x_{1} + a_{2} x_{2} + a_{3} x_{3} + \dots + a_{n} x_{n} = d$ where $d \in ℝ$ is the constant.

An $n$ -tuple $(s_{1}, s_{2}, \dots, s_{n}) \in ℝ^{n}$ is a solution of, or satisfies, that equation if substituting the numbers $s_{1}$ , …, $s_{n}$ for the variables gives a true statement: $a_{1} s_{1} + a_{2} s_{2} + \dots + a_{n} s_{n} = d$ . A system of linear equations

$\begin{array}{rcrcrcrcr} a_{1, 1} x_{1} & + & a_{1, 2} x_{2} & + & \dots & + & a_{1, n} x_{n} & = & d_{1} \\ a_{2, 1} x_{1} & + & a_{2, 2} x_{2} & + & \dots & + & a_{2, n} x_{n} & = & d_{2} \\ ⋮ \\ a_{m, 1} x_{1} & + & a_{m, 2} x_{2} & + & \dots & + & a_{m, n} x_{n} & = & d_{m} \end{array}$

has the solution $(s_{1}, s_{2}, \dots, s_{n})$ if that $n$ -tuple is a solution of all of the equations.

Example 1.2 The combination $3 x_{1} + 2 x_{2}$ of $x_{1}$ and $x_{2}$ is linear. The combination $3 x_{1}^{2} + 2 x_{2}$ is not a linear function of $x_{1}$ and $x_{2}$ , nor is $3 x_{1} + 2 \sin (x_{2})$ .

We usually take $x_{1}$ , …, $x_{n}$ to be unequal to each other because in a sum with repeats we can rearrange to make the elements unique, as with $2 x + 3 y + 4 x = 6 x + 3 y$ . We sometimes include terms with a zero coefficient, as in $x - 2 y + 0 z$ , and at other times omit them, depending on what is convenient.

Example 1.3 The ordered pair $(- 1, 5)$ is a solution of this system.

$\begin{array}{rcrcr} 3 x_{1} & + & 2 x_{2} & = & 7 \\ - x_{1} & + & x_{2} & = & 6 \end{array}$

In contrast, $(5, - 1)$ is not a solution.

Finding the set of all solutions is solving the system. We don’t need guesswork or good luck; there is an algorithm that always works. This algorithm is Gauss’s Method (or Gaussian elimination or linear elimination ).

Example 1.4 To solve this system

$\begin{array}{rcrcrcr} 3 x_{3} & = & 9 \\ x_{1} & + & 5 x_{2} & - & 2 x_{3} & = & 2 \\ \frac{1}{3} x_{1} & + & 2 x_{2} & = & 3 \end{array}$

we transform it, step by step, until it is in a form that we can easily solve.

The first transformation rewrites the system by interchanging the first and third row.

$\begin{aligned} ⟶_{\begin{matrix} \end{matrix}}^{swap row 1 with row 3} \begin{array}{rcrcrcr} \frac{1}{3} x_{1} & + & 2 x_{2} & = & 3 \\ x_{1} & + & 5 x_{2} & - & 2 x_{3} & = & 2 \\ 3 x_{3} & = & 9 \end{array} \end{aligned}$

The second transformation rescales the first row by a factor of $3$ .

$\begin{aligned} ⟶_{\begin{matrix} \end{matrix}}^{multiply row 1 by 3} \begin{array}{rcrcrcr} x_{1} & + & 6 x_{2} & = & 9 \\ x_{1} & + & 5 x_{2} & - & 2 x_{3} & = & 2 \\ 3 x_{3} & = & 9 \end{array} \end{aligned}$

The third transformation is the only nontrivial one in this example. We mentally multiply both sides of the first row by $- 1$ , mentally add that to the second row, and write the result in as the new second row.

$\begin{aligned} ⟶_{\begin{matrix} \end{matrix}}^{add - 1 times row 1 to row 2} \begin{array}{rcrcrcr} x_{1} & + & 6 x_{2} & = & 9 \\ - x_{2} & - & 2 x_{3} & = & - 7 \\ 3 x_{3} & = & 9 \end{array} \end{aligned}$

These steps have brought the system to a form where we can easily find the value of each variable. The bottom equation shows that $x_{3} = 3$ . Substituting $3$ for $x_{3}$ in the middle equation shows that $x_{2} = 1$ . Substituting those two into the top equation gives that $x_{1} = 3$ . Thus the system has a unique solution; the solution set is ${(3, 1, 3)}$ .

We will use Gauss’s Method throughout the book. It is fast and easy. We will now show that it is also safe: Gauss’s Method never loses solutions nor does it ever pick up extraneous solutions, so that a tuple is a solution to the system before we apply the method if and only if it is a solution after.

Theorem 1.5 (Gauss’s Method) If a linear system is changed to another by one of these operations

an equation is swapped with another
an equation has both sides multiplied by a nonzero constant
an equation is replaced by the sum of itself and a multiple of another

then the two systems have the same set of solutions.

Each of the three operations has a restriction. Multiplying a row by $0$ is not allowed because obviously that can change the solution set. Similarly, adding a multiple of a row to itself is not allowed because adding $- 1$ times the row to itself has the effect of multiplying the row by $0$ . And we disallow swapping a row with itself, to make some results in the fourth chapter easier. Besides, it’s pointless.

Proof We will cover the equation swap operation here. The other two cases are similar and are Exercise 1.33.

Consider a linear system.

$\begin{array}{rcrcrcrcr} a_{1, 1} x_{1} & + & a_{1, 2} x_{2} & + & \dots & + & a_{1, n} x_{n} & = & d_{1} \\ ⋮ \\ a_{i, 1} x_{1} & + & a_{i, 2} x_{2} & + & \dots & + & a_{i, n} x_{n} & = & d_{i} \\ ⋮ \\ a_{j, 1} x_{1} & + & a_{j, 2} x_{2} & + & \dots & + & a_{j, n} x_{n} & = & d_{j} \\ ⋮ \\ a_{m, 1} x_{1} & + & a_{m, 2} x_{2} & + & \dots & + & a_{m, n} x_{n} & = & d_{m} \end{array}$

The tuple $(s_{1}, \dots, s_{n})$ satisfies this system if and only if substituting the values for the variables, the $s$ ’s for the $x$ ’s, gives a conjunction of true statements: $a_{1, 1} s_{1} + a_{1, 2} s_{2} + \dots + a_{1, n} s_{n} = d_{1}$ and … $a_{i, 1} s_{1} + a_{i, 2} s_{2} + \dots + a_{i, n} s_{n} = d_{i}$ and … $a_{j, 1} s_{1} + a_{j, 2} s_{2} + \dots + a_{j, n} s_{n} = d_{j}$ and … $a_{m, 1} s_{1} + a_{m, 2} s_{2} + \dots + a_{m, n} s_{n} = d_{m}$ .

In a list of statements joined with ‘and’ we can rearrange the order of the statements. Thus this requirement is met if and only if $a_{1, 1} s_{1} + a_{1, 2} s_{2} + \dots + a_{1, n} s_{n} = d_{1}$ and … $a_{j, 1} s_{1} + a_{j, 2} s_{2} + \dots + a_{j, n} s_{n} = d_{j}$ and … $a_{i, 1} s_{1} + a_{i, 2} s_{2} + \dots + a_{i, n} s_{n} = d_{i}$ and … $a_{m, 1} s_{1} + a_{m, 2} s_{2} + \dots + a_{m, n} s_{n} = d_{m}$ . This is exactly the requirement that $(s_{1}, \dots, s_{n})$ solves the system after the row swap.

QED

Definition 1.6 The three operations from Theorem 1.5 are the elementary reduction operations, or row operations, or Gaussian operations. They are swapping , multiplying by a scalar (or rescaling ), and row combination .

When writing out the calculations, we will abbreviate ‘row $i$ ’ by ‘ $ρ_{i}$ ’ (this is the Greek letter rho, pronounced aloud as “row”). For instance, we will denote a row combination operation by $k ρ_{i} + ρ_{j}$ , with the row that changes written second. To save writing we will often combine addition steps when they use the same $ρ_{i}$ , as in the next example.

Example 1.7 Gauss’s Method systematically applies the row operations to solve a system. Here is a typical case.

$\begin{array}{rcrcrcr} x & + & y & = & 0 \\ 2 x & - & y & + & 3 z & = & 3 \\ x & - & 2 y & - & z & = & 3 \end{array}$

We begin by using the first row to eliminate the $2 x$ in the second row and the $x$ in the third. To get rid of the $2 x$ we mentally multiply the entire first row by $- 2$ , add that to the second row, and write the result in as the new second row. To eliminate the $x$ in the third row we multiply the first row by $- 1$ , add that to the third row, and write the result in as the new third row.

$\begin{aligned} ⟶_{\begin{matrix} - ρ_{1} + ρ_{3} \end{matrix}}^{- 2 ρ_{1} + ρ_{2}} \begin{array}{rcrcrcr} x & + & y & = & 0 \\ - 3 y & + & 3 z & = & 3 \\ - 3 y & - & z & = & 3 \end{array} \end{aligned}$

We finish by transforming the second system into a third, where the bottom equation involves only one unknown. We do that by using the second row to eliminate the $y$ term from the third row.

$⟶_{\begin{array}{l} \end{array}}^{- ρ_{2} + ρ_{3}} \begin{array}{rcrcrcr} x & + & y & = & 0 \\ - 3 y & + & 3 z & = & 3 \\ - 4 z & = & 0 \end{array}$

Now finding the system’s solution is easy. The third row gives $z = 0$ . Substitute that back into the second row to get $y = - 1$ . Then substitute back into the first row to get $x = 1$ .

Example 1.8 For the Physics problem from the start of this chapter, Gauss’s Method gives this.

$\begin{array}{rcrcr} 40 h & + & 15 c & = & 100 \\ - 50 h & + & 25 c & = & 50 \end{array} ⟶_{\begin{array}{l} \end{array}}^{5 / 4 ρ_{1} + ρ_{2}} \begin{array}{rcrcr} 40 h & + & 15 c & = & 100 \\ (175 / 4) c & = & 175 \end{array}$

So $c = 4$ , and back-substitution gives that $h = 1$ . (We will solve the Chemistry problem later.)

Example 1.9 The reduction

$\begin{aligned} \begin{array}{rcrcrcr} x & + & y & + & z & = & 9 \\ 2 x & + & 4 y & - & 3 z & = & 1 \\ 3 x & + & 6 y & - & 5 z & = & 0 \end{array} & ⟶_{\begin{matrix} - 3 ρ_{1} + ρ_{3} \end{matrix}}^{- 2 ρ_{1} + ρ_{2}} \begin{array}{rcrcrcr} x & + & y & + & z & = & 9 \\ 2 y & - & 5 z & = & - 17 \\ 3 y & - & 8 z & = & - 27 \end{array} \\ ⟶_{\begin{matrix} \end{matrix}}^{- (3 / 2) ρ_{2} + ρ_{3}} \begin{array}{rcrcrcr} x & + & y & + & z & = & 9 \\ 2 y & - & 5 z & = & - 17 \\ - (1 / 2) z & = & - (3 / 2) \end{array} \end{aligned}$

shows that $z = 3$ , $y = - 1$ , and $x = 7$ .

As illustrated above, the point of Gauss’s Method is to use the elementary reduction operations to set up back-substitution.

Definition 1.10 In each row of a system, the first variable with a nonzero coefficient is the row’s leading variable . A system is in echelon form if each leading variable is to the right of the leading variable in the row above it, except for the leading variable in the first row, and any rows with all-zero coefficients are at the bottom.

Example 1.11 The prior three examples only used the operation of row combination. This linear system requires the swap operation to get it into echelon form because after the first combination

$\begin{aligned} \begin{array}{rcrcrcrcr} x & - & y & = & 0 \\ 2 x & - & 2 y & + & z & + & 2 w & = & 4 \\ y & + & w & = & 0 \\ 2 z & + & w & = & 5 \end{array} & ⟶_{\begin{matrix} \end{matrix}}^{- 2 ρ_{1} + ρ_{2}} \begin{array}{rcrcrcrcr} x & - & y & = & 0 \\ z & + & 2 w & = & 4 \\ y & + & w & = & 0 \\ 2 z & + & w & = & 5 \end{array} \end{aligned}$

the second equation has no leading $y$ . We exchange it for a lower-down row that has a leading $y$ .

$\begin{aligned} ⟶_{\begin{matrix} \end{matrix}}^{ρ_{2} \leftrightarrow ρ_{3}} \begin{array}{rcrcrcrcr} x & - & y & = & 0 \\ y & + & w & = & 0 \\ z & + & 2 w & = & 4 \\ 2 z & + & w & = & 5 \end{array} \end{aligned}$

(Had there been more than one suitable row below the second then we could have used any one.) With that, Gauss’s Method proceeds as before.

$\begin{aligned} ⟶_{\begin{matrix} \end{matrix}}^{- 2 ρ_{3} + ρ_{4}} \begin{array}{rcrcrcrcr} x & - & y & = & 0 \\ y & + & w & = & 0 \\ z & + & 2 w & = & 4 \\ - 3 w & = & - 3 \end{array} \end{aligned}$

Back-substitution gives $w = 1$ , $z = 2$ , $y = - 1$ , and $x = - 1$ .

Strictly speaking, to solve linear systems we don’t need the row rescaling operation. We have introduced it here because it is convenient and because we will use it later in this chapter as part of a variation of Gauss’s Method, the Gauss-Jordan Method.

All of the systems so far have the same number of equations as unknowns. All of them have a solution and for all of them there is only one solution. We finish this subsection by seeing other things that can happen.

Example 1.12 This system has more equations than variables.

$\begin{array}{rcrcr} x & + & 3 y & = & 1 \\ 2 x & + & y & = & - 3 \\ 2 x & + & 2 y & = & - 2 \end{array}$

Gauss’s Method helps us understand this system also, since this

$\begin{aligned} ⟶_{\begin{matrix} - 2 ρ_{1} + ρ_{3} \end{matrix}}^{- 2 ρ_{1} + ρ_{2}} \begin{array}{rcrcr} x & + & 3 y & = & 1 \\ - 5 y & = & - 5 \\ - 4 y & = & - 4 \end{array} \end{aligned}$

shows that one of the equations is redundant. Echelon form

$⟶_{\begin{array}{l} \end{array}}^{- (4 / 5) ρ_{2} + ρ_{3}} \begin{array}{rcrcr} x & + & 3 y & = & 1 \\ - 5 y & = & - 5 \\ 0 & = & 0 \end{array}$

gives that $y = 1$ and $x = - 2$ . The ‘ $0 = 0$ ’ reflects the redundancy.

Gauss’s Method is also useful on systems with more variables than equations. The next subsection has many examples.

Another way that linear systems can differ from the examples shown above is that some linear systems do not have a unique solution. This can happen in two ways. The first is that a system can fail to have any solution at all.

Example 1.13 Contrast the system in the last example with this one.

$\begin{array}{rcrcr} x & + & 3 y & = & 1 \\ 2 x & + & y & = & - 3 \\ 2 x & + & 2 y & = & 0 \end{array} ⟶_{\begin{array}{l} - 2 ρ_{1} + ρ_{3} \end{array}}^{- 2 ρ_{1} + ρ_{2}} \begin{array}{rcrcr} x & + & 3 y & = & 1 \\ - 5 y & = & - 5 \\ - 4 y & = & - 2 \end{array}$

Here the system is inconsistent: no pair of numbers $(s_{1}, s_{2})$ satisfies all three equations simultaneously. Echelon form makes the inconsistency obvious.

$⟶_{\begin{array}{l} \end{array}}^{- (4 / 5) ρ_{2} + ρ_{3}} \begin{array}{rcrcr} x & + & 3 y & = & 1 \\ - 5 y & = & - 5 \\ 0 & = & 2 \end{array}$

The solution set is empty.

Example 1.14 The prior system has more equations than unknowns but that is not what causes the inconsistency— Example 1.12 has more equations than unknowns and yet is consistent. Nor is having more equations than unknowns necessary for inconsistency, as we see with this inconsistent system that has the same number of equations as unknowns.

$\begin{array}{rcrcr} x & + & 2 y & = & 8 \\ 2 x & + & 4 y & = & 8 \end{array} ⟶_{\begin{array}{l} \end{array}}^{- 2 ρ_{1} + ρ_{2}} \begin{array}{rcrcr} x & + & 2 y & = & 8 \\ 0 & = & - 8 \end{array}$

Instead, inconsistency has to do with the interaction of the left and right sides; in the first system above the left side’s second equation is twice the first but the right side’s second constant is not twice the first. Later we will have more to say about dependencies between a system’s parts.

The other way that a linear system can fail to have a unique solution, besides having no solutions, is to have many solutions.

Example 1.15 In this system

$\begin{array}{rcrcr} x & + & y & = & 4 \\ 2 x & + & 2 y & = & 8 \end{array}$

any pair of numbers satisfying the first equation also satisfies the second. The solution set ${(x, y) ∣ x + y = 4}$ is infinite; some example member pairs are $(0, 4)$ , $(- 1, 5)$ , and $(2.5, 1.5)$ .

The result of applying Gauss’s Method here contrasts with the prior example because we do not get a contradictory equation.

$⟶_{\begin{array}{l} \end{array}}^{- 2 ρ_{1} + ρ_{2}} \begin{array}{rcrcr} x & + & y & = & 4 \\ 0 & = & 0 \end{array}$

Don’t be fooled by that example: a $0 = 0$ equation is not the signal that a system has many solutions.

Example 1.16 The absence of a $0 = 0$ equation does not keep a system from having many different solutions. This system is in echelon form, has no $0 = 0$ , but has infinitely many solutions, including $(0, 1, - 1)$ , $(0, 1 / 2, - 1 / 2)$ , $(0, 0, 0)$ , and $(0, - π, π)$ (any triple whose first component is $0$ and whose second component is the negative of the third is a solution).

$\begin{array}{rcrcrcr} x & + & y & + & z & = & 0 \\ y & + & z & = & 0 \end{array}$

Nor does the presence of $0 = 0$ mean that the system must have many solutions. Example 1.12 shows that. So does this system, which does not have any solutions at all despite that in echelon form it has a $0 = 0$ row.

$\begin{aligned} \begin{array}{rcrcrcr} 2 x & - & 2 z & = & 6 \\ y & + & z & = & 1 \\ 2 x & + & y & - & z & = & 7 \\ 3 y & + & 3 z & = & 0 \end{array} & ⟶_{\begin{matrix} \end{matrix}}^{- ρ_{1} + ρ_{3}} \begin{array}{rcrcrcr} 2 x & - & 2 z & = & 6 \\ y & + & z & = & 1 \\ y & + & z & = & 1 \\ 3 y & + & 3 z & = & 0 \end{array} \\ ⟶_{\begin{matrix} - 3 ρ_{2} + ρ_{4} \end{matrix}}^{- ρ_{2} + ρ_{3}} \begin{array}{rcrcrcr} 2 x & - & 2 z & = & 6 \\ y & + & z & = & 1 \\ 0 & = & 0 \\ 0 & = & - 3 \end{array} \end{aligned}$

In summary, Gauss’s Method uses the row operations to set a system up for back substitution. If any step shows a contradictory equation then we can stop with the conclusion that the system has no solutions. If we reach echelon form without a contradictory equation, and each variable is a leading variable in its row, then the system has a unique solution and we find it by back substitution. Finally, if we reach echelon form without a contradictory equation, and there is not a unique solution— that is, at least one variable is not a leading variable— then the system has many solutions.

The next subsection explores the third case. We will see that such a system must have infinitely many solutions and we will describe the solution set.

Note. In the exercises here, and in the rest of the book, you must justify all of your answers. For instance, if a question asks whether a system has a solution then you must justify a yes response by producing the solution and must justify a no response by showing that no solution exists.

Exercises

Exercise 1.17 Worked answer

Recommended. Use Gauss’s Method to find the unique solution for each system.
1. $\begin{array}{rcrcr} 2 x & + & 3 y & = & 13 \\ x & - & y & = & - 1 \end{array}$
2. $\begin{array}{rcrcrcr} x & - & z & = & 0 \\ 3 x & + & y & = & 1 \\ - x & + & y & + & z & = & 4 \end{array}$
Back to Exercise 1.17

Answer.
1. Gauss’s Method
  
  $⟶_{\begin{array}{l} \end{array}}^{- (1 / 2) ρ_{1} + ρ_{2}} \begin{array}{rcrcr} 2 x & + & 3 y & = & 13 \\ - & (5 / 2) y & = & - 15 / 2 \end{array}$
  
  gives that the solution is $y = 3$ and $x = 2$ .
2. Gauss’s Method here
  
  $⟶_{\begin{array}{l} ρ_{1} + ρ_{3} \end{array}}^{- 3 ρ_{1} + ρ_{2}} \begin{array}{rcrcrcr} x & - & z & = & 0 \\ y & + & 3 z & = & 1 \\ y & = & 4 \end{array} ⟶_{\begin{array}{l} \end{array}}^{- ρ_{2} + ρ_{3}} \begin{array}{rcrcrcr} x & - & z & = & 0 \\ y & + & 3 z & = & 1 \\ - 3 z & = & 3 \end{array}$
  
  gives $x = - 1$ , $y = 4$ , and $z = - 1$ .
Exercise 1.18 Worked answer

Each system is in echelon form. For each, say whether the system has a unique solution, no solution, or infinitely many solutions.
1. $\begin{array}{rcrcr} - 3 x & + & 2 y & = & 0 \\ - 2 y & = & 0 \end{array}$
2. $\begin{array}{rcrcrcr} x & + & y & = & 4 \\ y & - & z & = & 0 \end{array}$
3. $\begin{array}{rcrcrcr} x & + & y & = & 4 \\ y & - z & = & 0 \\ 0 & = & 0 \end{array}$
4. $\begin{array}{rcrcr} x & + & y & = & 4 \\ 0 & = & 4 \end{array}$
5. $\begin{array}{rcrcrcr} 3 x & + & 6 y & + & z & = & - 0.5 \\ - z & = & 2.5 \end{array}$
6. $\begin{array}{rcrcr} x & - & 3 y & = & 2 \\ 0 & = & 0 \end{array}$
7. $\begin{array}{rcrcr} 2 x & + & 2 y & = & 4 \\ y & = & 1 \\ 0 & = & 4 \end{array}$
8. $\begin{array}{rcrcr} 2 x & + & y & = & 0 \end{array}$
9. $\begin{array}{rcrcr} x & - & y & = & - 1 \\ 0 & = & 0 \\ 0 & = & 4 \end{array}$
10. $\begin{array}{rcrcrcr} x & + & y & - & 3 z & = & - 1 \\ y & - & z & = & 2 \\ z & = & 0 \\ 0 & = & 0 \end{array}$
Back to Exercise 1.18

Answer. If a system has a contradictory equation then it has no solution. Otherwise, if there are any variables that are not leading a row then it has infinitely many solution. In the final case, where there is no contradictory equation and every variable leads some row, it has a unique solution.
1. Unique solution
2. Infinitely many solutions
3. Infinitely many solutions
4. No solution
5. Infinitely many solutions
6. Infinitely many solutions
7. No solution
8. Infinitely many solutions
9. No solution
10. Unique solution
Exercise 1.19 Worked answer

Recommended. Use Gauss’s Method to solve each system or conclude ‘many solutions’ or ‘no solutions’.
1. $\begin{array}{rcrcr} 2 x & + & 2 y & = & 5 \\ x & - & 4 y & = & 0 \end{array}$
2. $\begin{array}{rcrcr} - x & + & y & = & 1 \\ x & + & y & = & 2 \end{array}$
3. $\begin{array}{rcrcrcr} x & - & 3 y & + & z & = & 1 \\ x & + & y & + & 2 z & = & 14 \end{array}$
4. $\begin{array}{rcrcr} - x & - & y & = & 1 \\ - 3 x & - & 3 y & = & 2 \end{array}$
5. $\begin{array}{rcrcrcr} 4 y & + & z & = & 20 \\ 2 x & - & 2 y & + & z & = & 0 \\ x & + & z & = & 5 \\ x & + & y & - & z & = & 10 \end{array}$
6. $\begin{array}{rcrcrcrcr} 2 x & + & z & + & w & = & 5 \\ y & - & w & = & - 1 \\ 3 x & - & z & - & w & = & 0 \\ 4 x & + & y & + & 2 z & + & w & = & 9 \end{array}$
Back to Exercise 1.19

Answer.
1. Gaussian reduction
  
  $⟶_{\begin{array}{l} \end{array}}^{- (1 / 2) ρ_{1} + ρ_{2}} \begin{array}{rcrcr} 2 x & + & 2 y & = & 5 \\ - 5 y & = & - 5 / 2 \end{array}$
  
  shows that $y = 1 / 2$ and $x = 2$ is the unique solution.
2. Gauss’s Method
  
  $⟶_{\begin{array}{l} \end{array}}^{ρ_{1} + ρ_{2}} \begin{array}{rcrcr} - x & + & y & = & 1 \\ 2 y & = & 3 \end{array}$
  
  gives $y = 3 / 2$ and $x = 1 / 2$ as the only solution.
3. Row reduction
  
  $⟶_{\begin{array}{l} \end{array}}^{- ρ_{1} + ρ_{2}} \begin{array}{rcrcrcr} x & - & 3 y & + & z & = & 1 \\ 4 y & + & z & = & 13 \end{array}$
  
  shows, because the variable $z$ is not a leading variable in any row, that there are many solutions.
4. Row reduction
  
  $⟶_{\begin{array}{l} \end{array}}^{- 3 ρ_{1} + ρ_{2}} \begin{array}{rcrcr} - x & - & y & = & 1 \\ 0 & = & - 1 \end{array}$
  
  shows that there is no solution.
5. Gauss’s Method
  
  $\begin{aligned} ⟶_{\begin{array}{l} \end{array}}^{ρ_{1} \leftrightarrow ρ_{4}} \begin{array}{rcrcrcr} x & + & y & - & z & = & 10 \\ 2 x & - & 2 y & + & z & = & 0 \\ x & + & z & = & 5 \\ 4 y & + & z & = & 20 \end{array} & ⟶_{\begin{matrix} - ρ_{1} + ρ_{3} \end{matrix}}^{- 2 ρ_{1} + ρ_{2}} \begin{array}{rcrcrcr} x & + & y & - & z & = & 10 \\ - 4 y & + & 3 z & = & - 20 \\ - y & + & 2 z & = & - 5 \\ 4 y & + & z & = & 20 \end{array} \\ ⟶_{\begin{matrix} ρ_{2} + ρ_{4} \end{matrix}}^{- (1 / 4) ρ_{2} + ρ_{3}} \begin{array}{rcrcrcr} x & + & y & - & z & = & 10 \\ - 4 y & + & 3 z & = & - 20 \\ (5 / 4) z & = & 0 \\ 4 z & = & 0 \end{array} \end{aligned}$
  
  gives the unique solution $(x, y, z) = (5, 5, 0)$ .
6. Here Gauss’s Method gives
  
  $\begin{aligned} ⟶_{\begin{matrix} - 2 ρ_{1} + ρ_{4} \end{matrix}}^{- (3 / 2) ρ_{1} + ρ_{3}} \begin{array}{rcrcrcrcr} 2 x & + & z & + & w & = & 5 \\ y & - & w & = & - 1 \\ - & (5 / 2) z & - & (5 / 2) w & = & - 15 / 2 \\ y & - & w & = & - 1 \end{array} \\ ⟶_{\begin{matrix} \end{matrix}}^{- ρ_{2} + ρ_{4}} \begin{array}{rcrcrcrcr} 2 x & + & z & + & w & = & 5 \\ y & - & w & = & - 1 \\ - & (5 / 2) z & - & (5 / 2) w & = & - 15 / 2 \\ 0 & = & 0 \end{array} \end{aligned}$
  
  which shows that there are many solutions.
Exercise 1.20 Worked answer

Solve each system or conclude ‘many solutions’ or ‘no solutions’. Use Gauss’s Method.
1. $\begin{array}{rcrcrcr} x & + & y & + & z & = & 5 \\ x & - & y & = & 0 \\ y & + & 2 z & = & 7 \end{array}$
2. $\begin{array}{rcrcrcr} 3 x & + & z & = & 7 \\ x & - & y & + & 3 z & = & 4 \\ x & + & 2 y & - & 5 z & = & - 1 \end{array}$
3. $\begin{array}{rcrcrcr} x & + & 3 y & + & z & = & 0 \\ - x & - & y & = & 2 \\ - x & + & y & + & 2 z & = & 8 \end{array}$
Back to Exercise 1.20

Answer.
1. Gauss’s Method
  
  $\begin{aligned} \begin{array}{rcrcrcr} x & + & y & + & z & = & 5 \\ x & - & y & = & 0 \\ y & + & 2 z & = & 7 \end{array} & ⟶_{\begin{matrix} \end{matrix}}^{- ρ_{1} + ρ_{2}} \begin{array}{rcrcrcr} x & + & y & + & z & = & 5 \\ 0 & - 2 y & - & z & = & - 5 \\ y & + & 2 z & = & 7 \end{array} \\ ⟶_{\begin{matrix} \end{matrix}}^{(1 / 2) ρ_{2} + ρ_{3}} \begin{array}{rcrcrcr} x & + & y & + & z & = & 5 \\ 0 & - 2 y & - & 1 & = & - 5 \\ + & (3 / 2) z & = & 7 / 2 \end{array} \end{aligned}$
  
  followed by back-substitution gives $x = 1$ , $y = 1$ , and $z = 3$ .
2. Here Gauss’s Method
  
  $\begin{aligned} \begin{array}{rcrcrcr} 3 x & + & z & = & 7 \\ x & - & y & + & 3 z & = & 4 \\ x & + & 2 y & - & 5 z & = & - 1 \end{array} & ⟶_{\begin{matrix} - (1 / 3) ρ_{1} + ρ_{3} \end{matrix}}^{- (1 / 3) ρ_{1} + ρ_{2}} \begin{array}{rcrcrcr} 3 x & + & z & = & 7 \\ - y & + & (8 / 3) z & = & 5 / 3 \\ 2 y & - & (16 / 3) z & = & - (10 / 3) \end{array} \\ ⟶_{\begin{matrix} \end{matrix}}^{2 ρ_{2} + ρ_{3}} \begin{array}{rcrcrcr} 3 x & + & z & = & 7 \\ - y & + & (8 / 3) z & = & 5 / 3 \\ 0 & = & 0 \end{array} \end{aligned}$
  
  finds that the variable $z$ does not lead a row. There are infinitely many solutions.
3. The steps
  
  $\begin{array}{rcrcrcr} x & + & 3 y & + & z & = & 0 \\ - x & - & y & = & 2 \\ - x & + & y & + & 2 z & = & 8 \end{array} ⟶_{\begin{array}{l} ρ_{1} + ρ_{3} \end{array}}^{ρ_{1} + ρ_{2}} \begin{array}{rcrcrcr} x & + & 3 y & + & z & = & 0 \\ 2 y & + & z & = & 2 \\ 4 y & + & 3 z & = & 8 \end{array} ⟶_{\begin{array}{l} \end{array}}^{- 2 ρ_{2} + ρ_{3}} \begin{array}{rcrcrcr} x & + & 3 y & + & z & = & 0 \\ 2 y & + & z & = & 2 \\ z & = & 4 \end{array}$
  
  give $(x, y, z) = (- 1, - 1, 4)$ .
Exercise 1.21 Worked answer

Recommended. We can solve linear systems by methods other than Gauss’s. One often taught in high school is to solve one of the equations for a variable, then substitute the resulting expression into other equations. Then we repeat that step until there is an equation with only one variable. From that we get the first number in the solution and then we get the rest with back-substitution. This method takes longer than Gauss’s Method, since it involves more arithmetic operations, and is also more likely to lead to errors. To illustrate how it can lead to wrong conclusions, we will use the system

$\begin{array}{rcrcr} x & + & 3 y & = & 1 \\ 2 x & + & y & = & - 3 \\ 2 x & + & 2 y & = & 0 \end{array}$

from Example 1.13.
1. Solve the first equation for $x$ and substitute that expression into the second equation. Find the resulting $y$ .
2. Again solve the first equation for $x$ , but this time substitute that expression into the third equation. Find this $y$ .
What extra step must a user of this method take to avoid erroneously concluding a system has a solution?
Back to Exercise 1.21

Answer.
1. From $x = 1 - 3 y$ we get that $2 (1 - 3 y) + y = - 3$ , giving $y = 1$ .
2. From $x = 1 - 3 y$ we get that $2 (1 - 3 y) + 2 y = 0$ , leading to the conclusion that $y = 1 / 2$ .
Users of this method must check any potential solutions by substituting back into all the equations.
Exercise 1.22 Worked answer

Recommended. For which values of $k$ are there no solutions, many solutions, or a unique solution to this system?

$\begin{array}{rcrcr} x & - & y & = & 1 \\ 3 x & - & 3 y & = & k \end{array}$

Back to Exercise 1.22

Answer. Do the reduction

$⟶_{\begin{array}{l} \end{array}}^{- 3 ρ_{1} + ρ_{2}} \begin{array}{rcrcr} x & - & y & = & 1 \\ 0 & = & - 3 + k \end{array}$

to conclude this system has no solutions if $k \neq 3$ and if $k = 3$ then it has infinitely many solutions. It never has a unique solution.
Exercise 1.23 Worked answer

This system is not linear in that it says $\sin α$ instead of $α$

$\begin{array}{rcrcrcr} 2 \sin α & - & \cos β & + & 3 \tan γ & = & 3 \\ 4 \sin α & + & 2 \cos β & - & 2 \tan γ & = & 10 \\ 6 \sin α & - & 3 \cos β & + & \tan γ & = & 9 \end{array}$

and yet we can apply Gauss’s Method. Do so. Does the system have a solution?

Back to Exercise 1.23

Answer. Let $x = \sin α$ , $y = \cos β$ , and $z = \tan γ$ :

$\begin{array}{rcrcrcr} 2 x & - & y & + & 3 z & = & 3 \\ 4 x & + & 2 y & - & 2 z & = & 10 \\ 6 x & - & 3 y & + & z & = & 9 \end{array} ⟶_{\begin{array}{l} - 3 ρ_{1} + ρ_{3} \end{array}}^{- 2 ρ_{1} + ρ_{2}} \begin{array}{rcrcrcr} 2 x & - & y & + & 3 z & = & 3 \\ 4 y & - & 8 z & = & 4 \\ - 8 z & = & 0 \end{array}$

gives $z = 0$ , $y = 1$ , and $x = 2$ . Note that no $α \in ℝ$ satisfies that $\sin α = 2$ .
Exercise 1.24 Worked answer

Recommended. What conditions must the constants, the $b$ ’s, satisfy so that each of these systems has a solution? Hint. Apply Gauss’s Method and see what happens to the right side.
1. $\begin{array}{rcrcr} x & - & 3 y & = & b_{1} \\ 3 x & + & y & = & b_{2} \\ x & + & 7 y & = & b_{3} \\ 2 x & + & 4 y & = & b_{4} \end{array}$
2. $\begin{array}{rcrcrcr} x_{1} & + & 2 x_{2} & + & 3 x_{3} & = & b_{1} \\ 2 x_{1} & + & 5 x_{2} & + & 3 x_{3} & = & b_{2} \\ x_{1} & + & 8 x_{3} & = & b_{3} \end{array}$
Back to Exercise 1.24

Answer.
1. Gauss’s Method
  
  $⟶_{\begin{array}{l} - ρ_{1} + ρ_{3} \\ - 2 ρ_{1} + ρ_{4} \end{array}}^{- 3 ρ_{1} + ρ_{2}} \begin{array}{rcrcr} x & - & 3 y & = & b_{1} \\ 10 y & = & - 3 b_{1} + b_{2} \\ 10 y & = & - b_{1} + b_{3} \\ 10 y & = & - 2 b_{1} + b_{4} \end{array} ⟶_{\begin{array}{l} - ρ_{2} + ρ_{4} \end{array}}^{- ρ_{2} + ρ_{3}} \begin{array}{rcrcr} x & - & 3 y & = & b_{1} \\ 10 y & = & - 3 b_{1} + b_{2} \\ 0 & = & 2 b_{1} - b_{2} + b_{3} \\ 0 & = & b_{1} - b_{2} + b_{4} \end{array}$
  
  shows that this system is consistent if and only if both $b_{3} = - 2 b_{1} + b_{2}$ and $b_{4} = - b_{1} + b_{2}$ .
2. Reduction
  
  $\begin{aligned} ⟶_{\begin{matrix} - ρ_{1} + ρ_{3} \end{matrix}}^{- 2 ρ_{1} + ρ_{2}} \begin{array}{rcrcrcr} x_{1} & + & 2 x_{2} & + & 3 x_{3} & = & b_{1} \\ x_{2} & - & 3 x_{3} & = & - 2 b_{1} + b_{2} \\ - 2 x_{2} & + & 5 x_{3} & = & - b_{1} + b_{3} \end{array} \\ ⟶_{\begin{matrix} \end{matrix}}^{2 ρ_{2} + ρ_{3}} \begin{array}{rcrcrcr} x_{1} & + & 2 x_{2} & + & 3 x_{3} & = & b_{1} \\ x_{2} & - & 3 x_{3} & = & - 2 b_{1} + b_{2} \\ - x_{3} & = & - 5 b_{1} + 2 b_{2} + b_{3} \end{array} \end{aligned}$
  
  shows that each of $b_{1}$ , $b_{2}$ , and $b_{3}$ can be any real number—this system always has a unique solution.
Exercise 1.25 Worked answer

True or false: a system with more unknowns than equations has at least one solution. (As always, to say ‘true’ you must prove it, while to say ‘false’ you must produce a counterexample.)

Back to Exercise 1.25

Answer. This system with more unknowns than equations

$\begin{array}{rcrcrcr} x & + & y & + & z & = & 0 \\ x & + & y & + & z & = & 1 \end{array}$

has no solution.
Exercise 1.26 Worked answer

Must any Chemistry problem like the one that starts this subsection— a balance the reaction problem—have infinitely many solutions?

Back to Exercise 1.26

Answer. Yes. For example, the fact that we can have the same reaction in two different flasks shows that twice any solution is another, different, solution (if a physical reaction occurs then there must be at least one nonzero solution).
Exercise 1.27 Worked answer

Recommended. Find the coefficients $a$ , $b$ , and $c$ so that the graph of $f (x) = a x^{2} + b x + c$ passes through the points $(1, 2)$ , $(- 1, 6)$ , and $(2, 3)$ .

Back to Exercise 1.27

Answer. Because $f (1) = 2$ , $f (- 1) = 6$ , and $f (2) = 3$ we get a linear system.

$\begin{array}{rcrcrcr} 1 a & + & 1 b & + & c & = & 2 \\ 1 a & - & 1 b & + & c & = & 6 \\ 4 a & + & 2 b & + & c & = & 3 \end{array}$

Gauss’s Method

$⟶_{\begin{array}{l} - 4 ρ_{1} + ρ_{3} \end{array}}^{- ρ_{1} + ρ_{2}} \begin{array}{rcrcrcr} a & + & b & + & c & = & 2 \\ - 2 b & = & 4 \\ - 2 b & - & 3 c & = & - 5 \end{array} ⟶_{\begin{array}{l} \end{array}}^{- ρ_{2} + ρ_{3}} \begin{array}{rcrcrcr} a & + & b & + & c & = & 2 \\ - 2 b & = & 4 \\ - 3 c & = & - 9 \end{array}$

shows that the solution is $f (x) = 1 x^{2} - 2 x + 3$ .
Exercise 1.28 Worked answer

After Theorem 1.5 we note that multiplying a row by $0$ is not allowed because that could change a solution set. Give an example of a system with solution set $S_{0}$ where after multiplying a row by $0$ the new system has a solution set $S_{1}$ and $S_{0}$ is a proper subset of $S_{1}$ , that is, $S_{0} \neq S_{1}$ . Give an example where $S_{0} = S_{1}$ .

Back to Exercise 1.28

Answer. Here $S_{0} = {(1, 1)}$

$\begin{array}{rcrcr} x & + & y & = & 2 \\ x & - & y & = & 0 \end{array} ⟶_{\begin{array}{l} \end{array}}^{0 ρ_{2}} \begin{array}{rcrcr} x & + & y & = & 2 \\ 0 & = & 0 \end{array}$

while $S_{1}$ is a proper superset because it contains at least two points: $(1, 1)$ and $(2, 0)$ . In this example the solution set does not change.

$\begin{array}{rcrcr} x & + & y & = & 2 \\ 2 x & + & 2 y & = & 4 \end{array} ⟶_{\begin{array}{l} \end{array}}^{0 ρ_{2}} \begin{array}{rcrcr} x & + & y & = & 2 \\ 0 & = & 0 \end{array}$
Exercise 1.29 Worked answer

Gauss’s Method works by combining the equations in a system to make new equations.
1. Can we derive the equation $3 x - 2 y = 5$ by a sequence of Gaussian reduction steps from the equations in this system?
  
  $\begin{array}{rcrcr} x & + & y & = & 1 \\ 4 x & - & y & = & 6 \end{array}$
2. Can we derive the equation $5 x - 3 y = 2$ with a sequence of Gaussian reduction steps from the equations in this system?
  
  $\begin{array}{rcrcr} 2 x & + & 2 y & = & 5 \\ 3 x & + & y & = & 4 \end{array}$
3. Can we derive $6 x - 9 y + 5 z = - 2$ by a sequence of Gaussian reduction steps from the equations in the system?
  
  $\begin{array}{rcrcrcr} 2 x & + & y & - & z & = & 4 \\ 6 x & - & 3 y & + & z & = & 5 \end{array}$
Back to Exercise 1.29

Answer.
1. Yes, by inspection the given equation results from $- ρ_{1} + ρ_{2}$ .
2. No. The pair $(1, 1)$ satisfies the given equation. However, that pair does not satisfy the first equation in the system.
3. Yes. To see if the given row is $c_{1} ρ_{1} + c_{2} ρ_{2}$ , solve the system of equations relating the coefficients of $x$ , $y$ , $z$ , and the constants:
  
  $\begin{array}{rcrcr} 2 c_{1} & + & 6 c_{2} & = & 6 \\ c_{1} & - & 3 c_{2} & = & - 9 \\ - c_{1} & + & c_{2} & = & 5 \\ 4 c_{1} & + & 5 c_{2} & = & - 2 \end{array}$
  
  and get $c_{1} = - 3$ and $c_{2} = 2$ , so the given row is $- 3 ρ_{1} + 2 ρ_{2}$ .
Exercise 1.30 Worked answer

Prove that, where $a, b, c, d, e$ are real numbers with $a \neq 0$ , if this linear equation

$a x + b y = c$

has the same solution set as this one

$a x + d y = e$

then they are the same equation. What if $a = 0$ ?

Back to Exercise 1.30

Answer. If $a \neq 0$ then the solution set of the first equation is this.

$\begin{matrix} {(x, y) \in ℝ^{2} ∣ x = (c - b y) / a = (c / a) - (b / a) \cdot y} & (*) \end{matrix}$

Thus, given $y$ we can compute the associated $x$ . Taking $y = 0$ gives the solution $(c / a, 0)$ , and since the second equation $a x + d y = e$ is supposed to have the same solution set, substituting into it gives that $a (c / a) + d \cdot 0 = e$ , so $c = e$ . Taking $y = 1$ in ( $*$ ) gives $a ((c - b) / a) + d \cdot 1 = e$ , and so $b = d$ . Hence they are the same equation.

When $a = 0$ the equations can be different and still have the same solution set: e.g., $0 x + 3 y = 6$ and $0 x + 6 y = 12$ .
Exercise 1.31 Worked answer

Show that if $a d - b c \neq 0$ then

$\begin{array}{rcrcr} a x & + & b y & = & j \\ c x & + & d y & = & k \end{array}$

has a unique solution.

Back to Exercise 1.31

Answer. We take three cases: that $a \neq 0$ , that $a = 0$ and $c \neq 0$ , and that both $a = 0$ and $c = 0$ .

For the first, we assume that $a \neq 0$ . Then the reduction

$⟶_{\begin{array}{l} \end{array}}^{- (c / a) ρ_{1} + ρ_{2}} \begin{array}{rcrcr} a x & + & b y & = & j \\ (- (c b / a) + d) y & = & - (c j / a) + k \end{array}$

shows that this system has a unique solution if and only if $- (c b / a) + d \neq 0$ ; remember that $a \neq 0$ so that back substitution yields a unique $x$ (observe, by the way, that $j$ and $k$ play no role in the conclusion that there is a unique solution, although if there is a unique solution then they contribute to its value). But $- (c b / a) + d = (a d - b c) / a$ and a fraction is not equal to $0$ if and only if its numerator is not equal to $0$ . Thus, in this first case, there is a unique solution if and only if $a d - b c \neq 0$ .

In the second case, if $a = 0$ but $c \neq 0$ , then we swap

$\begin{array}{rcrcr} c x & + & d y & = & k \\ b y & = & j \end{array}$

to conclude that the system has a unique solution if and only if $b \neq 0$ (we use the case assumption that $c \neq 0$ to get a unique $x$ in back substitution). But—where $a = 0$ and $c \neq 0$ — the condition “ $b \neq 0$ ” is equivalent to the condition “ $a d - b c \neq 0$ ”. That finishes the second case.

Finally, for the third case, if both $a$ and $c$ are $0$ then the system

$\begin{array}{rcrcr} 0 x & + & b y & = & j \\ 0 x & + & d y & = & k \end{array}$

might have no solutions (if the second equation is not a multiple of the first) or it might have infinitely many solutions (if the second equation is a multiple of the first then for each $y$ satisfying both equations, any pair $(x, y)$ will do), but it never has a unique solution. Note that $a = 0$ and $c = 0$ gives that $a d - b c = 0$ .
Exercise 1.32 Worked answer

Recommended. In the system

$\begin{array}{rcrcr} a x & + & b y & = & c \\ d x & + & e y & = & f \end{array}$

each of the equations describes a line in the $x y$ -plane. By geometrical reasoning, show that there are three possibilities: there is a unique solution, there is no solution, and there are infinitely many solutions.

Back to Exercise 1.32

Answer. Recall that if a pair of lines share two distinct points then they are the same line. That’s because two points determine a line, so these two points determine each of the two lines, and so they are the same line.

Thus the lines can share one point (giving a unique solution), share no points (giving no solutions), or share at least two points (which makes them the same line).
Exercise 1.33 Worked answer

Finish the proof of Theorem 1.5.

Back to Exercise 1.33

Answer. For the reduction operation of multiplying $ρ_{i}$ by a nonzero real number $k$ , we have that $(s_{1}, \dots, s_{n})$ satisfies this system

$\begin{array}{rcrcrcrcr} a_{1, 1} x_{1} & + & a_{1, 2} x_{2} & + & \dots & + & a_{1, n} x_{n} & = & d_{1} \\ ⋮ \\ k a_{i, 1} x_{1} & + & k a_{i, 2} x_{2} & + & \dots & + & k a_{i, n} x_{n} & = & k d_{i} \\ ⋮ \\ a_{m, 1} x_{1} & + & a_{m, 2} x_{2} & + & \dots & + & a_{m, n} x_{n} & = & d_{m} \end{array}$

if and only if $a_{1, 1} s_{1} + a_{1, 2} s_{2} + \dots + a_{1, n} s_{n} = d_{1}$ and … $k a_{i, 1} s_{1} + k a_{i, 2} s_{2} + \dots + k a_{i, n} s_{n} = k d_{i}$ and … $a_{m, 1} s_{1} + a_{m, 2} s_{2} + \dots + a_{m, n} s_{n} = d_{m}$ by the definition of ‘satisfies’. Because $k \neq 0$ , that’s true if and only if $a_{1, 1} s_{1} + a_{1, 2} s_{2} + \dots + a_{1, n} s_{n} = d_{1}$ and … $a_{i, 1} s_{1} + a_{i, 2} s_{2} + \dots + a_{i, n} s_{n} = d_{i}$ and … $a_{m, 1} s_{1} + a_{m, 2} s_{2} + \dots + a_{m, n} s_{n} = d_{m}$ (this is straightforward canceling on both sides of the $i$ -th equation), which says that $(s_{1}, \dots, s_{n})$ solves

$\begin{array}{rcrcrcrcr} a_{1, 1} x_{1} & + & a_{1, 2} x_{2} & + & \dots & + & a_{1, n} x_{n} & = & d_{1} \\ ⋮ \\ a_{i, 1} x_{1} & + & a_{i, 2} x_{2} & + & \dots & + & a_{i, n} x_{n} & = & d_{i} \\ ⋮ \\ a_{m, 1} x_{1} & + & a_{m, 2} x_{2} & + & \dots & + & a_{m, n} x_{n} & = & d_{m} \end{array}$

as required.

For the combination operation $k ρ_{i} + ρ_{j}$ , the tuple $(s_{1}, \dots, s_{n})$ satisfies

$\begin{array}{rcrcrcrcr} a_{1, 1} x_{1} & + & \dots & + & a_{1, n} x_{n} & = & d_{1} \\ ⋮ \\ a_{i, 1} x_{1} & + & \dots & + & a_{i, n} x_{n} & = & d_{i} \\ ⋮ \\ (k a_{i, 1} + a_{j, 1}) x_{1} & + & \dots & + & (k a_{i, n} + a_{j, n}) x_{n} & = & k d_{i} + d_{j} \\ ⋮ \\ a_{m, 1} x_{1} & + & \dots & + & a_{m, n} x_{n} & = & d_{m} \end{array}$

if and only if $a_{1, 1} s_{1} + \dots + a_{1, n} s_{n} = d_{1}$ and … $a_{i, 1} s_{1} + \dots + a_{i, n} s_{n} = d_{i}$ and … $(k a_{i, 1} + a_{j, 1}) s_{1} + \dots + (k a_{i, n} + a_{j, n}) s_{n} = k d_{i} + d_{j}$ and … $a_{m, 1} s_{1} + a_{m, 2} s_{2} + \dots + a_{m, n} s_{n} = d_{m}$ again by the definition of ‘satisfies’. Subtract $k$ times the equation $i$ from equation $j$ . (Here is where we need $i \neq j$ ; if $i = j$ then the two $d_{i}$ ’s above are not equal.) The previous compound statement holds if and only if $a_{1, 1} s_{1} + \dots + a_{1, n} s_{n} = d_{1}$ and … $a_{i, 1} s_{1} + \dots + a_{i, n} s_{n} = d_{i}$ and… $(k a_{i, 1} + a_{j, 1}) s_{1} + \dots + (k a_{i, n} + a_{j, n}) s_{n} - (k a_{i, 1} s_{1} + \dots + k a_{i, n} s_{n}) = k d_{i} + d_{j} - k d_{i}$ and… $a_{m, 1} s_{1} + \dots + a_{m, n} s_{n} = d_{m}$ , which after cancellation says that $(s_{1}, \dots, s_{n})$ solves

$\begin{array}{rcrcrcrcr} a_{1, 1} x_{1} & + & \dots & + & a_{1, n} x_{n} & = & d_{1} \\ ⋮ \\ a_{i, 1} x_{1} & + & \dots & + & a_{i, n} x_{n} & = & d_{i} \\ ⋮ \\ a_{j, 1} x_{1} & + & \dots & + & a_{j, n} x_{n} & = & d_{j} \\ ⋮ \\ a_{m, 1} x_{1} & + & \dots & + & a_{m, n} x_{n} & = & d_{m} \end{array}$

as required.
Exercise 1.34 Worked answer

Is there a two-unknowns linear system whose solution set is all of $ℝ^{2}$ ?

Back to Exercise 1.34

Answer. Yes, this one-equation system:

$0 x + 0 y = 0$

is satisfied by every $(x, y) \in ℝ^{2}$ .
Exercise 1.35 Worked answer

Recommended. Are any of the operations used in Gauss’s Method redundant? That is, can we make any of the operations from a combination of the others?

Back to Exercise 1.35

Answer. Yes. This sequence of operations swaps rows $i$ and $j$

$⟶_{\begin{array}{l} \end{array}}^{ρ_{i} + ρ_{j}} ⟶_{\begin{array}{l} \end{array}}^{- ρ_{j} + ρ_{i}} ⟶_{\begin{array}{l} \end{array}}^{ρ_{i} + ρ_{j}} ⟶_{\begin{array}{l} \end{array}}^{- 1 ρ_{i}}$

so the row-swap operation is redundant in the presence of the other two.
Exercise 1.36 Worked answer

Prove that each operation of Gauss’s Method is reversible. That is, show that if two systems are related by a row operation $S_{1} \to S_{2}$ then there is a row operation to go back $S_{2} \to S_{1}$ .

Back to Exercise 1.36

Answer. Reverse a row swap $ρ_{i} \leftrightarrow ρ_{j}$ by swapping back $ρ_{j} \leftrightarrow ρ_{i}$ . Reverse the $k ρ_{i}$ step of multiplying $k \neq 0$ on both sides of a row by dividing through $(1 / k) ρ_{i}$ .

The row combination case is the nontrivial one. The operation $k ρ_{i} + ρ_{j}$ results in this $j$ -th row.

$k \cdot a_{i, 1} + a_{j, 1} + \dots + k \cdot a_{i, n} + a_{j, n} = k \cdot d_{i} + d_{j}$

The $i$ -th row unchanged because of the $i \neq j$ restriction. Because the $i$ -th row is unchanged, the operation $- k ρ_{i} + ρ_{j}$ returns the $j$ -th row to its original state.

(Observe that the $i = j$ condition on the $k ρ_{i} + ρ_{j}$ is needed, or else this could happen

$\begin{array}{rcrcr} 3 x & + & 2 y & = & 7 \end{array} ⟶_{\begin{array}{l} \end{array}}^{2 ρ_{1} + ρ_{1}} \begin{array}{rcrcr} 9 x & + & 6 y & = & 21 \end{array} ⟶_{\begin{array}{l} \end{array}}^{- 2 ρ_{1} + ρ_{1}} \begin{array}{rcrcr} - 9 x & - & 6 y & = & - 21 \end{array}$

and so the result wouldn’t hold.)
Exercise 1.37 Worked answer

Puzzle. [Anton] A box holding pennies, nickels and dimes contains thirteen coins with a total value of $83$ cents. How many coins of each type are in the box? (These are US coins; a penny is $1$ cent, a nickel is $5$ cents, and a dime is $10$ cents.)

Back to Exercise 1.37

Answer. Let $p$ , $n$ , and $d$ be the number of pennies, nickels, and dimes. For variables that are real numbers, this system

$\begin{array}{rcrcrcr} p & + & n & + & d & = & 13 \\ p & + & 5 n & + & 10 d & = & 83 \end{array} ⟶_{\begin{array}{l} \end{array}}^{- ρ_{1} + ρ_{2}} \begin{array}{rcrcrcr} p & + & n & + & d & = & 13 \\ 4 n & + & 9 d & = & 70 \end{array}$

has more than one solution; in fact, it has infinitely many of them. However, it has a limited number of solutions in which $p$ , $n$ , and $d$ are non-negative integers. Running through $d = 0$ , …, $d = 8$ shows that $(p, n, d) = (3, 4, 6)$ is the only solution using natural numbers.
Exercise 1.38 Worked answer

Puzzle. [Con. Prob. 1955] Four positive integers are given. Select any three of the integers, find their arithmetic average, and add this result to the fourth integer. Thus the numbers 29, 23, 21, and 17 are obtained. One of the original integers is:
1. 19
2. 21
3. 23
4. 29
5. 17
Back to Exercise 1.38

Answer. Solving the system

$\begin{array}{rcrcr} (1 / 3) (a + b + c) & + & d & = & 29 \\ (1 / 3) (b + c + d) & + & a & = & 23 \\ (1 / 3) (c + d + a) & + & b & = & 21 \\ (1 / 3) (d + a + b) & + & c & = & 17 \end{array}$

we obtain $a = 12$ , $b = 9$ , $c = 3$ , $d = 21$ . Thus the second item, 21, is the correct answer.
Exercise 1.39 Worked answer

Puzzle. [Am. Math. Mon., Jan. 1935] Laugh at this: $AHAHA + TEHE = TEHAW$ . It resulted from substituting a code letter for each digit of a simple example in addition, and it is required to identify the letters and prove the solution unique.

Back to Exercise 1.39

Answer. This is how the answer was given in the cited source. A comparison of the units and hundreds columns of this addition shows that there must be a carry from the tens column. The tens column then tells us that $A < H$ , so there can be no carry from the units or hundreds columns. The five columns then give the following five equations.

$\begin{aligned} A + E & = W \\ 2 H & = A + 10 \\ H & = W + 1 \\ H + T & = E + 10 \\ A + 1 & = T \end{aligned}$

The five linear equations in five unknowns, if solved simultaneously, produce the unique solution: $A = 4$ , $T = 5$ , $H = 7$ , $W = 6$ and $E = 2$ , so that the original example in addition was $47474 + 5272 = 52746$ .
Exercise 1.40 Worked answer

Puzzle. [Wohascum no. 2] The Wohascum County Board of Commissioners, which has 20 members, recently had to elect a President. There were three candidates ( $A$ , $B$ , and $C$ ); on each ballot the three candidates were to be listed in order of preference, with no abstentions. It was found that 11 members, a majority, preferred $A$ over $B$ (thus the other 9 preferred $B$ over $A$ ). Similarly, it was found that 12 members preferred $C$ over $A$ . Given these results, it was suggested that $B$ should withdraw, to enable a runoff election between $A$ and $C$ . However, $B$ protested, and it was then found that 14 members preferred $B$ over $C$ ! The Board has not yet recovered from the resulting confusion. Given that every possible order of $A$ , $B$ , $C$ appeared on at least one ballot, how many members voted for $B$ as their first choice?

Back to Exercise 1.40

Answer. This is how the answer was given in the cited source. Some additional material was added from [joriki]. Eight commissioners voted for $B$ . To see this, we will use the given information to study how many voters chose each order of $A$ , $B$ , $C$ .

The six orders of preference are $A B C$ , $A C B$ , $B A C$ , $B C A$ , $C A B$ , $C B A$ ; assume they receive $a$ , $b$ , $c$ , $d$ , $e$ , $f$ votes respectively. We know that

$\begin{array}{rcrcrcr} a & + & b & + & e & = & 11 \\ d & + & e & + & f & = & 12 \\ a & + & c & + & d & = & 14 \end{array}$

from the number preferring $A$ over $B$ , the number preferring $C$ over $A$ , and the number preferring $B$ over $C$ . Because 20 votes were cast, $a + b + \dots + f = 20$ . Subtracting the sum of the three above equations from twice the prior equation gives $b + c + f = 3$ . We’ve specified that each preference order got at least one vote, so that means $b = c = f = 1$ .

From the above three equations the complete solution is then $a = 6$ , $b = 1$ , $c = 1$ , $d = 7$ , $e = 4$ , and $f = 1$ , as we can find with Gauss’s Method. The number of commissioners voting for $B$ as their first choice is therefore $c + d = 1 + 7 = 8$ .

Comments. The answer to this question would have been the same had we known only that at least 14 commissioners preferred $B$ over $C$ .

The seemingly paradoxical nature of the commissioner’s preferences ( $A$ is preferred to $B$ , and $B$ is preferred to $C$ , and $C$ is preferred to $A$ ), an example of “non-transitive dominance”, is common when individual choices are pooled.
Exercise 1.41 Worked answer

Puzzle. [Am. Math. Mon., Jan. 1963] “This system of $n$ linear equations with $n$ unknowns,” said the Great Mathematician, “has a curious property.”

“Good heavens!” said the Poor Nut, “What is it?”

“Note,” said the Great Mathematician, “that the constants are in arithmetic progression.”

“It’s all so clear when you explain it!” said the Poor Nut. “Do you mean like $6 x + 9 y = 12$ and $15 x + 18 y = 21$ ?”

“Quite so,” said the Great Mathematician, pulling out his bassoon. “Indeed, the system has a unique solution. Can you find it?”

“Good heavens!” cried the Poor Nut, “I am baffled.”

Are you?

Back to Exercise 1.41

Answer. This is how the answer was given in the cited source. We have not used “dependent” yet; it means here that Gauss’s Method shows that there is not a unique solution. If $n \geq 3$ the system is dependent and the solution is not unique. Hence $n < 3$ . But the term “system” implies $n > 1$ . Hence $n = 2$ . If the equations are

$\begin{array}{rcrcr} a x & + & (a + d) y & = & a + 2 d \\ (a + 3 d) x & + & (a + 4 d) y & = & a + 5 d \end{array}$

then $x = - 1$ , $y = 2$ .

Describing the Solution Set

A linear system with a unique solution has a solution set with one element. A linear system with no solution has a solution set that is empty. In these cases the solution set is easy to describe. Solution sets are a challenge to describe only when they contain many elements.

Example 2.1 This system has many solutions because in echelon form

$\begin{aligned} \begin{array}{rcrcrcr} 2 x & + & z & = & 3 \\ x & - & y & - & z & = & 1 \\ 3 x & - & y & = & 4 \end{array} & ⟶_{\begin{matrix} - (3 / 2) ρ_{1} + ρ_{3} \end{matrix}}^{- (1 / 2) ρ_{1} + ρ_{2}} \begin{array}{rcrcrcr} 2 x & + & z & = & 3 \\ - y & - & (3 / 2) z & = & - 1 / 2 \\ - y & - & (3 / 2) z & = & - 1 / 2 \end{array} \\ ⟶_{\begin{matrix} \end{matrix}}^{- ρ_{2} + ρ_{3}} \begin{array}{rcrcrcr} 2 x & + & z & = & 3 \\ - y & - & (3 / 2) z & = & - 1 / 2 \\ 0 & = & 0 \end{array} \end{aligned}$

not all of the variables are leading variables. Theorem 1.5 shows that an $(x, y, z)$ satisfies the first system if and only if it satisfies the third. So we can describe the solution set ${(x, y, z) ∣ 2 x + z = 3 and x - y - z = 1 and 3 x - y = 4}$ in this way.

$\begin{matrix} {(x, y, z) ∣ 2 x + z = 3 and - y - 3 z / 2 = - 1 / 2} & (*) \end{matrix}$

This description is better because it has two equations instead of three but it is not optimal because it still has some hard to understand interactions among the variables.

To improve it, use the variable that does not lead any equation, $z$ , to describe the variables that do lead, $x$ and $y$ . The second equation gives $y = (1 / 2) - (3 / 2) z$ and the first equation gives $x = (3 / 2) - (1 / 2) z$ . Thus we can describe the solution set as this set of triples.

$\begin{matrix} {((3 / 2) - (1 / 2) z, (1 / 2) - (3 / 2) z, z) ∣ z \in ℝ} & (* *) \end{matrix}$

Compared with ( $*$ ), the advantage of ( $* *$ ) is that $z$ can be any real number. This makes the job of deciding which tuples are in the solution set much easier. For instance, taking $z = 2$ shows that $(1 / 2, - 5 / 2, 2)$ is a solution.

Definition 2.2 In an echelon form linear system the variables that are not leading are free.

Example 2.3 Reduction of a linear system can end with more than one variable free. Gauss’s Method on this system

$\begin{aligned} \begin{array}{rcrcrcrcr} x & + & y & + & z & - & w & = & 1 \\ y & - & z & + & w & = & - 1 \\ 3 x & + & 6 z & - & 6 w & = & 6 \\ - y & + & z & - & w & = & 1 \end{array} & ⟶_{\begin{matrix} \end{matrix}}^{- 3 ρ_{1} + ρ_{3}} \begin{array}{rcrcrcrcr} x & + & y & + & z & - & w & = & 1 \\ y & - & z & + & w & = & - 1 \\ - 3 y & + & 3 z & - & 3 w & = & 3 \\ - y & + & z & - & w & = & 1 \end{array} \\ ⟶_{\begin{matrix} ρ_{2} + ρ_{4} \end{matrix}}^{3 ρ_{2} + ρ_{3}} \begin{array}{rcrcrcrcr} x & + & y & + & z & - & w & = & 1 \\ y & - & z & + & w & = & - 1 \\ 0 & = & 0 \\ 0 & = & 0 \end{array} \end{aligned}$

leaves $x$ and $y$ leading and both $z$ and $w$ free. To get the description that we prefer, we work from the bottom. We first express the leading variable $y$ in terms of $z$ and $w$ , as $y = - 1 + z - w$ . Moving up to the top equation, substituting for $y$ gives $x + (- 1 + z - w) + z - w = 1$ and solving for $x$ leaves $x = 2 - 2 z + 2 w$ . The solution set

$\begin{matrix} {(2 - 2 z + 2 w, - 1 + z - w, z, w) ∣ z, w \in ℝ} & (* *) \end{matrix}$

has the leading variables expressed in terms of the variables that are free.

Example 2.4 The list of leading variables may skip over some columns. After this reduction

$\begin{aligned} \begin{array}{rcrcrcrcr} 2 x & - & 2 y & = & 0 \\ z & + & 3 w & = & 2 \\ 3 x & - & 3 y & = & 0 \\ x & - & y & + & 2 z & + & 6 w & = & 4 \end{array} & ⟶_{\begin{matrix} - (1 / 2) ρ_{1} + ρ_{4} \end{matrix}}^{- (3 / 2) ρ_{1} + ρ_{3}} \begin{array}{rcrcrcrcr} 2 x & - & 2 y & = & 0 \\ z & + & 3 w & = & 2 \\ 0 & = & 0 \\ 2 z & + & 6 w & = & 4 \end{array} \\ ⟶_{\begin{matrix} \end{matrix}}^{- 2 ρ_{2} + ρ_{4}} \begin{array}{rcrcrcrcr} 2 x & - & 2 y & = & 0 \\ z & + & 3 w & = & 2 \\ 0 & = & 0 \\ 0 & = & 0 \end{array} \end{aligned}$

$x$ and $z$ are the leading variables, not $x$ and $y$ . The free variables are $y$ and $w$ and so we can describe the solution set as ${(y, y, 2 - 3 w, w) ∣ y, w \in ℝ}$ . For instance, $(1, 1, 2, 0)$ satisfies the system—take $y = 1$ and $w = 0$ . The four-tuple $(1, 0, 5, 4)$ is not a solution since its first coordinate does not equal its second.

A variable that we use to describe a family of solutions is a parameter. We say that the solution set in the prior example is parametrized with $y$ and $w$ .

The terms ‘parameter’ and ‘free variable’ do not mean the same thing. In the prior example $y$ and $w$ are free because in the echelon form system they do not lead. They are parameters because we used them to describe the set of solutions. Had we instead rewritten the second equation as $w = 2 / 3 - (1 / 3) z$ then the free variables would still be $y$ and $w$ but the parameters would be $y$ and $z$ .

In the rest of this book we will solve linear systems by bringing them to echelon form and then parametrizing with the free variables.

Example 2.5 This is another system with infinitely many solutions.

$\begin{aligned} \begin{array}{rcrcrcrcr} x & + & 2 y & = & 1 \\ 2 x & + & z & = & 2 \\ 3 x & + & 2 y & + & z & - & w & = & 4 \end{array} & ⟶_{\begin{matrix} - 3 ρ_{1} + ρ_{3} \end{matrix}}^{- 2 ρ_{1} + ρ_{2}} \begin{array}{rcrcrcrcr} x & + & 2 y & = & 1 \\ - 4 y & + & z & = & 0 \\ - 4 y & + & z & - & w & = & 1 \end{array} \\ ⟶_{\begin{matrix} \end{matrix}}^{- ρ_{2} + ρ_{3}} \begin{array}{rcrcrcrcr} x & + & 2 y & = & 1 \\ - 4 y & + & z & = & 0 \\ - w & = & 1 \end{array} \end{aligned}$

The leading variables are $x$ , $y$ , and $w$ . The variable $z$ is free. Notice that, although there are infinitely many solutions, the value of $w$ doesn’t vary but is constant $w = - 1$ . To parametrize, write $w$ in terms of $z$ with $w = - 1 + 0 z$ . Then $y = (1 / 4) z$ . Substitute for $y$ in the first equation to get $x = 1 - (1 / 2) z$ . The solution set is ${(1 - (1 / 2) z, (1 / 4) z, z, - 1) ∣ z \in ℝ}$ .

Parametrizing solution sets shows that systems with free variables have infinitely many solutions. For instance, above $z$ takes on all of infinitely many real number values, each associated with a different solution.

We finish this subsection by developing a streamlined notation for linear systems and their solution sets.

Definition 2.6 An $m \times n$ matrix is a rectangular array of numbers with $m$ rows and $n$ columns. Each number in the matrix is an entry.

We usually denote a matrix with an upper case roman letter. For instance,

$A = (\begin{matrix} 1 & 2.2 & 5 \\ 3 & 4 & - 7 \end{matrix})$

has $2$ rows and $3$ columns and so is a $2 \times 3$ matrix. Read that aloud as “two-by-three”; the number of rows is always stated first. (The matrix has parentheses around it so that when two matrices are adjacent we can tell where one ends and the other begins.) We name matrix entries with the corresponding lower-case letter, so that the entry in the second row and first column of the above array is $a_{2, 1} = 3$ . Note that the order of the subscripts matters: $a_{1, 2} \neq a_{2, 1}$ since $a_{1, 2} = 2.2$ . We denote the set of all $m \times n$ matrices by $ℳ_{m \times n}$ .

We do Gauss’s Method using matrices in essentially the same way that we did it for systems of equations: a matrix row’s leading entry is its first nonzero entry (if it has one) and we perform row operations to arrive at matrix echelon form, where the leading entry in lower rows are to the right of those in the rows above. We like matrix notation because it lightens the clerical load, the copying of variables and the writing of $+$ ’s and $=$ ’s.

Example 2.7 We can abbreviate this linear system

$\begin{array}{rcrcrcr} x & + & 2 y & = & 4 \\ y & - & z & = & 0 \\ x & + & 2 z & = & 4 \end{array}$

with this matrix.

$(\begin{array}{cccr} 1 & 2 & 0 & 4 \\ 0 & 1 & - 1 & 0 \\ 1 & 0 & 2 & 4 \end{array})$

The vertical bar reminds a reader of the difference between the coefficients on the system’s left hand side and the constants on the right. With a bar, this is an augmented matrix.

$(\begin{array}{cccr} 1 & 2 & 0 & 4 \\ 0 & 1 & - 1 & 0 \\ 1 & 0 & 2 & 4 \end{array}) ⟶_{\begin{array}{l} \end{array}}^{- ρ_{1} + ρ_{3}} (\begin{array}{cccr} 1 & 2 & 0 & 4 \\ 0 & 1 & - 1 & 0 \\ 0 & - 2 & 2 & 0 \end{array}) ⟶_{\begin{array}{l} \end{array}}^{2 ρ_{2} + ρ_{3}} (\begin{array}{cccr} 1 & 2 & 0 & 4 \\ 0 & 1 & - 1 & 0 \\ 0 & 0 & 0 & 0 \end{array})$

The second row stands for $y - z = 0$ and the first row stands for $x + 2 y = 4$ so the solution set is ${(4 - 2 z, z, z) ∣ z \in ℝ}$ .

Matrix notation also clarifies the descriptions of solution sets. Example 2.3’s ${(2 - 2 z + 2 w, - 1 + z - w, z, w) ∣ z, w \in ℝ}$ is hard to read. We will rewrite it to group all of the constants together, all of the coefficients of $z$ together, and all of the coefficients of $w$ together. We write them vertically, in one-column matrices.

${(\begin{matrix} 2 \\ - 1 \\ 0 \\ 0 \end{matrix}) + (\begin{matrix} - 2 \\ 1 \\ 1 \\ 0 \end{matrix}) \cdot z + (\begin{matrix} 2 \\ - 1 \\ 0 \\ 1 \end{matrix}) \cdot w ∣ z, w \in ℝ}$

For instance, the top line says that $x = 2 - 2 z + 2 w$ and the second line says that $y = - 1 + z - w$ . (Our next section gives a geometric interpretation that will help us picture the solution sets.)

Definition 2.8 A column vector, often just called a vector, is a matrix with a single column. A matrix with a single row is a row vector. The entries of a vector are sometimes called components. A column or row vector whose components are all zeros is a zero vector.

Vectors are an exception to the convention of representing matrices with capital roman letters. We use lower-case roman or greek letters overlined with an arrow: $\vec{a}$ , $\vec{b}$ , … or $\vec{α}$ , $\vec{β}$ , … (boldface is also common: $𝒂$ or $𝜶$ ). For instance, this is a column vector with a third component of $7$ .

$\vec{v} = (\begin{matrix} 1 \\ 3 \\ 7 \end{matrix})$

A zero vector is denoted $\vec{0}$ . There are many different zero vectors—the one-tall zero vector, the two-tall zero vector, etc.—but nonetheless we will often say “the” zero vector, expecting that the size will be clear from the context.

Definition 2.9 The linear equation $a_{1} x_{1} + a_{2} x_{2} + \dots + a_{n} x_{n} = d$ with unknowns $x_{1}, \dots, x_{n}$ is satisfied by

$\vec{s} = (\begin{matrix} s_{1} \\ ⋮ \\ s_{n} \end{matrix})$

if $a_{1} s_{1} + a_{2} s_{2} + \dots + a_{n} s_{n} = d$ . A vector satisfies a linear system if it satisfies each equation in the system.

The style of description of solution sets that we use involves adding the vectors, and also multiplying them by real numbers. Before we give the examples showing the style we first need to define these operations.

Definition 2.10 The vector sum of $\vec{u}$ and $\vec{v}$ is the vector of the sums.

$\vec{u} + \vec{v} = (\begin{matrix} u_{1} \\ ⋮ \\ u_{n} \end{matrix}) + (\begin{matrix} v_{1} \\ ⋮ \\ v_{n} \end{matrix}) = (\begin{matrix} u_{1} + v_{1} \\ ⋮ \\ u_{n} + v_{n} \end{matrix})$

Note that for the addition to be defined the vectors must have the same number of entries. This entry-by-entry addition works for any pair of matrices, not just vectors, provided that they have the same number of rows and columns.

Definition 2.11 The scalar multiplication of the real number $r$ and the vector $\vec{v}$ is the vector of the multiples.

$r \cdot \vec{v} = r \cdot (\begin{matrix} v_{1} \\ ⋮ \\ v_{n} \end{matrix}) = (\begin{matrix} r v_{1} \\ ⋮ \\ r v_{n} \end{matrix})$

As with the addition operation, the entry-by-entry scalar multiplication operation extends beyond vectors to apply to any matrix.

We write scalar multiplication either as $r \cdot \vec{v}$ or $\vec{v} \cdot r$ , and sometimes even omit the ‘ $\cdot$ ’ symbol: $r \vec{v}$ . (Do not refer to scalar multiplication as ‘scalar product’ because that name is for a different operation.)

Example 2.12 $(\begin{matrix} 2 \\ 3 \\ 1 \end{matrix}) + (\begin{matrix} 3 \\ - 1 \\ 4 \end{matrix}) = (\begin{matrix} 2 + 3 \\ 3 - 1 \\ 1 + 4 \end{matrix}) = (\begin{matrix} 5 \\ 2 \\ 5 \end{matrix}) 7 \cdot (\begin{matrix} 1 \\ 4 \\ - 1 \\ - 3 \end{matrix}) = (\begin{matrix} 7 \\ 28 \\ - 7 \\ - 21 \end{matrix})$

Observe that the definitions of addition and scalar multiplication agree where they overlap; for instance, $\vec{v} + \vec{v} = 2 \vec{v}$ .

With these definitions, we are set to use matrix and vector notation to both solve systems and express the solution.

Example 2.13 This system

$\begin{array}{rcrcrcrcrcr} 2 x & + & y & - & w & = & 4 \\ y & + & w & + & u & = & 4 \\ x & - & z & + & 2 w & = & 0 \end{array}$

reduces in this way.

$\begin{aligned} (\begin{array}{cccccr} 2 & 1 & 0 & - 1 & 0 & 4 \\ 0 & 1 & 0 & 1 & 1 & 4 \\ 1 & 0 & - 1 & 2 & 0 & 0 \end{array}) & ⟶_{\begin{matrix} \end{matrix}}^{- (1 / 2) ρ_{1} + ρ_{3}} (\begin{array}{cccccr} 2 & 1 & 0 & - 1 & 0 & 4 \\ 0 & 1 & 0 & 1 & 1 & 4 \\ 0 & - 1 / 2 & - 1 & 5 / 2 & 0 & - 2 \end{array}) \\ ⟶_{\begin{matrix} \end{matrix}}^{(1 / 2) ρ_{2} + ρ_{3}} (\begin{array}{cccccr} 2 & 1 & 0 & - 1 & 0 & 4 \\ 0 & 1 & 0 & 1 & 1 & 4 \\ 0 & 0 & - 1 & 3 & 1 / 2 & 0 \end{array}) \end{aligned}$

The solution set is ${(w + (1 / 2) u, 4 - w - u, 3 w + (1 / 2) u, w, u) ∣ w, u \in ℝ}$ . We write that in vector form.

${(\begin{matrix} x \\ y \\ z \\ w \\ u \end{matrix}) = (\begin{matrix} 0 \\ 4 \\ 0 \\ 0 \\ 0 \end{matrix}) + (\begin{matrix} 1 \\ - 1 \\ 3 \\ 1 \\ 0 \end{matrix}) w + (\begin{matrix} 1 / 2 \\ - 1 \\ 1 / 2 \\ 0 \\ 1 \end{matrix}) u ∣ w, u \in ℝ}$

Note how well vector notation sets off the coefficients of each parameter. For instance, the third row of the vector form shows plainly that if $u$ is fixed then $z$ increases three times as fast as $w$ . Another thing shown plainly is that setting both $w$ and $u$ to zero gives that

$(\begin{matrix} x \\ y \\ z \\ w \\ u \end{matrix}) = (\begin{matrix} 0 \\ 4 \\ 0 \\ 0 \\ 0 \end{matrix})$

is a particular solution of the linear system.

Example 2.14 In the same way, the system

$\begin{array}{rcrcrcr} x & - & y & + & z & = & 1 \\ 3 x & + & z & = & 3 \\ 5 x & - & 2 y & + & 3 z & = & 5 \end{array}$

reduces

$\begin{aligned} (\begin{array}{cccr} 1 & - 1 & 1 & 1 \\ 3 & 0 & 1 & 3 \\ 5 & - 2 & 3 & 5 \end{array}) & ⟶_{\begin{matrix} - 5 ρ_{1} + ρ_{3} \end{matrix}}^{- 3 ρ_{1} + ρ_{2}} (\begin{array}{cccr} 1 & - 1 & 1 & 1 \\ 0 & 3 & - 2 & 0 \\ 0 & 3 & - 2 & 0 \end{array}) \\ ⟶_{\begin{matrix} \end{matrix}}^{- ρ_{2} + ρ_{3}} (\begin{array}{cccr} 1 & - 1 & 1 & 1 \\ 0 & 3 & - 2 & 0 \\ 0 & 0 & 0 & 0 \end{array}) \end{aligned}$

to give a one-parameter solution set.

${(\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}) + (\begin{matrix} - 1 / 3 \\ 2 / 3 \\ 1 \end{matrix}) z ∣ z \in ℝ}$

As in the prior example, the vector not associated with the parameter

$(\begin{matrix} 1 \\ 0 \\ 0 \end{matrix})$

is a particular solution of the system.

Before the exercises, we will consider what we have accomplished and what we will do in the remainder of the chapter. So far we have done the mechanics of Gauss’s Method. We have not stopped to consider any of the questions that arise, except for proving Theorem 1.5—which justifies the method by showing that it gives the right answers.

For example, can we always describe solution sets as above, with a particular solution vector added to an unrestricted linear combination of some other vectors? We’ve noted that the solution sets described in this way have infinitely many members so answering this question would tell us about the size of solution sets. The following subsection shows that the answer is “yes.” This chapter’s second section then uses that answer to describe the geometry of solution sets.

Other questions arise from the observation that we can do Gauss’s Method in more than one way (for instance, when swapping rows we may have a choice of rows to swap with). Theorem 1.5 says that we must get the same solution set no matter how we proceed but if we do Gauss’s Method in two ways must we get the same number of free variables in each echelon form system? Must those be the same variables, that is, is it impossible to solve a problem one way to get $y$ and $w$ free and solve it another way to get $y$ and $z$ free? The third section of this chapter answers “yes,” that from any starting linear system, all derived echelon form versions have the same free variables.

Thus, by the end of the chapter we will not only have a solid grounding in the practice of Gauss’s Method but we will also have a solid grounding in the theory. We will know exactly what can and cannot happen in a reduction.

Exercises

Exercise 2.15 Worked answer

Recommended. Find the indicated entry of the matrix, if it is defined.

$A = (\begin{matrix} 1 & 3 & 1 \\ 2 & - 1 & 4 \end{matrix})$
1. $a_{2, 1}$
2. $a_{1, 2}$
3. $a_{2, 2}$
4. $a_{3, 1}$
Back to Exercise 2.15

Answer.
1. $2$
2. $3$
3. $- 1$
4. Not defined.
Exercise 2.16 Worked answer

Recommended. Give the size of each matrix.
1. $(\begin{matrix} 1 & 0 & 4 \\ 2 & 1 & 5 \end{matrix})$
2. $(\begin{matrix} 1 & 1 \\ - 1 & 1 \\ 3 & - 1 \end{matrix})$
3. $(\begin{matrix} 5 & 10 \\ 10 & 5 \end{matrix})$
Back to Exercise 2.16

Answer.
1. $2 \times 3$
2. $3 \times 2$
3. $2 \times 2$
Exercise 2.17 Worked answer

Recommended. Do the indicated vector operation, if it is defined.
1. $(\begin{matrix} 2 \\ 1 \\ 1 \end{matrix}) + (\begin{matrix} 3 \\ 0 \\ 4 \end{matrix})$
2. $5 (\begin{matrix} 4 \\ - 1 \end{matrix})$
3. $(\begin{matrix} 1 \\ 5 \\ 1 \end{matrix}) - (\begin{matrix} 3 \\ 1 \\ 1 \end{matrix})$
4. $7 (\begin{matrix} 2 \\ 1 \end{matrix}) + 9 (\begin{matrix} 3 \\ 5 \end{matrix})$
5. $(\begin{matrix} 1 \\ 2 \end{matrix}) + (\begin{matrix} 1 \\ 2 \\ 3 \end{matrix})$
6. $6 (\begin{matrix} 3 \\ 1 \\ 1 \end{matrix}) - 4 (\begin{matrix} 2 \\ 0 \\ 3 \end{matrix}) + 2 (\begin{matrix} 1 \\ 1 \\ 5 \end{matrix})$
Back to Exercise 2.17

Answer.
1. $(\begin{matrix} 5 \\ 1 \\ 5 \end{matrix})$
2. $(\begin{matrix} 20 \\ - 5 \end{matrix})$
3. $(\begin{matrix} - 2 \\ 4 \\ 0 \end{matrix})$
4. $(\begin{matrix} 41 \\ 52 \end{matrix})$
5. Not defined.
6. $(\begin{matrix} 12 \\ 8 \\ 4 \end{matrix})$
Exercise 2.18 Worked answer

Recommended. Solve each system using matrix notation. Express the solution using vectors.
1. $\begin{array}{rcrcr} 3 x & + & 6 y & = & 18 \\ x & + & 2 y & = & 6 \end{array}$
2. $\begin{array}{rcrcr} x & + & y & = & 1 \\ x & - & y & = & - 1 \end{array}$
3. $\begin{array}{rcrcrcr} x_{1} & + & x_{3} & = & 4 \\ x_{1} & - & x_{2} & + & 2 x_{3} & = & 5 \\ 4 x_{1} & - & x_{2} & + & 5 x_{3} & = & 17 \end{array}$
4. $\begin{array}{rcrcrcr} 2 a & + & b & - & c & = & 2 \\ 2 a & + & c & = & 3 \\ a & - & b & = & 0 \end{array}$
5. $\begin{array}{rcrcrcrcr} x & + & 2 y & - & z & = & 3 \\ 2 x & + & y & + & w & = & 4 \\ x & - & y & + & z & + & w & = & 1 \end{array}$
6. $\begin{array}{rcrcrcrcr} x & + & z & + & w & = & 4 \\ 2 x & + & y & - & w & = & 2 \\ 3 x & + & y & + & z & = & 7 \end{array}$
Back to Exercise 2.18

Answer.
1. This reduction
  
  $(\begin{array}{ccr} 3 & 6 & 18 \\ 1 & 2 & 6 \end{array}) ⟶_{\begin{array}{l} \end{array}}^{(- 1 / 3) ρ_{1} + ρ_{2}} (\begin{array}{ccr} 3 & 6 & 18 \\ 0 & 0 & 0 \end{array})$
  
  leaves $x$ leading and $y$ free. Making $y$ the parameter, gives $x = 6 - 2 y$ and this solution set.
  
  ${(\begin{matrix} 6 \\ 0 \end{matrix}) + (\begin{matrix} - 2 \\ 1 \end{matrix}) y ∣ y \in ℝ}$
2. A reduction
  
  $(\begin{array}{ccr} 1 & 1 & 1 \\ 1 & - 1 & - 1 \end{array}) ⟶_{\begin{array}{l} \end{array}}^{- ρ_{1} + ρ_{2}} (\begin{array}{ccr} 1 & 1 & 1 \\ 0 & - 2 & - 2 \end{array})$
  
  gives the unique solution $y = 1$ , $x = 0$ . The solution set is this.
  
  ${(\begin{matrix} 0 \\ 1 \end{matrix})}$
3. Gauss’s Method
  
  $(\begin{array}{cccr} 1 & 0 & 1 & 4 \\ 1 & - 1 & 2 & 5 \\ 4 & - 1 & 5 & 17 \end{array}) ⟶_{\begin{array}{l} - 4 ρ_{1} + ρ_{3} \end{array}}^{- ρ_{1} + ρ_{2}} (\begin{array}{cccr} 1 & 0 & 1 & 4 \\ 0 & - 1 & 1 & 1 \\ 0 & - 1 & 1 & 1 \end{array}) ⟶_{\begin{array}{l} \end{array}}^{- ρ_{2} + ρ_{3}} (\begin{array}{cccr} 1 & 0 & 1 & 4 \\ 0 & - 1 & 1 & 1 \\ 0 & 0 & 0 & 0 \end{array})$
  
  leaves $x_{1}$ and $x_{2}$ leading with $x_{3}$ free. The solution set is this.
  
  ${(\begin{matrix} 4 \\ - 1 \\ 0 \end{matrix}) + (\begin{matrix} - 1 \\ 1 \\ 1 \end{matrix}) x_{3} ∣ x_{3} \in ℝ}$
4. This reduction
  
  $\begin{aligned} (\begin{array}{cccr} 2 & 1 & - 1 & 2 \\ 2 & 0 & 1 & 3 \\ 1 & - 1 & 0 & 0 \end{array}) & ⟶_{\begin{matrix} - (1 / 2) ρ_{1} + ρ_{3} \end{matrix}}^{- ρ_{1} + ρ_{2}} (\begin{array}{cccr} 2 & 1 & - 1 & 2 \\ 0 & - 1 & 2 & 1 \\ 0 & - 3 / 2 & 1 / 2 & - 1 \end{array}) \\ ⟶_{\begin{matrix} \end{matrix}}^{(- 3 / 2) ρ_{2} + ρ_{3}} (\begin{array}{cccr} 2 & 1 & - 1 & 2 \\ 0 & - 1 & 2 & 1 \\ 0 & 0 & - 5 / 2 & - 5 / 2 \end{array}) \end{aligned}$
  
  shows that the solution set is a singleton set.
  
  ${(\begin{matrix} 1 \\ 1 \\ 1 \end{matrix})}$
5. This reduction is easy
  
  $\begin{aligned} (\begin{array}{ccccr} 1 & 2 & - 1 & 0 & 3 \\ 2 & 1 & 0 & 1 & 4 \\ 1 & - 1 & 1 & 1 & 1 \end{array}) & ⟶_{\begin{matrix} - ρ_{1} + ρ_{3} \end{matrix}}^{- 2 ρ_{1} + ρ_{2}} (\begin{array}{ccccr} 1 & 2 & - 1 & 0 & 3 \\ 0 & - 3 & 2 & 1 & - 2 \\ 0 & - 3 & 2 & 1 & - 2 \end{array}) \\ ⟶_{\begin{matrix} \end{matrix}}^{- ρ_{2} + ρ_{3}} (\begin{array}{ccccr} 1 & 2 & - 1 & 0 & 3 \\ 0 & - 3 & 2 & 1 & - 2 \\ 0 & 0 & 0 & 0 & 0 \end{array}) \end{aligned}$
  
  and ends with $x$ and $y$ leading while $z$ and $w$ are free. Solving for $y$ gives $y = (2 + 2 z + w) / 3$ and substitution shows that $x + 2 (2 + 2 z + w) / 3 - z = 3$ so $x = (5 / 3) - (1 / 3) z - (2 / 3) w$ , making this the solution set.
  
  ${(\begin{matrix} 5 / 3 \\ 2 / 3 \\ 0 \\ 0 \end{matrix}) + (\begin{matrix} - 1 / 3 \\ 2 / 3 \\ 1 \\ 0 \end{matrix}) z + (\begin{matrix} - 2 / 3 \\ 1 / 3 \\ 0 \\ 1 \end{matrix}) w ∣ z, w \in ℝ}$
6. The reduction
  
  $\begin{aligned} (\begin{array}{ccccr} 1 & 0 & 1 & 1 & 4 \\ 2 & 1 & 0 & - 1 & 2 \\ 3 & 1 & 1 & 0 & 7 \end{array}) & ⟶_{\begin{matrix} - 3 ρ_{1} + ρ_{3} \end{matrix}}^{- 2 ρ_{1} + ρ_{2}} (\begin{array}{ccccr} 1 & 0 & 1 & 1 & 4 \\ 0 & 1 & - 2 & - 3 & - 6 \\ 0 & 1 & - 2 & - 3 & - 5 \end{array}) \\ ⟶_{\begin{matrix} \end{matrix}}^{- ρ_{2} + ρ_{3}} (\begin{array}{ccccr} 1 & 0 & 1 & 1 & 4 \\ 0 & 1 & - 2 & - 3 & - 6 \\ 0 & 0 & 0 & 0 & 1 \end{array}) \end{aligned}$
  
  shows that there is no solution—the solution set is empty.
Exercise 2.19 Worked answer

Solve each system using matrix notation. Give each solution set in vector notation.
1. $\begin{array}{rcrcrcr} 2 x & + & y & - & z & = & 1 \\ 4 x & - & y & = & 3 \end{array}$
2. $\begin{array}{rcrcrcrcr} x & - & z & = & 1 \\ y & + & 2 z & - & w & = & 3 \\ x & + & 2 y & + & 3 z & - & w & = & 7 \end{array}$
3. $\begin{array}{rcrcrcrcr} x & - & y & + & z & = & 0 \\ y & + & w & = & 0 \\ 3 x & - & 2 y & + & 3 z & + & w & = & 0 \\ - y & - & w & = & 0 \end{array}$
4. $\begin{array}{rcrcrcrcrcr} a & + & 2 b & + & 3 c & + & d & - & e & = & 1 \\ 3 a & - & b & + & c & + & d & + & e & = & 3 \end{array}$
Back to Exercise 2.19

Answer.
1. The reduction
  
  $(\begin{array}{cccr} 2 & 1 & - 1 & 1 \\ 4 & - 1 & 0 & 3 \end{array}) ⟶_{\begin{array}{l} \end{array}}^{- 2 ρ_{1} + ρ_{2}} (\begin{array}{cccr} 2 & 1 & - 1 & 1 \\ 0 & - 3 & 2 & 1 \end{array})$
  
  ends with $x$ and $y$ leading, and with $z$ free. Solving for $y$ gives $y = (1 - 2 z) / (- 3)$ , and then substitution $2 x + (1 - 2 z) / (- 3) - z = 1$ shows that $x = ((4 / 3) + (1 / 3) z) / 2$ . Hence the solution set is this.
  
  ${(\begin{matrix} 2 / 3 \\ - 1 / 3 \\ 0 \end{matrix}) + (\begin{matrix} 1 / 6 \\ 2 / 3 \\ 1 \end{matrix}) z ∣ z \in ℝ}$
2. This application of Gauss’s Method
  
  $\begin{aligned} (\begin{array}{ccccr} 1 & 0 & - 1 & 0 & 1 \\ 0 & 1 & 2 & - 1 & 3 \\ 1 & 2 & 3 & - 1 & 7 \end{array}) & ⟶_{\begin{matrix} \end{matrix}}^{- ρ_{1} + ρ_{3}} (\begin{array}{ccccr} 1 & 0 & - 1 & 0 & 1 \\ 0 & 1 & 2 & - 1 & 3 \\ 0 & 2 & 4 & - 1 & 6 \end{array}) \\ ⟶_{\begin{matrix} \end{matrix}}^{- 2 ρ_{2} + ρ_{3}} (\begin{array}{ccccr} 1 & 0 & - 1 & 0 & 1 \\ 0 & 1 & 2 & - 1 & 3 \\ 0 & 0 & 0 & 1 & 0 \end{array}) \end{aligned}$
  
  leaves $x$ , $y$ , and $w$ leading. The solution set is here.
  
  ${(\begin{matrix} 1 \\ 3 \\ 0 \\ 0 \end{matrix}) + (\begin{matrix} 1 \\ - 2 \\ 1 \\ 0 \end{matrix}) z ∣ z \in ℝ}$
3. This row reduction
  
  $\begin{aligned} (\begin{array}{ccccr} 1 & - 1 & 1 & 0 & 0 \\ 0 & 1 & 0 & 1 & 0 \\ 3 & - 2 & 3 & 1 & 0 \\ 0 & - 1 & 0 & - 1 & 0 \end{array}) & ⟶_{\begin{matrix} \end{matrix}}^{- 3 ρ_{1} + ρ_{3}} (\begin{array}{ccccr} 1 & - 1 & 1 & 0 & 0 \\ 0 & 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 & 0 \\ 0 & - 1 & 0 & - 1 & 0 \end{array}) \\ ⟶_{\begin{matrix} ρ_{2} + ρ_{4} \end{matrix}}^{- ρ_{2} + ρ_{3}} (\begin{array}{ccccr} 1 & - 1 & 1 & 0 & 0 \\ 0 & 1 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array}) \end{aligned}$
  
  ends with $z$ and $w$ free. We have this solution set.
  
  ${(\begin{matrix} 0 \\ 0 \\ 0 \\ 0 \end{matrix}) + (\begin{matrix} - 1 \\ 0 \\ 1 \\ 0 \end{matrix}) z + (\begin{matrix} - 1 \\ - 1 \\ 0 \\ 1 \end{matrix}) w ∣ z, w \in ℝ}$
4. Gauss’s Method done in this way
  
  $(\begin{array}{cccccr} 1 & 2 & 3 & 1 & - 1 & 1 \\ 3 & - 1 & 1 & 1 & 1 & 3 \end{array}) ⟶_{\begin{array}{l} \end{array}}^{- 3 ρ_{1} + ρ_{2}} (\begin{array}{cccccr} 1 & 2 & 3 & 1 & - 1 & 1 \\ 0 & - 7 & - 8 & - 2 & 4 & 0 \end{array})$
  
  ends with $c$ , $d$ , and $e$ free. Solving for $b$ shows that $b = (8 c + 2 d - 4 e) / (- 7)$ and then substitution $a + 2 (8 c + 2 d - 4 e) / (- 7) + 3 c + 1 d - 1 e = 1$ shows that $a = 1 - (5 / 7) c - (3 / 7) d - (1 / 7) e$ and we have the solution set.
  
  ${(\begin{matrix} 1 \\ 0 \\ 0 \\ 0 \\ 0 \end{matrix}) + (\begin{matrix} - 5 / 7 \\ - 8 / 7 \\ 1 \\ 0 \\ 0 \end{matrix}) c + (\begin{matrix} - 3 / 7 \\ - 2 / 7 \\ 0 \\ 1 \\ 0 \end{matrix}) d + (\begin{matrix} - 1 / 7 \\ 4 / 7 \\ 0 \\ 0 \\ 1 \end{matrix}) e ∣ c, d, e \in ℝ}$
Exercise 2.20 Worked answer

Solve each system using matrix notation. Express the solution set using vectors.
1. $\begin{array}{rcrcrcr} 3 x & + & 2 y & + & z & = & 1 \\ x & - & y & + & z & = & 2 \\ 5 x & + & 5 y & + & z & = & 0 \end{array}$
2. $\begin{array}{rcrcrcrcr} x & + & y & - & 2 z & = & 0 \\ x & - & y & = & - 3 \\ 3 x & - & y & - & 2 z & = & - 6 \\ 2 y & - & 2 z & = & 3 \end{array}$
3. $\begin{array}{rcrcrcrcrcr} 2 x & - & y & - & z & + & w & = & 4 \\ x & + & y & + & z & = & - 1 \end{array}$
4. $\begin{array}{rcrcrcr} x & + & y & - & 2 z & = & 0 \\ x & - & y & = & - 3 \\ 3 x & - & y & - & 2 z & = & 0 \end{array}$
Back to Exercise 2.20

Answer.
1. This reduction
  
  $\begin{aligned} (\begin{array}{cccc} 3 & 2 & 1 & 1 \\ 1 & - 1 & 1 & 2 \\ 5 & 5 & 1 & 0 \end{array}) & ⟶_{\begin{matrix} - (5 / 3) ρ_{1} + ρ_{3} \end{matrix}}^{- (1 / 3) ρ_{1} + ρ_{2}} (\begin{array}{cccc} 3 & 2 & 1 & 1 \\ 0 & - 5 / 3 & 2 / 3 & 5 / 3 \\ 0 & 5 / 3 & - 2 / 3 & - 5 / 3 \end{array}) \\ ⟶_{\begin{matrix} \end{matrix}}^{ρ_{2} + ρ_{3}} (\begin{array}{cccc} 3 & 2 & 1 & 1 \\ 0 & - 5 / 3 & 2 / 3 & 5 / 3 \\ 0 & 0 & 0 & 0 \end{array}) \end{aligned}$
  
  gives this solution set.
  
  ${(\begin{matrix} x \\ y \\ z \end{matrix}) = (\begin{matrix} 1 \\ - 1 \\ 0 \end{matrix}) + (\begin{matrix} - 3 / 5 \\ 2 / 5 \\ 1 \end{matrix}) z ∣ z \in ℝ}$
2. This is the reduction.
  
  $\begin{aligned} (\begin{array}{cccc} 1 & 1 & - 2 & 0 \\ 1 & - 1 & 0 & 3 \\ 3 & - 1 & - 2 & - 6 \\ 0 & 2 & - 2 & 3 \end{array}) & ⟶_{\begin{matrix} - 3 ρ_{1} + ρ_{3} \end{matrix}}^{- ρ_{1} + ρ_{2}} (\begin{array}{cccc} 1 & 1 & - 2 & 0 \\ 0 & - 2 & 2 & - 3 \\ 0 & - 4 & 4 & - 6 \\ 0 & 2 & - 2 & 3 \end{array}) \\ ⟶_{\begin{matrix} ρ_{2} + ρ_{4} \end{matrix}}^{- 2 ρ_{2} + ρ_{3}} (\begin{array}{cccc} 1 & 1 & - 2 & 0 \\ 0 & - 2 & 2 & - 3 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}) \end{aligned}$
  
  The solution set is this.
  
  ${(\begin{matrix} - 3 / 2 \\ 3 / 2 \\ 0 \end{matrix}) + (\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}) z ∣ z \in ℝ}$
3. Gauss’s Method
  
  $(\begin{array}{ccccc} 2 & - 1 & - 1 & 1 & 4 \\ 1 & 1 & 1 & 0 & - 1 \end{array}) ⟶_{\begin{array}{l} \end{array}}^{- (1 / 2) ρ_{1} + ρ_{2}} (\begin{array}{ccccc} 2 & - 1 & - 1 & 1 & 4 \\ 0 & 3 / 2 & 3 / 2 & - 1 / 2 & - 3 \end{array})$
  
  gives the solution set.
  
  ${(\begin{matrix} 1 \\ - 2 \\ 0 \\ 0 \end{matrix}) + (\begin{matrix} 0 \\ - 1 \\ 1 \\ 0 \end{matrix}) z (\begin{matrix} - 1 / 3 \\ 1 / 3 \\ 0 \\ 1 \end{matrix}) w ∣ z, w \in ℝ}$
4. Here is the reduction.
  
  $(\begin{array}{cccc} 1 & 1 & - 2 & 0 \\ 1 & - 1 & 0 & - 3 \\ 3 & - 1 & - 2 & 0 \end{array}) ⟶_{\begin{array}{l} - 3 ρ_{1} + ρ_{3} \end{array}}^{- ρ_{1} + ρ_{2}} (\begin{array}{cccc} 1 & 1 & - 2 & 0 \\ 0 & - 2 & 2 & - 3 \\ 0 & - 4 & 4 & 0 \end{array}) ⟶_{\begin{array}{l} \end{array}}^{- 2 ρ_{2} + ρ_{3}} (\begin{array}{cccc} 1 & 1 & - 2 & 0 \\ 0 & - 2 & 2 & - 3 \\ 0 & 0 & 0 & 6 \end{array})$
  
  The solution set is empty ${}$ .
Exercise 2.21 Worked answer

Recommended. The vector is in the set. What value of the parameters produces that vector?
1. $(\begin{matrix} 5 \\ - 5 \end{matrix})$ , ${(\begin{matrix} 1 \\ - 1 \end{matrix}) k ∣ k \in ℝ}$
2. $(\begin{matrix} - 1 \\ 2 \\ 1 \end{matrix})$ , ${(\begin{matrix} - 2 \\ 1 \\ 0 \end{matrix}) i + (\begin{matrix} 3 \\ 0 \\ 1 \end{matrix}) j ∣ i, j \in ℝ}$
3. $(\begin{matrix} 0 \\ - 4 \\ 2 \end{matrix})$ , ${(\begin{matrix} 1 \\ 1 \\ 0 \end{matrix}) m + (\begin{matrix} 2 \\ 0 \\ 1 \end{matrix}) n ∣ m, n \in ℝ}$
Back to Exercise 2.21

Answer. For each problem we get a system of linear equations by looking at the equations of components.
1. $k = 5$
2. The second components show that $i = 2$ , the third components show that $j = 1$ .
3. $m = - 4$ , $n = 2$
Exercise 2.22 Worked answer

Decide if the vector is in the set.
1. $(\begin{matrix} 3 \\ - 1 \end{matrix})$ , ${(\begin{matrix} - 6 \\ 2 \end{matrix}) k ∣ k \in ℝ}$
2. $(\begin{matrix} 5 \\ 4 \end{matrix})$ , ${(\begin{matrix} 5 \\ - 4 \end{matrix}) j ∣ j \in ℝ}$
3. $(\begin{matrix} 2 \\ 1 \\ - 1 \end{matrix})$ , ${(\begin{matrix} 0 \\ 3 \\ - 7 \end{matrix}) + (\begin{matrix} 1 \\ - 1 \\ 3 \end{matrix}) r ∣ r \in ℝ}$
4. $(\begin{matrix} 1 \\ 0 \\ 1 \end{matrix})$ , ${(\begin{matrix} 2 \\ 0 \\ 1 \end{matrix}) j + (\begin{matrix} - 3 \\ - 1 \\ 1 \end{matrix}) k ∣ j, k \in ℝ}$
Back to Exercise 2.22

Answer. For each problem we get a system of linear equations by looking at the equations of components.
1. Yes; take $k = - 1 / 2$ .
2. No; the system with equations $5 = 5 \cdot j$ and $4 = - 4 \cdot j$ has no solution.
3. Yes; take $r = 2$ .
4. No. The second components give $k = 0$ . Then the third components give $j = 1$ . But the first components don’t check.
Exercise 2.23 Worked answer

[Cleary] A farmer with 1200 acres is considering planting three different crops, corn, soybeans, and oats. The farmer wants to use all $1200$ acres. Seed corn costs $ $20$ per acre, while soybean and oat seed cost $ $50$ and $ $12$ per acre respectively. The farmer has $ $40 000$ available to buy seed and intends to spend it all.
1. Use the information above to formulate two linear equations with three unknowns and solve it.
2. Solutions to the system are choices that the farmer can make. Write down two reasonable solutions.
3. Suppose that in the fall when the crops mature, the farmer can bring in revenue of $ $100$ per acre for corn, $ $300$ per acre for soybeans and $ $80$ per acre for oats. Which of your two solutions in the prior part would have resulted in a larger revenue?
Back to Exercise 2.23

Answer.
1. Let $c$ be the number of acres of corn, $s$ be the number of acres of soy, and $a$ be the number of acres of oats.
  
  $\begin{array}{rcrcrcr} c & + & s & + & a & = & 1200 \\ 20 c & + & 50 s & + & 12 a & = & 40 000 \end{array} ⟶_{\begin{array}{l} \end{array}}^{- 20 ρ_{1} + ρ_{2}} \begin{array}{rcrcrcr} c & + & s & + & a & = & 1200 \\ 30 s & - & 8 a & = & 16 000 \end{array}$
  
  To describe the solution set we can parametrize using $a$ .
  
  ${(\begin{matrix} c \\ s \\ a \end{matrix}) = (\begin{matrix} 20 000 / 30 \\ 16 000 / 30 \\ 0 \end{matrix}) + (\begin{matrix} - 38 / 30 \\ 8 / 30 \\ 1 \end{matrix}) a ∣ a \in ℝ}$
2. There are many answers possible here. For instance we can take $a = 0$ to get $c = 20 000 / 30 \approx 666.66$ and $s = 16000 / 30 \approx 533.33$ . Another example is to take $a = 20 000 / 38 \approx 526.32$ , giving $c = 0$ and $s = 7360 / 38 \approx 193.68$ .
3. Plug your answers from the prior part into $100 c + 300 s + 80 a$ .
Exercise 2.24 Worked answer

Parametrize the solution set of this one-equation system.

$x_{1} + x_{2} + \dots + x_{n} = 0$

Back to Exercise 2.24

Answer. This system has one equation. The leading variable is $x_{1}$ , the other variables are free.

${(\begin{matrix} - 1 \\ 1 \\ ⋮ \\ 0 \end{matrix}) x_{2} + \dots + (\begin{matrix} - 1 \\ 0 \\ ⋮ \\ 1 \end{matrix}) x_{n} ∣ x_{2}, \dots, x_{n} \in ℝ}$
Exercise 2.25 Worked answer

Recommended.
1. Apply Gauss’s Method to the left-hand side to solve
  
  $\begin{array}{rcrcrcrcr} x & + & 2 y & - & w & = & a \\ 2 x & + & z & = & b \\ x & + & y & + & 2 w & = & c \end{array}$
  
  for $x$ , $y$ , $z$ , and $w$ , in terms of the constants $a$ , $b$ , and $c$ .
2. Use your answer from the prior part to solve this.
  
  $\begin{array}{rcrcrcrcr} x & + & 2 y & - & w & = & 3 \\ 2 x & + & z & = & 1 \\ x & + & y & + & 2 w & = & - 2 \end{array}$
Back to Exercise 2.25

Answer.
1. Gauss’s Method here gives
  
  $\begin{aligned} (\begin{array}{ccccc} 1 & 2 & 0 & - 1 & a \\ 2 & 0 & 1 & 0 & b \\ 1 & 1 & 0 & 2 & c \end{array}) & ⟶_{\begin{matrix} - ρ_{1} + ρ_{3} \end{matrix}}^{- 2 ρ_{1} + ρ_{2}} (\begin{array}{ccccc} 1 & 2 & 0 & - 1 & a \\ 0 & - 4 & 1 & 2 & - 2 a + b \\ 0 & - 1 & 0 & 3 & - a + c \end{array}) \\ ⟶_{\begin{matrix} \end{matrix}}^{- (1 / 4) ρ_{2} + ρ_{3}} (\begin{array}{ccccc} 1 & 2 & 0 & - 1 & a \\ 0 & - 4 & 1 & 2 & - 2 a + b \\ 0 & 0 & - 1 / 4 & 5 / 2 & - (1 / 2) a - (1 / 4) b + c \end{array}) \end{aligned}$
  
  leaving $w$ free. Solve: $z = 2 a + b - 4 c + 10 w$ , and $- 4 y = - 2 a + b - (2 a + b - 4 c + 10 w) - 2 w$ so $y = a - c + 3 w$ , and $x = a - 2 (a - c + 3 w) + w = - a + 2 c - 5 w .$ Therefore the solution set is this.
  
  ${(\begin{matrix} - a + 2 c \\ a - c \\ 2 a + b - 4 c \\ 0 \end{matrix}) + (\begin{matrix} - 5 \\ 3 \\ 10 \\ 1 \end{matrix}) w ∣ w \in ℝ}$
2. Plug in with $a = 3$ , $b = 1$ , and $c = - 2$ .
  
  ${(\begin{matrix} - 7 \\ 5 \\ 15 \\ 0 \end{matrix}) + (\begin{matrix} - 5 \\ 3 \\ 10 \\ 1 \end{matrix}) w ∣ w \in ℝ}$
Exercise 2.26 Worked answer

Why is the comma needed in the notation ‘ $a_{i, j}$ ’ for matrix entries?

Back to Exercise 2.26

Answer. Leaving the comma out, say by writing $a_{123}$ , is ambiguous because it could mean $a_{1, 23}$ or $a_{12, 3}$ .
Exercise 2.27 Worked answer

Recommended. Give the $4 \times 4$ matrix whose $i, j$ -th entry is
1. $i + j$ ;
2. $- 1$ to the $i + j$ power.
Back to Exercise 2.27

Answer.
1. $(\begin{matrix} 2 & 3 & 4 & 5 \\ 3 & 4 & 5 & 6 \\ 4 & 5 & 6 & 7 \\ 5 & 6 & 7 & 8 \end{matrix})$
2. $(\begin{matrix} 1 & - 1 & 1 & - 1 \\ - 1 & 1 & - 1 & 1 \\ 1 & - 1 & 1 & - 1 \\ - 1 & 1 & - 1 & 1 \end{matrix})$
Exercise 2.28 Worked answer

For any matrix $A$ , the transpose of $A$ , written $A^{𝖳}$ , is the matrix whose columns are the rows of $A$ . Find the transpose of each of these.
1. $(\begin{matrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{matrix})$
2. $(\begin{matrix} 2 & - 3 \\ 1 & 1 \end{matrix})$
3. $(\begin{matrix} 5 & 10 \\ 10 & 5 \end{matrix})$
4. $(\begin{matrix} 1 \\ 1 \\ 0 \end{matrix})$
Back to Exercise 2.28

Answer.
1. $(\begin{matrix} 1 & 4 \\ 2 & 5 \\ 3 & 6 \end{matrix})$
2. $(\begin{matrix} 2 & 1 \\ - 3 & 1 \end{matrix})$
3. $(\begin{matrix} 5 & 10 \\ 10 & 5 \end{matrix})$
4. $(\begin{matrix} 1 & 1 & 0 \end{matrix})$
Exercise 2.29 Worked answer

Recommended.
1. Describe all functions $f (x) = a x^{2} + b x + c$ such that $f (1) = 2$ and $f (- 1) = 6$ .
2. Describe all functions $f (x) = a x^{2} + b x + c$ such that $f (1) = 2$ .
Back to Exercise 2.29

Answer.
1. Plugging in $x = 1$ and $x = - 1$ gives
  
  $\begin{array}{rcrcrcr} a & + & b & + & c & = & 2 \\ a & - & b & + & c & = & 6 \end{array} ⟶_{\begin{array}{l} \end{array}}^{- ρ_{1} + ρ_{2}} \begin{array}{rcrcrcr} a & + & b & + & c & = & 2 \\ - 2 b & = & 4 \end{array}$
  
  so the set of functions is ${f (x) = (4 - c) x^{2} - 2 x + c ∣ c \in ℝ}$ .
2. Putting in $x = 1$ gives
  
  $\begin{array}{rcrcrcr} a & + & b & + & c & = & 2 \end{array}$
  
  so the set of functions is ${f (x) = (2 - b - c) x^{2} + b x + c ∣ b, c \in ℝ}$ .
Exercise 2.30 Worked answer

Show that any set of five points from the plane $ℝ^{2}$ lie on a common conic section, that is, they all satisfy some equation of the form $a x^{2} + b y^{2} + c x y + d x + e y + f = 0$ where some of $a, \dots, f$ are nonzero.

Back to Exercise 2.30

Answer. On plugging in the five pairs $(x, y)$ we get a system with the five equations and six unknowns $a$ , …, $f$ . Because there are more unknowns than equations, if no inconsistency exists among the equations then there are infinitely many solutions (at least one variable will end up free).

But no inconsistency can exist because $a = 0$ , …, $f = 0$ is a solution (we are only using this zero solution to show that the system is consistent—the prior paragraph shows that there are nonzero solutions).
Exercise 2.31 Worked answer

Make up a four equations/four unknowns system having
1. a one-parameter solution set;
2. a two-parameter solution set;
3. a three-parameter solution set.
Back to Exercise 2.31

Answer.
1. Here is one—the fourth equation is redundant but still OK.
  
  $\begin{array}{rcrcrcrcr} x & + & y & - & z & + & w & = & 0 \\ y & - & z & = & 0 \\ 2 z & + & 2 w & = & 0 \\ z & + & w & = & 0 \end{array}$
2. Here is one.
  
  $\begin{array}{rcrcrcrcr} x & + & y & - & z & + & w & = & 0 \\ w & = & 0 \\ w & = & 0 \\ w & = & 0 \end{array}$
3. This is one.
  
  $\begin{array}{rcrcrcrcr} x & + & y & - & z & + & w & = & 0 \\ x & + & y & - & z & + & w & = & 0 \\ x & + & y & - & z & + & w & = & 0 \\ x & + & y & - & z & + & w & = & 0 \end{array}$
Exercise 2.32 Worked answer

Puzzle. [Shepelev] This puzzle is from a Russian web-site http://www.arbuz.uz/ and there are many solutions to it, but mine uses linear algebra and is very naive. There’s a planet inhabited by arbuzoids (watermeloners, to translate from Russian). Those creatures are found in three colors: red, green and blue. There are $13$ red arbuzoids, $15$ blue ones, and $17$ green. When two differently colored arbuzoids meet, they both change to the third color.

The question is, can it ever happen that all of them assume the same color?

Back to Exercise 2.32

Answer. This is how the answer was given in the cited source. My solution was to define the numbers of arbuzoids as $3$ -dimensional vectors, and express all possible elementary transitions as such vectors, too:

R: $13$

G: $15$

B: $17$

Operations: $(\begin{matrix} - 1 \\ - 1 \\ 2 \end{matrix})$ , $(\begin{matrix} - 1 \\ 2 \\ - 1 \end{matrix})$ , and $(\begin{matrix} 2 \\ - 1 \\ - 1 \end{matrix})$

Now, it is enough to check whether the solution to one of the following systems of linear equations exists:

$(\begin{matrix} 13 \\ 15 \\ 17 \end{matrix}) + x (\begin{matrix} - 1 \\ - 1 \\ 2 \end{matrix}) + y (\begin{matrix} - 1 \\ 2 \\ - 1 \end{matrix}) + z (\begin{matrix} 2 \\ - 1 \\ - 1 \end{matrix}) = (\begin{matrix} 0 \\ 0 \\ 45 \end{matrix}) (or (\begin{matrix} 0 \\ 45 \\ 0 \end{matrix}) or (\begin{matrix} 45 \\ 0 \\ 0 \end{matrix}))$

Solving

$(\begin{array}{cccr} - 1 & - 1 & 2 & - 13 \\ - 1 & 2 & - 1 & - 15 \\ 2 & - 1 & - 1 & 28 \end{array}) ⟶_{\begin{array}{l} 2 ρ_{1} + ρ_{3} \end{array}}^{- ρ_{1} + ρ_{2}} ⟶_{\begin{array}{l} \end{array}}^{ρ_{2} + ρ_{3}} (\begin{array}{cccr} - 1 & - 1 & 2 & - 13 \\ 0 & 3 & - 3 & - 2 \\ 0 & 0 & 0 & 0 \end{array})$

gives $y + 2 / 3 = z$ so if the number of transformations $z$ is an integer then $y$ is not. The other two systems give similar conclusions so there is no solution.
Exercise 2.33 Worked answer

Puzzle. [USSR Olympiad no. 174]
1. Solve the system of equations.
  
  $\begin{array}{rcrcr} a x & + & y & = & a^{2} \\ x & + & a y & = & 1 \end{array}$
  
  For what values of $a$ does the system fail to have solutions, and for what values of $a$ are there infinitely many solutions?
2. Answer the above question for the system.
  
  $\begin{array}{rcrcr} a x & + & y & = & a^{3} \\ x & + & a y & = & 1 \end{array}$
Back to Exercise 2.33

Answer. This is how the answer was given in the cited source.
1. Formal solution of the system yields
  
  $x = \frac{a^{3} - 1}{a^{2} - 1} y = \frac{- a^{2} + a}{a^{2} - 1} .$
  
  If $a + 1 \neq 0$ and $a - 1 \neq 0$ , then the system has the single solution
  
  $x = \frac{a^{2} + a + 1}{a + 1} y = \frac{- a}{a + 1} .$
  
  If $a = - 1$ , or if $a = + 1$ , then the formulas are meaningless; in the first instance we arrive at the system
  
  ${\begin{array}{rcrcr} - x & + & y & = & 1 \\ x & - & y & = & 1 \end{array}$
  
  which is a contradictory system. In the second instance we have
  
  ${\begin{array}{rcrcr} x & + & y & = & 1 \\ x & + & y & = & 1 \end{array}$
  
  which has an infinite number of solutions (for example, for $x$ arbitrary, $y = 1 - x$ ).
2. Solution of the system yields
  
  $x = \frac{a^{4} - 1}{a^{2} - 1} y = \frac{- a^{3} + a}{a^{2} - 1} .$
  
  Here, is $a^{2} - 1 \neq 0$ , the system has the single solution $x = a^{2} + 1$ , $y = - a$ . For $a = - 1$ and $a = 1$ , we obtain the systems
  
  ${\begin{array}{rcrcr} - x & + & y & = & - 1 \\ x & - & y & = & 1 \end{array} {\begin{array}{rcrcr} x & + & y & = & 1 \\ x & + & y & = & 1 \end{array}$
  
  both of which have an infinite number of solutions.
Exercise 2.34 Worked answer

Puzzle. [Math. Mag., Sept. 1952] In air a gold-surfaced sphere weighs $7588$ grams. It is known that it may contain one or more of the metals aluminum, copper, silver, or lead. When weighed successively under standard conditions in water, benzene, alcohol, and glycerin its respective weights are $6588$ , $6688$ , $6778$ , and $6328$ grams. How much, if any, of the forenamed metals does it contain if the specific gravities of the designated substances are taken to be as follows?

Aluminum $2.7$ Alcohol 0.81

Copper $8.9$ Benzene $0.90$

Gold $19.3$ Glycerin $1.26$

Lead $11.3$ Water $1.00$

Silver $10.8$

Back to Exercise 2.34

Answer. This is how the answer was given in the cited source. Let $u$ , $v$ , $x$ , $y$ , $z$ be the volumes in ${c m}^{3}$ of Al, Cu, Pb, Ag, and Au, respectively, contained in the sphere, which we assume to be not hollow. Since the loss of weight in water (specific gravity $1.00$ ) is $1000$ grams, the volume of the sphere is $1000 {cm}^{3}$ . Then the data, some of which is superfluous, though consistent, leads to only $2$ independent equations, one relating volumes and the other, weights.

$\begin{array}{rcrcrcrcrcr} u & + & v & + & x & + & y & + & z & = & 1000 \\ 2.7 u & + & 8.9 v & + & 11.3 x & + & 10.5 y & + & 19.3 z & = & 7558 \end{array}$

Clearly the sphere must contain some aluminum to bring its mean specific gravity below the specific gravities of all the other metals. There is no unique result to this part of the problem, for the amounts of three metals may be chosen arbitrarily, provided that the choices will not result in negative amounts of any metal.

If the ball contains only aluminum and gold, there are $294.5 {cm}^{3}$ of gold and $705.5 {cm}^{3}$ of aluminum. Another possibility is $124.7 {cm}^{3}$ each of Cu, Au, Pb, and Ag and $501.2 {cm}^{3}$ of Al.

$General = Particular + Homogeneous$

In the prior subsection the descriptions of solution sets all fit a pattern. They have a vector that is a particular solution of the system added to an unrestricted combination of some other vectors. The solution set from Example 2.13 illustrates.

${\underset{\begin{matrix} particular \\ solution \end{matrix}}{\underset{}{(\begin{matrix} 0 \\ 4 \\ 0 \\ 0 \\ 0 \end{matrix})}} + \underset{\begin{matrix} unrestricted \\ combination \end{matrix}}{\underset{}{w (\begin{matrix} 1 \\ - 1 \\ 3 \\ 1 \\ 0 \end{matrix}) + u (\begin{matrix} 1 / 2 \\ - 1 \\ 1 / 2 \\ 0 \\ 1 \end{matrix})}} ∣ w, u \in ℝ}$

The combination is unrestricted in that $w$ and $u$ can be any real numbers—there is no condition like “such that $2 w - u = 0$ ” to restrict which pairs $w, u$ we can use.

That example shows an infinite solution set fitting the pattern. The other two kinds of solution sets also fit. A one-element solution set fits because it has a particular solution and the unrestricted combination part is trivial. That is, instead of being a combination of two vectors or of one vector, it is a combination of no vectors. (By convention the sum of an empty set of vectors is the zero vector.) An empty solution set fits the pattern because there is no particular solution and thus there are no sums of that form.

Theorem 3.1 Any linear system’s solution set has the form

${\vec{p} + c_{1} {\vec{β}}_{1} + \dots + c_{k} {\vec{β}}_{k} ∣ c_{1}, \dots, c_{k} \in ℝ}$

where $\vec{p}$ is any particular solution and where the number of vectors ${\vec{β}}_{1}$ , …, ${\vec{β}}_{k}$ equals the number of free variables that the system has after a Gaussian reduction.

The solution description has two parts, the particular solution $\vec{p}$ and the unrestricted linear combination of the $\vec{β}$ ’s. We shall prove the theorem with two corresponding lemmas.

We will focus first on the unrestricted combination. For that we consider systems that have the vector of zeroes as a particular solution so that we can shorten $\vec{p} + c_{1} {\vec{β}}_{1} + \dots + c_{k} {\vec{β}}_{k}$ to $c_{1} {\vec{β}}_{1} + \dots + c_{k} {\vec{β}}_{k}$ .

Definition 3.2 A linear equation is homogeneous if it has a constant of zero, so that it can be written as $a_{1} x_{1} + a_{2} x_{2} + \dots + a_{n} x_{n} = 0$ .

Example 3.3 With any linear system like

$\begin{array}{rcrcr} 3 x & + & 4 y & = 3 \\ 2 x & - & y & = 1 \end{array}$

we associate a system of homogeneous equations by setting the right side to zeros.

$\begin{array}{rcrcr} 3 x & + & 4 y & = 0 \\ 2 x & - & y & = 0 \end{array}$

Compare the reduction of the original system

$\begin{array}{rcrcr} 3 x & + & 4 y & = 3 \\ 2 x & - & y & = 1 \end{array} ⟶_{\begin{array}{l} \end{array}}^{- (2 / 3) ρ_{1} + ρ_{2}} \begin{array}{rcrcr} 3 x & + & 4 y & = 3 \\ - (11 / 3) y & = - 1 \end{array}$

with the reduction of the associated homogeneous system.

$\begin{array}{rcrcr} 3 x & + & 4 y & = 0 \\ 2 x & - & y & = 0 \end{array} ⟶_{\begin{array}{l} \end{array}}^{- (2 / 3) ρ_{1} + ρ_{2}} \begin{array}{rcrcr} 3 x & + & 4 y & = 0 \\ - (11 / 3) y & = 0 \end{array}$

Obviously the two reductions go in the same way. We can study how to reduce a linear system by instead studying how to reduce the associated homogeneous system.

Studying the associated homogeneous system has a great advantage over studying the original system. Nonhomogeneous systems can be inconsistent. But a homogeneous system must be consistent since there is always at least one solution, the zero vector.

Example 3.4 Some homogeneous systems have the zero vector as their only solution.

$\begin{array}{rcrcrcr} 3 x & + & 2 y & + & z & = & 0 \\ 6 x & + & 4 y & = & 0 \\ y & + & z & = & 0 \end{array} ⟶_{\begin{array}{l} \end{array}}^{- 2 ρ_{1} + ρ_{2}} \begin{array}{rcrcrcr} 3 x & + & 2 y & + & z & = & 0 \\ - 2 z & = & 0 \\ y & + & z & = & 0 \end{array} ⟶_{\begin{array}{l} \end{array}}^{ρ_{2} \leftrightarrow ρ_{3}} \begin{array}{rcrcrcr} 3 x & + & 2 y & + & z & = & 0 \\ y & + & z & = & 0 \\ - 2 z & = & 0 \end{array}$

Example 3.5 Some homogeneous systems have many solutions. One is the Chemistry problem from the first page of the first subsection.

$\begin{aligned} \begin{array}{rcrcrcrcr} 7 x & - & 7 z & = & 0 \\ 8 x & + & y & - & 5 z & - & 2 w & = & 0 \\ y & - & 3 z & = & 0 \\ 3 y & - & 6 z & - & w & = & 0 \end{array} & ⟶_{\begin{matrix} \end{matrix}}^{- (8 / 7) ρ_{1} + ρ_{2}} \begin{array}{rcrcrcrcr} 7 x & - & 7 z & = & 0 \\ y & + & 3 z & - & 2 w & = & 0 \\ y & - & 3 z & = & 0 \\ 3 y & - & 6 z & - & w & = & 0 \end{array} \\ ⟶_{\begin{matrix} - 3 ρ_{2} + ρ_{4} \end{matrix}}^{- ρ_{2} + ρ_{3}} \begin{array}{rcrcrcrcr} 7 x & - & 7 z & = & 0 \\ y & + & 3 z & - & 2 w & = & 0 \\ - 6 z & + & 2 w & = & 0 \\ - 15 z & + & 5 w & = & 0 \end{array} \\ ⟶_{\begin{matrix} \end{matrix}}^{- (5 / 2) ρ_{3} + ρ_{4}} \begin{array}{rcrcrcrcr} 7 x & - & 7 z & = & 0 \\ y & + & 3 z & - & 2 w & = & 0 \\ - 6 z & + & 2 w & = & 0 \\ 0 & = & 0 \end{array} \end{aligned}$

The solution set

${(\begin{matrix} 1 / 3 \\ 1 \\ 1 / 3 \\ 1 \end{matrix}) w ∣ w \in ℝ}$

has many vectors besides the zero vector (if we take $w$ to be a number of molecules then solutions make sense only when $w$ is a nonnegative multiple of $3$ ).

Lemma 3.6 For any homogeneous linear system there exist vectors ${\vec{β}}_{1}$ , …, ${\vec{β}}_{k}$ such that the solution set of the system is

${c_{1} {\vec{β}}_{1} + \dots + c_{k} {\vec{β}}_{k} ∣ c_{1}, \dots, c_{k} \in ℝ}$

where $k$ is the number of free variables in an echelon form version of the system.

Before the proof, we discuss a couple of points about it. First, the proof verifies that, given a homogeneous system in echelon form, we can use back substitution to express all of the leading variables in terms of the free variables. To see why this suffices to establish the lemma, we give an example (we will also use this walk-through as a chance to illustrate the proof’s notation).

Consider this system of homogeneous equations in echelon form.

$\begin{array}{rcrcrcrcrcr} x & + & y & + & 2 z & + & u & + & v & = & 0 \\ y & + & z & + & u & - & v & = & 0 \\ u & + & v & = & 0 \end{array}$

It has five variables: $x_{1} = x$ , …, $x_{5} = v$ . To do back substitution, we start with the bottom row, equation $m = 3$ . We solve for its leading variable, $x_{ℓ_{3}} = x_{4} = u$ , using its leading term $a_{m, ℓ_{m}} x_{m} = a_{3, 4} x_{4} = 1 \cdot u$ (the notation’s ‘ $ℓ$ ’ stands for “leading”). We get $u = - v$ .

Now back substitution enters its iteration phase. We will use the variable $t$ to count how many rows up from the bottom the current row is, so we take $t = 1$ and look at row $m - t = 3 - 1 = 2$ . We come to this row having so far expressed the leading variable in terms of free variables for row $3$ . We substitute for $x_{ℓ_{3}} = x_{4} = u$ its expression in terms of free variables, getting $y + z + (- v) - v = 0$ . Then solving for this row’s leading variable gives $x_{ℓ_{m - t}} = x_{2} = y = - z + 2 v$ .

Next is the row that is $t = 2$ up from the bottom, row $m - t = 1$ . We have so far expressed the leading variable in terms of free variables for rows $3$ and $2$ . Substitute for $x_{ℓ_{3}} = u$ and $x_{ℓ_{2}} = y$ , giving $x + (- z + 2 v) + 2 z + (- v) + v = 0$ . Then solve for the leading variable in terms of the free variables as $x_{ℓ_{1}} = x = - z - 2 v$ .

The point of this example is that now we are done. Just write the solution in vector notation

$(\begin{matrix} x \\ y \\ z \\ u \\ v \end{matrix}) = (\begin{matrix} - 1 \\ - 1 \\ 1 \\ 0 \\ 0 \end{matrix}) z + (\begin{matrix} - 2 \\ 2 \\ 0 \\ - 1 \\ 1 \end{matrix}) v where z, v \in ℝ$

to recognize that ${\vec{β}}_{1}$ and ${\vec{β}}_{2}$ of the lemma’s statement are the vectors associated with the free variables $z$ and $v$ .

The second point about the proof follows from that example also. Back substitution moves up the system, using the conclusions from lower rows to do the next row. This suggests the proof strategy of mathematical induction.

Induction is an important and non-obvious proof technique that will appear a number of times in this book. Our proofs by induction will come in two steps, a base step and an inductive step. In the base step, we verify that the statement is true for some first instance, here that for an echelon form linear system we can write the bottom equation’s leading variable in terms of free variables. In the inductive step, we must establish an implication, namely that if the statement is true for all prior cases then it follows that the statement holds for the present case also. Specifically, here the inductive hypothesis is that if for the lower-down rows we can express the leading variables in terms of the free variables, then for the next row up we can express the leading variable in those terms also.

Those two steps together prove the statement for all the rows because by the base step it is true for the bottom equation, and by the inductive step the fact that it is true for the bottom equation shows that it is true for the next one up. Then, another application of the inductive step implies that it is true for the third equation up, etc.¹

Proof Apply Gauss’s Method to get to echelon form. There may be some $0 = 0$ equations; we ignore these (if the system consists only of $0 = 0$ equations then the lemma is trivially true because there are no leading variables). But because the system is homogeneous there are no contradictory equations.

We will use induction to verify that each leading variable can be expressed in terms of free variables. That will finish the proof because we can use the free variables as parameters and the $\vec{β}$ ’s are the vectors of coefficients of those free variables.

For the base step, consider the bottom-most equation,

$a_{m, ℓ_{m}} x_{ℓ_{m}} + a_{m, 1 + ℓ_{m}} x_{1 + ℓ_{m}} + \dots + a_{m, n} x_{n} = 0$

where $a_{m, ℓ_{m}} \neq 0$ . This is the bottom row, so any variables after the leading one must be free. Move these to the right hand side and divide by $a_{m, ℓ_{m}}$

$x_{ℓ_{m}} = (- a_{m, 1 + ℓ_{m}} / a_{m, ℓ_{m}}) x_{1 + ℓ_{m}} + \dots + (- a_{m, n} / a_{m, ℓ_{m}}) x_{n}$

to express the leading variable in terms of free variables. (A technical point: if in the bottom equation there are no variables to the right of $x_{ℓ_{m}}$ then $x_{ℓ_{m}} = 0$ . This satisfies the statement that we are verifying because, as alluded to at the start of this subsection, it has $x_{ℓ_{m}}$ written as a sum of a number of the free variables, namely as the sum of zero many, under the convention that a trivial sum totals to $0$ .)

For the inductive step, assume that assume that for the $m$ -th equation, and the $(m - 1)$ -th equation, etc., up to and including the $m - (t - 1)$ -th equation, we can express the leading variable of that equation in terms of free variables. That is, assume that the statement that we can express the leading variable of the row in terms of the free variables holds for the bottom-most $t$ rows, with $1 \leq t < m$ . We must verify that the statement therefore also holds for the next equation up, the $(m - t)$ -th equation.

For each of $x_{ℓ_{m}}$ , …, $x_{ℓ_{m - (t - 1)}}$ , substitute its expression in terms of free variables. The result has a leading term of $a_{m - t, ℓ_{m - t}} x_{ℓ_{m - t}}$ with $a_{m - t, ℓ_{m - t}} \neq 0$ , and the rest of the left hand side is a combination of free variables. Move the free variables to the right side and divide by $a_{m - t, ℓ_{m - t}}$ , to end with $x_{ℓ_{m - t}}$ expressed in terms of free variables.

We have done both the base step and the inductive step, so by the principle of mathematical induction the proposition is true.

QED

This shows, as discussed between the lemma and its proof, that we can parametrize solution sets using the free variables. We say that the set of vectors ${c_{1} {\vec{β}}_{1} + \dots + c_{k} {\vec{β}}_{k} ∣ c_{1}, \dots, c_{k} \in ℝ}$ is generated by or spanned by the set ${{\vec{β}}_{1}, \dots, {\vec{β}}_{k}}$ .

To finish the proof of Theorem 3.1 the next lemma considers the particular solution part of the solution set’s description.

Lemma 3.7 For a linear system and for any particular solution $\vec{p}$ , the solution set equals ${\vec{p} + \vec{h} ∣ \vec{h} satisfies the associated homogeneous system}$ .

Proof We will show mutual set inclusion, that any solution to the system is in the above set and that anything in the set is a solution of the system.²

For set inclusion the first way, that if a vector solves the system then it is in the set described above, assume that $\vec{s}$ solves the system. Then $\vec{s} - \vec{p}$ solves the associated homogeneous system since for each equation index $i$ ,

$\begin{matrix} a_{i, 1} (s_{1} - p_{1}) + \dots + a_{i, n} (s_{n} - p_{n}) \\ = (a_{i, 1} s_{1} + \dots + a_{i, n} s_{n}) - (a_{i, 1} p_{1} + \dots + a_{i, n} p_{n}) = d_{i} - d_{i} = 0 \end{matrix}$

where $p_{j}$ and $s_{j}$ are the $j$ -th components of $\vec{p}$ and $\vec{s}$ . Express $\vec{s}$ in the required $\vec{p} + \vec{h}$ form by writing $\vec{s} - \vec{p}$ as $\vec{h}$ .

For set inclusion the other way, take a vector of the form $\vec{p} + \vec{h}$ , where $\vec{p}$ solves the system and $\vec{h}$ solves the associated homogeneous system and note that $\vec{p} + \vec{h}$ solves the given system since for any equation index $i$ ,

$\begin{matrix} a_{i, 1} (p_{1} + h_{1}) + \dots + a_{i, n} (p_{n} + h_{n}) \\ = (a_{i, 1} p_{1} + \dots + a_{i, n} p_{n}) + (a_{i, 1} h_{1} + \dots + a_{i, n} h_{n}) = d_{i} + 0 = d_{i} \end{matrix}$

where as earlier $p_{j}$ and $h_{j}$ are the $j$ -th components of $\vec{p}$ and $\vec{h}$ .

QED

The two lemmas together establish Theorem 3.1. Remember that theorem with the slogan, “ $General = Particular + Homogeneous$ ”.

Example 3.8 This system illustrates Theorem 3.1.

$\begin{array}{rcrcrcr} x & + & 2 y & - & z & = & 1 \\ 2 x & + & 4 y & = & 2 \\ y & - & 3 z & = & 0 \end{array}$

Gauss’s Method

$⟶_{\begin{array}{l} \end{array}}^{- 2 ρ_{1} + ρ_{2}} \begin{array}{rcrcrcr} x & + & 2 y & - & z & = & 1 \\ 2 z & = & 0 \\ y & - & 3 z & = & 0 \end{array} ⟶_{\begin{array}{l} \end{array}}^{ρ_{2} \leftrightarrow ρ_{3}} \begin{array}{rcrcrcr} x & + & 2 y & - & z & = & 1 \\ y & - & 3 z & = & 0 \\ 2 z & = & 0 \end{array}$

shows that the general solution is a singleton set.

${(\begin{matrix} 1 \\ 0 \\ 0 \end{matrix})}$

That single vector is obviously a particular solution. The associated homogeneous system reduces via the same row operations

$\begin{array}{rcrcrcr} x & + & 2 y & - & z & = & 0 \\ 2 x & + & 4 y & = & 0 \\ y & - & 3 z & = & 0 \end{array} ⟶_{\begin{array}{l} \end{array}}^{- 2 ρ_{1} + ρ_{2}} ⟶_{\begin{array}{l} \end{array}}^{ρ_{2} \leftrightarrow ρ_{3}} \begin{array}{rcrcrcr} x & + & 2 y & - & z & = & 0 \\ y & - & 3 z & = & 0 \\ 2 z & = & 0 \end{array}$

to also give a singleton set.

${(\begin{matrix} 0 \\ 0 \\ 0 \end{matrix})}$

So, as discussed at the start of this subsection, in this single-solution case the general solution results from taking the particular solution and adding to it the unique solution of the associated homogeneous system.

Example 3.9 The start of this subsection also discusses that the case where the general solution set is empty fits the $General = Particular + Homogeneous$ pattern too. This system illustrates.

$\begin{array}{rcrcrcrcr} x & + & z & + & w & = & - 1 \\ 2 x & - & y & + & w & = & 3 \\ x & + & y & + & 3 z & + & 2 w & = & 1 \end{array} ⟶_{\begin{array}{l} - ρ_{1} + ρ_{3} \end{array}}^{- 2 ρ_{1} + ρ_{2}} \begin{array}{rcrcrcrcr} x & + & z & + & w & = & - 1 \\ - y & - & 2 z & - & w & = & 5 \\ y & + & 2 z & + & w & = & 2 \end{array}$

It has no solutions because the final two equations conflict. But the associated homogeneous system does have a solution, as do all homogeneous systems.

$\begin{array}{rcrcrcrcr} x & + & z & + & w & = & 0 \\ 2 x & - & y & + & w & = & 0 \\ x & + & y & + & 3 z & + & 2 w & = & 0 \end{array} ⟶_{\begin{array}{l} - ρ_{1} + ρ_{3} \end{array}}^{- 2 ρ_{1} + ρ_{2}} ⟶_{\begin{array}{l} \end{array}}^{ρ_{2} + ρ_{3}} \begin{array}{rcrcrcrcr} x & + & z & + & w & = & 0 \\ - y & - & 2 z & - & w & = & 0 \\ 0 & = & 0 \end{array}$

In fact, the solution set is infinite.

${(\begin{matrix} - 1 \\ - 2 \\ 1 \\ 0 \end{matrix}) z + (\begin{matrix} - 1 \\ - 1 \\ 0 \\ 1 \end{matrix}) w ∣ z, w \in ℝ}$

Nonetheless, because the original system has no particular solution, its general solution set is empty—there are no vectors of the form $\vec{p} + \vec{h}$ because there are no $\vec{p}$ ’s.

Corollary 3.10 Solution sets of linear systems are either empty, have one element, or have infinitely many elements.

Proof We’ve seen examples of all three happening so we need only prove that there are no other possibilities.

First observe a homogeneous system with at least one non- $\vec{0}$ solution $\vec{v}$ has infinitely many solutions. This is because any scalar multiple of $\vec{v}$ also solves the homogeneous system and there are infinitely many vectors in the set of scalar multiples of $\vec{v}$ : if $s, t \in ℝ$ are unequal then $s \vec{v} \neq t \vec{v}$ , since $s \vec{v} - t \vec{v} = (s - t) \vec{v}$ is non- $\vec{0}$ as any non- $0$ component of $\vec{v}$ , when rescaled by the non- $0$ factor $s - t$ , will give a non- $0$ value.

Now apply Lemma 3.7 to conclude that a solution set

${\vec{p} + \vec{h} ∣ \vec{h} solves the associated homogeneous system}$

is either empty (if there is no particular solution $\vec{p}$ ), or has one element (if there is a $\vec{p}$ and the homogeneous system has the unique solution $\vec{0}$ ), or is infinite (if there is a $\vec{p}$ and the homogeneous system has a non- $\vec{0}$ solution, and thus by the prior paragraph has infinitely many solutions).

QED

This table summarizes the factors affecting the size of a general solution.

number of solutions of the

homogeneous system

particular

solution

exists?

one

infinitely many

yes

unique

solution

infinitely many

solutions

The dimension on the top of the table is the simpler one. When we perform Gauss’s Method on a linear system, ignoring the constants on the right side and so paying attention only to the coefficients on the left-hand side, we either end with every variable leading some row or else we find some variable that does not lead a row, that is, we find some variable that is free. (We formalize “ignoring the constants on the right” by considering the associated homogeneous system.)

A notable special case is systems having the same number of equations as unknowns. Such a system will have a solution, and that solution will be unique, if and only if it reduces to an echelon form system where every variable leads its row (since there are the same number of variables as rows), which will happen if and only if the associated homogeneous system has a unique solution.

Definition 3.11 A square matrix is nonsingular if it is the matrix of coefficients of a homogeneous system with a unique solution. It is singular otherwise, that is, if it is the matrix of coefficients of a homogeneous system with infinitely many solutions.

Example 3.12 The first of these matrices is nonsingular while the second is singular

$(\begin{matrix} 1 & 2 \\ 3 & 4 \end{matrix}) (\begin{matrix} 1 & 2 \\ 3 & 6 \end{matrix})$

because the first of these homogeneous systems has a unique solution while the second has infinitely many solutions.

$\begin{array}{rcrcr} x & + & 2 y & = & 0 \\ 3 x & + & 4 y & = & 0 \end{array} \begin{array}{rcrcr} x & + & 2 y & = & 0 \\ 3 x & + & 6 y & = & 0 \end{array}$

We have made the distinction in the definition because a system with the same number of equations as variables behaves in one of two ways, depending on whether its matrix of coefficients is nonsingular or singular. Where the matrix of coefficients is nonsingular the system has a unique solution for any constants on the right side: for instance, Gauss’s Method shows that this system

$\begin{array}{rcrcr} x & + & 2 y & = & a \\ 3 x & + & 4 y & = & b \end{array}$

has the unique solution $x = b - 2 a$ and $y = (3 a - b) / 2$ . On the other hand, where the matrix of coefficients is singular the system never has a unique solution—it has either no solutions or else has infinitely many, as with these.

$\begin{array}{rcrcr} x & + & 2 y & = & 1 \\ 3 x & + & 6 y & = & 2 \end{array} \begin{array}{rcrcr} x & + & 2 y & = & 1 \\ 3 x & + & 6 y & = & 3 \end{array}$

The definition uses the word ‘singular’ because it means “departing from general expectation.” People often, naively, expect that systems with the same number of variables as equations will have a unique solution. Thus, we can think of the word as connoting “troublesome,” or at least “not ideal.” (That ‘singular’ applies to those systems that never have exactly one solution is ironic but it is the standard term.)

Example 3.13 The systems from Example 3.3, Example 3.4, and Example 3.8 each have an associated homogeneous system with a unique solution. Thus these matrices are nonsingular.

$(\begin{matrix} 3 & 4 \\ 2 & - 1 \end{matrix}) (\begin{matrix} 3 & 2 & 1 \\ 6 & - 4 & 0 \\ 0 & 1 & 1 \end{matrix}) (\begin{matrix} 1 & 2 & - 1 \\ 2 & 4 & 0 \\ 0 & 1 & - 3 \end{matrix})$

The Chemistry problem from Example 3.5 is a homogeneous system with more than one solution so its matrix is singular.

$(\begin{matrix} 7 & 0 & - 7 & 0 \\ 8 & 1 & - 5 & - 2 \\ 0 & 1 & - 3 & 0 \\ 0 & 3 & - 6 & - 1 \end{matrix})$

The table above has two dimensions. We have considered the one on top: we can tell into which column a given linear system goes solely by considering the system’s left-hand side; the constants on the right-hand side play no role in this.

The table’s other dimension, determining whether a particular solution exists, is tougher. Consider these two systems with the same left side but different right sides.

$\begin{array}{rcrcr} 3 x & + & 2 y & = & 5 \\ 3 x & + & 2 y & = & 5 \end{array} \begin{array}{rcrcr} 3 x & + & 2 y & = & 5 \\ 3 x & + & 2 y & = & 4 \end{array}$

The first has a solution while the second does not, so here the constants on the right side decide if the system has a solution. We could conjecture that the left side of a linear system determines the number of solutions while the right side determines if solutions exist but that guess is not correct. Compare these two, with the same right sides but different left sides.

$\begin{array}{rcrcr} 3 x & + & 2 y & = & 5 \\ 4 x & + & 2 y & = & 4 \end{array} \begin{array}{rcrcr} 3 x & + & 2 y & = & 5 \\ 3 x & + & 2 y & = & 4 \end{array}$

The first has a solution but the second does not. Thus the constants on the right side of a system don’t alone determine whether a solution exists. Rather, that depends on some interaction between the left and right.

For some intuition about that interaction, consider this system with one of the coefficients left unspecified, as the variable $c$ .

$\begin{array}{rcrcrcr} x & + & 2 y & + & 3 z & = & 1 \\ x & + & y & + & z & = & 1 \\ c x & + & 3 y & + & 4 z & = & 0 \end{array}$

If $c = 2$ then this system has no solution because the left-hand side has the third row as the sum of the first two, while the right-hand does not. If $c \neq 2$ then this system has a unique solution (try it with $c = 1$ ). For a system to have a solution, if one row of the matrix of coefficients on the left is a linear combination of other rows then on the right the constant from that row must be the same combination of constants from the same rows.

More intuition about the interaction comes from studying linear combinations. That will be our focus in the second chapter, after we finish the study of Gauss’s Method itself in the rest of this chapter.

Exercises

Exercise 3.14 Worked answer

Solve this system. Then solve the associated homogeneous system.

$\begin{array}{rcrcrcrcr} x & + & y & - & 2 z & = & 0 \\ x & - & y & = & - 3 \\ 3 x & - & y & - & 2 z & = & - 6 \\ 2 y & - & 2 z & = & 3 \end{array}$

Back to Exercise 3.14

Answer. This reduction solves the system.

$\begin{aligned} (\begin{array}{cccc} 1 & 1 & - 2 & 0 \\ 1 & - 1 & 0 & 3 \\ 3 & - 1 & - 2 & - 6 \\ 0 & 2 & - 2 & 3 \end{array}) & ⟶_{\begin{matrix} - 3 ρ_{1} + ρ_{3} \end{matrix}}^{- ρ_{1} + ρ_{2}} (\begin{array}{cccc} 1 & 1 & - 2 & 0 \\ 0 & - 2 & 2 & - 3 \\ 0 & - 4 & 4 & - 6 \\ 0 & 2 & - 2 & 3 \end{array}) \\ ⟶_{\begin{matrix} ρ_{2} + ρ_{4} \end{matrix}}^{- 2 ρ_{2} + ρ_{3}} (\begin{array}{cccc} 1 & 1 & - 2 & 0 \\ 0 & - 2 & 2 & - 3 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}) \end{aligned}$

The solution set is this.

${(\begin{matrix} - 3 / 2 \\ 3 / 2 \\ 0 \end{matrix}) + (\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}) z ∣ z \in ℝ}$

Similarly we can reduce the associated homogeneous system

$\begin{aligned} (\begin{array}{cccc} 1 & 1 & - 2 & 0 \\ 1 & - 1 & 0 & 0 \\ 3 & - 1 & - 2 & 0 \\ 0 & 2 & - 2 & 0 \end{array}) & ⟶_{\begin{matrix} - 3 ρ_{1} + ρ_{3} \end{matrix}}^{- ρ_{1} + ρ_{2}} (\begin{array}{cccc} 1 & 1 & - 2 & 0 \\ 0 & - 2 & 2 & 0 \\ 0 & - 4 & 4 & 0 \\ 0 & 2 & - 2 & 0 \end{array}) \\ ⟶_{\begin{matrix} ρ_{2} + ρ_{4} \end{matrix}}^{- 2 ρ_{2} + ρ_{3}} (\begin{array}{cccc} 1 & 1 & - 2 & 0 \\ 0 & - 2 & 2 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}) \end{aligned}$

to get its solution set.

${(\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}) z ∣ z \in ℝ}$
Exercise 3.15 Worked answer

Recommended. Solve each system. Express the solution set using vectors. Identify a particular solution and the solution set of the homogeneous system.
1. $\begin{array}{rcrcr} 3 x & + & 6 y & = & 18 \\ x & + & 2 y & = & 6 \end{array}$
2. $\begin{array}{rcrcr} x & + & y & = & 1 \\ x & - & y & = & - 1 \end{array}$
3. $\begin{array}{rcrcrcr} x_{1} & + & x_{3} & = & 4 \\ x_{1} & - & x_{2} & + & 2 x_{3} & = & 5 \\ 4 x_{1} & - & x_{2} & + & 5 x_{3} & = & 17 \end{array}$
4. $\begin{array}{rcrcrcr} 2 a & + & b & - & c & = & 2 \\ 2 a & + & c & = & 3 \\ a & - & b & = & 0 \end{array}$
5. $\begin{array}{rcrcrcrcr} x & + & 2 y & - & z & = & 3 \\ 2 x & + & y & + & w & = & 4 \\ x & - & y & + & z & + & w & = & 1 \end{array}$
6. $\begin{array}{rcrcrcrcr} x & + & z & + & w & = & 4 \\ 2 x & + & y & - & w & = & 2 \\ 3 x & + & y & + & z & = & 7 \end{array}$
Back to Exercise 3.15

Answer. For the arithmetic to these, see the answers from the prior subsection.
1. This is the solution set.
  
  $S = {(\begin{matrix} 6 \\ 0 \end{matrix}) + (\begin{matrix} - 2 \\ 1 \end{matrix}) y ∣ y \in ℝ}$
  
  Here are the particular solution and the solution set for the associated homogeneous system.
  
  $(\begin{matrix} 6 \\ 0 \end{matrix}) and {(\begin{matrix} - 2 \\ 1 \end{matrix}) y ∣ y \in ℝ}$
  
  Note. There are two possible points of confusion here. First, the set $S$ given above is equal to this set
  
  $T = {(\begin{matrix} 4 \\ 1 \end{matrix}) + (\begin{matrix} - 2 \\ 1 \end{matrix}) y ∣ y \in ℝ}$
  
  because the two sets contain the same members. All of these are correct answers to, “What is a particular solution?”
  
  $(\begin{matrix} 6 \\ 0 \end{matrix}), (\begin{matrix} 4 \\ 1 \end{matrix}), (\begin{matrix} 2 \\ 2 \end{matrix}), (\begin{matrix} 1 \\ 2.5 \end{matrix})$
  
  The second point of confusion is that the letter we use in the set doesn’t matter. This set also equals $S$ .
  
  $U = {(\begin{matrix} 6 \\ 0 \end{matrix}) + (\begin{matrix} - 2 \\ 1 \end{matrix}) u ∣ u \in ℝ}$
2. This is the solution set.
  
  ${(\begin{matrix} 0 \\ 1 \end{matrix})}$
  
  These are a particular solution, and the solution set for the associated homogeneous system.
  
  $(\begin{matrix} 0 \\ 1 \end{matrix}) {(\begin{matrix} 0 \\ 0 \end{matrix})}$
3. The solution set is infinite.
  
  ${(\begin{matrix} 4 \\ - 1 \\ 0 \end{matrix}) + (\begin{matrix} - 1 \\ 1 \\ 1 \end{matrix}) x_{3} ∣ x_{3} \in ℝ}$
  
  This is a particular solution and the solution set for the associated homogeneous system.
  
  $(\begin{matrix} 4 \\ - 1 \\ 0 \end{matrix}) {(\begin{matrix} - 1 \\ 1 \\ 1 \end{matrix}) x_{3} ∣ x_{3} \in ℝ}$
4. The solution set is a singleton.
  
  ${(\begin{matrix} 1 \\ 1 \\ 1 \end{matrix})}$
  
  A particular solution and the solution set for the associated homogeneous system are here.
  
  $(\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}) {(\begin{matrix} 0 \\ 0 \\ 0 \end{matrix})}$
5. The solution set is infinite.
  
  ${(\begin{matrix} 5 / 3 \\ 2 / 3 \\ 0 \\ 0 \end{matrix}) + (\begin{matrix} - 1 / 3 \\ 2 / 3 \\ 1 \\ 0 \end{matrix}) z + (\begin{matrix} - 2 / 3 \\ 1 / 3 \\ 0 \\ 1 \end{matrix}) w ∣ z, w \in ℝ}$
  
  A particular solution and the solution set for the associated homogeneous system are here.
  
  $(\begin{matrix} 5 / 3 \\ 2 / 3 \\ 0 \\ 0 \end{matrix}) {(\begin{matrix} - 1 / 3 \\ 2 / 3 \\ 1 \\ 0 \end{matrix}) z + (\begin{matrix} - 2 / 3 \\ 1 / 3 \\ 0 \\ 1 \end{matrix}) w ∣ z, w \in ℝ}$
6. This system’s solution set is empty. Thus, there is no particular solution. The solution set of the associated homogeneous system is this.
  
  ${(\begin{matrix} - 1 \\ 2 \\ 1 \\ 0 \end{matrix}) z + (\begin{matrix} - 1 \\ 3 \\ 0 \\ 1 \end{matrix}) w ∣ z, w \in ℝ}$
Exercise 3.16 Worked answer

Solve each system, giving the solution set in vector notation. Identify a particular solution and the solution of the homogeneous system.
1. $\begin{array}{rcrcrcr} 2 x & + & y & - & z & = & 1 \\ 4 x & - & y & = & 3 \end{array}$
2. $\begin{array}{rcrcrcrcr} x & - & z & = & 1 \\ y & + & 2 z & - & w & = & 3 \\ x & + & 2 y & + & 3 z & - & w & = & 7 \end{array}$
3. $\begin{array}{rcrcrcrcr} x & - & y & + & z & = & 0 \\ y & + & w & = & 0 \\ 3 x & - & 2 y & + & 3 z & + & w & = & 0 \\ - y & - & w & = & 0 \end{array}$
4. $\begin{array}{rcrcrcrcrcr} a & + & 2 b & + & 3 c & + & d & - & e & = & 1 \\ 3 a & - & b & + & c & + & d & + & e & = & 3 \end{array}$
Back to Exercise 3.16

Answer. The answers from the prior subsection show the row operations. Each answer here just lists the solution set, the particular solution, and the homogeneous solution.
1. The solution set is this.
  
  ${(\begin{matrix} 2 / 3 \\ - 1 / 3 \\ 0 \end{matrix}) + (\begin{matrix} 1 / 6 \\ 2 / 3 \\ 1 \end{matrix}) z ∣ z \in ℝ}$
  
  A particular solution and the solution set for the associated homogeneous system are here.
  
  $(\begin{matrix} 2 / 3 \\ - 1 / 3 \\ 0 \end{matrix}) {(\begin{matrix} 1 / 6 \\ 2 / 3 \\ 1 \end{matrix}) z ∣ z \in ℝ}$
2. The solution set is infinite.
  
  ${(\begin{matrix} 1 \\ 3 \\ 0 \\ 0 \end{matrix}) + (\begin{matrix} 1 \\ - 2 \\ 1 \\ 0 \end{matrix}) z ∣ z \in ℝ}$
  
  Here are a particular solution and the solution set for the associated homogeneous system.
  
  $(\begin{matrix} 1 \\ 3 \\ 0 \\ 0 \end{matrix}) {(\begin{matrix} 1 \\ - 2 \\ 1 \\ 0 \end{matrix}) z ∣ z \in ℝ}$
3. This is the solution set.
  
  ${(\begin{matrix} 0 \\ 0 \\ 0 \\ 0 \end{matrix}) + (\begin{matrix} - 1 \\ 0 \\ 1 \\ 0 \end{matrix}) z + (\begin{matrix} - 1 \\ - 1 \\ 0 \\ 1 \end{matrix}) w ∣ z, w \in ℝ}$
  
  Here is a particular solution and the solution set for the associated homogeneous system.
  
  $(\begin{matrix} 0 \\ 0 \\ 0 \\ 0 \end{matrix}) {(\begin{matrix} - 1 \\ 0 \\ 1 \\ 0 \end{matrix}) z + (\begin{matrix} - 1 \\ - 1 \\ 0 \\ 1 \end{matrix}) w ∣ z, w \in ℝ}$
4. The solution set is this.
  
  ${(\begin{matrix} 1 \\ 0 \\ 0 \\ 0 \\ 0 \end{matrix}) + (\begin{matrix} - 5 / 7 \\ - 8 / 7 \\ 1 \\ 0 \\ 0 \end{matrix}) c + (\begin{matrix} - 3 / 7 \\ - 2 / 7 \\ 0 \\ 1 \\ 0 \end{matrix}) d + (\begin{matrix} - 1 / 7 \\ 4 / 7 \\ 0 \\ 0 \\ 1 \end{matrix}) e ∣ c, d, e \in ℝ}$
  
  And, this is a particular solution and the solution set for the associated homogeneous system.
  
  $(\begin{matrix} 1 \\ 0 \\ 0 \\ 0 \\ 0 \end{matrix}) {(\begin{matrix} - 5 / 7 \\ - 8 / 7 \\ 1 \\ 0 \\ 0 \end{matrix}) c + (\begin{matrix} - 3 / 7 \\ - 2 / 7 \\ 0 \\ 1 \\ 0 \end{matrix}) d + (\begin{matrix} - 1 / 7 \\ 4 / 7 \\ 0 \\ 0 \\ 1 \end{matrix}) e ∣ c, d, e \in ℝ}$
Exercise 3.17 Worked answer

Recommended. For the system

$\begin{array}{rcrcrcrcr} 2 x & - & y & - & w & = & 3 \\ y & + & z & + & 2 w & = & 2 \\ x & - & 2 y & - & z & = & - 1 \end{array}$

which of these can be used as the particular solution part of some general solution?
1. $(\begin{matrix} 0 \\ - 3 \\ 5 \\ 0 \end{matrix})$
2. $(\begin{matrix} 2 \\ 1 \\ 1 \\ 0 \end{matrix})$
3. $(\begin{matrix} - 1 \\ - 4 \\ 8 \\ - 1 \end{matrix})$
Back to Exercise 3.17

Answer. Just plug them in and see if they satisfy all three equations.
1. No.
2. Yes.
3. Yes.
Exercise 3.18 Worked answer

Recommended. Lemma 3.7 says that we can use any particular solution for $\vec{p}$ . Find, if possible, a general solution to this system

$\begin{array}{rcrcrcrcr} x & - & y & + & w & = & 4 \\ 2 x & + & 3 y & - & z & = & 0 \\ y & + & z & + & w & = & 4 \end{array}$

that uses the given vector as its particular solution.
1. $(\begin{matrix} 0 \\ 0 \\ 0 \\ 4 \end{matrix})$
2. $(\begin{matrix} - 5 \\ 1 \\ - 7 \\ 10 \end{matrix})$
3. $(\begin{matrix} 2 \\ - 1 \\ 1 \\ 1 \end{matrix})$
Back to Exercise 3.18

Answer. Gauss’s Method on the associated homogeneous system

$\begin{aligned} (\begin{array}{ccccr} 1 & - 1 & 0 & 1 & 0 \\ 2 & 3 & - 1 & 0 & 0 \\ 0 & 1 & 1 & 1 & 0 \end{array}) & ⟶_{\begin{matrix} \end{matrix}}^{- 2 ρ_{1} + ρ_{2}} (\begin{array}{ccccr} 1 & - 1 & 0 & 1 & 0 \\ 0 & 5 & - 1 & - 2 & 0 \\ 0 & 1 & 1 & 1 & 0 \end{array}) \\ ⟶_{\begin{matrix} \end{matrix}}^{- (1 / 5) ρ_{2} + ρ_{3}} (\begin{array}{ccccr} 1 & - 1 & 0 & 1 & 0 \\ 0 & 5 & - 1 & - 2 & 0 \\ 0 & 0 & 6 / 5 & 7 / 5 & 0 \end{array}) \end{aligned}$

gives this is the solution to the homogeneous problem.

${(\begin{matrix} - 5 / 6 \\ 1 / 6 \\ - 7 / 6 \\ 1 \end{matrix}) w ∣ w \in ℝ}$
1. That vector is indeed a particular solution, so the required general solution is this.
  
  ${(\begin{matrix} 0 \\ 0 \\ 0 \\ 4 \end{matrix}) + (\begin{matrix} - 5 / 6 \\ 1 / 6 \\ - 7 / 6 \\ 1 \end{matrix}) w ∣ w \in ℝ}$
2. That vector is a particular solution so the required general solution is this.
  
  ${(\begin{matrix} - 5 \\ 1 \\ - 7 \\ 10 \end{matrix}) + (\begin{matrix} - 5 / 6 \\ 1 / 6 \\ - 7 / 6 \\ 1 \end{matrix}) w ∣ w \in ℝ}$
3. That vector is not a solution of the system since it does not satisfy the third equation. No such general solution exists.
Exercise 3.19 Worked answer

One is nonsingular while the other is singular. Which is which?
1. $(\begin{matrix} 1 & 3 \\ 4 & - 12 \end{matrix})$
2. $(\begin{matrix} 1 & 3 \\ 4 & 12 \end{matrix})$
Back to Exercise 3.19

Answer. The first is nonsingular while the second is singular. Just do Gauss’s Method and see if the echelon form result has non- $0$ numbers in each entry on the diagonal.
Exercise 3.20 Worked answer

Recommended. Singular or nonsingular?
1. $(\begin{matrix} 1 & 2 \\ 1 & 3 \end{matrix})$
2. $(\begin{matrix} 1 & 2 \\ - 3 & - 6 \end{matrix})$
3. $(\begin{matrix} 1 & 2 & 1 \\ 1 & 3 & 1 \end{matrix})$
4. $(\begin{matrix} 1 & 2 & 1 \\ 1 & 1 & 3 \\ 3 & 4 & 7 \end{matrix})$
5. $(\begin{matrix} 2 & 2 & 1 \\ 1 & 0 & 5 \\ - 1 & 1 & 4 \end{matrix})$
Back to Exercise 3.20

Answer.
1. Nonsingular:
  
  $⟶_{\begin{array}{l} \end{array}}^{- ρ_{1} + ρ_{2}} (\begin{matrix} 1 & 2 \\ 0 & 1 \end{matrix})$
  
  ends with each row containing a leading entry.
2. Singular:
  
  $⟶_{\begin{array}{l} \end{array}}^{3 ρ_{1} + ρ_{2}} (\begin{matrix} 1 & 2 \\ 0 & 0 \end{matrix})$
  
  ends with row $2$ without a leading entry.
3. Neither. A matrix must be square for either word to apply.
4. Singular.
5. Nonsingular.
Exercise 3.21 Worked answer

Recommended. Is the given vector in the set generated by the given set?
1. $(\begin{matrix} 2 \\ 3 \end{matrix}),$ ${(\begin{matrix} 1 \\ 4 \end{matrix}), (\begin{matrix} 1 \\ 5 \end{matrix})}$
2. $(\begin{matrix} - 1 \\ 0 \\ 1 \end{matrix}),$ ${(\begin{matrix} 2 \\ 1 \\ 0 \end{matrix}), (\begin{matrix} 1 \\ 0 \\ 1 \end{matrix})}$
3. $(\begin{matrix} 1 \\ 3 \\ 0 \end{matrix}),$ ${(\begin{matrix} 1 \\ 0 \\ 4 \end{matrix}), (\begin{matrix} 2 \\ 1 \\ 5 \end{matrix}), (\begin{matrix} 3 \\ 3 \\ 0 \end{matrix}), (\begin{matrix} 4 \\ 2 \\ 1 \end{matrix})}$
4. $(\begin{matrix} 1 \\ 0 \\ 1 \\ 1 \end{matrix}),$ ${(\begin{matrix} 2 \\ 1 \\ 0 \\ 1 \end{matrix}), (\begin{matrix} 3 \\ 0 \\ 0 \\ 2 \end{matrix})}$
Back to Exercise 3.21

Answer. In each case we must decide if the vector is a linear combination of the vectors in the set.
1. Yes. Solve
  
  $c_{1} (\begin{matrix} 1 \\ 4 \end{matrix}) + c_{2} (\begin{matrix} 1 \\ 5 \end{matrix}) = (\begin{matrix} 2 \\ 3 \end{matrix})$
  
  with
  
  $(\begin{array}{ccr} 1 & 1 & 2 \\ 4 & 5 & 3 \end{array}) ⟶_{\begin{array}{l} \end{array}}^{- 4 ρ_{1} + ρ_{2}} (\begin{array}{ccr} 1 & 1 & 2 \\ 0 & 1 & - 5 \end{array})$
  
  to conclude that there are $c_{1}$ and $c_{2}$ giving the combination.
2. No. The reduction
  
  $(\begin{array}{ccr} 2 & 1 & - 1 \\ 1 & 0 & 0 \\ 0 & 1 & 1 \end{array}) ⟶_{\begin{array}{l} \end{array}}^{- (1 / 2) ρ_{1} + ρ_{2}} (\begin{array}{ccr} 2 & 1 & - 1 \\ 0 & - 1 / 2 & 1 / 2 \\ 0 & 1 & 1 \end{array}) ⟶_{\begin{array}{l} \end{array}}^{2 ρ_{2} + ρ_{3}} (\begin{array}{ccr} 2 & 1 & - 1 \\ 0 & - 1 / 2 & 1 / 2 \\ 0 & 0 & 2 \end{array})$
  
  shows that
  
  $c_{1} (\begin{matrix} 2 \\ 1 \\ 0 \end{matrix}) + c_{2} (\begin{matrix} 1 \\ 0 \\ 1 \end{matrix}) = (\begin{matrix} - 1 \\ 0 \\ 1 \end{matrix})$
  
  has no solution.
3. Yes. The reduction
  
  $\begin{aligned} (\begin{array}{ccccr} 1 & 2 & 3 & 4 & 1 \\ 0 & 1 & 3 & 2 & 3 \\ 4 & 5 & 0 & 1 & 0 \end{array}) & ⟶_{\begin{matrix} \end{matrix}}^{- 4 ρ_{1} + ρ_{3}} (\begin{array}{ccccr} 1 & 2 & 3 & 4 & 1 \\ 0 & 1 & 3 & 2 & 3 \\ 0 & - 3 & - 12 & - 15 & - 4 \end{array}) \\ ⟶_{\begin{matrix} \end{matrix}}^{3 ρ_{2} + ρ_{3}} (\begin{array}{ccccr} 1 & 2 & 3 & 4 & 1 \\ 0 & 1 & 3 & 2 & 3 \\ 0 & 0 & - 3 & - 9 & 5 \end{array}) \end{aligned}$
  
  shows that there are infinitely many ways
  
  ${(\begin{matrix} c_{1} \\ c_{2} \\ c_{3} \\ c_{4} \end{matrix}) = (\begin{matrix} - 10 \\ 8 \\ - 5 / 3 \\ 0 \end{matrix}) + (\begin{matrix} - 9 \\ 7 \\ - 3 \\ 1 \end{matrix}) c_{4} ∣ c_{4} \in ℝ}$
  
  to write a combination.
  
  $(\begin{matrix} 1 \\ 3 \\ 0 \end{matrix}) = c_{1} (\begin{matrix} 1 \\ 0 \\ 4 \end{matrix}) + c_{2} (\begin{matrix} 2 \\ 1 \\ 5 \end{matrix}) + c_{3} (\begin{matrix} 3 \\ 3 \\ 0 \end{matrix}) + c_{4} (\begin{matrix} 4 \\ 2 \\ 1 \end{matrix})$
4. No. Look at the third components.
Exercise 3.22 Worked answer

Prove that any linear system with a nonsingular matrix of coefficients has a solution, and that the solution is unique.

Back to Exercise 3.22

Answer. Because the matrix of coefficients is nonsingular, Gauss’s Method ends with an echelon form where each variable leads an equation. Back substitution gives a unique solution.

(Another way to see that the solution is unique is to note that with a nonsingular matrix of coefficients the associated homogeneous system has a unique solution, by definition. Since the general solution is the sum of a particular solution with each homogeneous solution, the general solution has at most one element.)
Exercise 3.23 Worked answer

In the proof of Lemma 3.6, what happens if there are no non- $0 = 0$ equations?

Back to Exercise 3.23

Answer. In this case the solution set is all of $ℝ^{n}$ and we can express it in the required form.

${c_{1} (\begin{matrix} 1 \\ 0 \\ ⋮ \\ 0 \end{matrix}) + c_{2} (\begin{matrix} 0 \\ 1 \\ ⋮ \\ 0 \end{matrix}) + \dots + c_{n} (\begin{matrix} 0 \\ 0 \\ ⋮ \\ 1 \end{matrix}) ∣ c_{1}, \dots, c_{n} \in ℝ}$
Exercise 3.24 Worked answer

Recommended. Prove that if $\vec{s}$ and $\vec{t}$ satisfy a homogeneous system then so do these vectors.
1. $\vec{s} + \vec{t}$
2. $3 \vec{s}$
3. $k \vec{s} + m \vec{t}$ for $k, m \in ℝ$
What’s wrong with this argument: “These three show that if a homogeneous system has one solution then it has many solutions—any multiple of a solution is another solution, and any sum of solutions is a solution also—so there are no homogeneous systems with exactly one solution.”?
Back to Exercise 3.24

Answer. Assume $\vec{s}, \vec{t} \in ℝ^{n}$ and write them as here.

$\vec{s} = (\begin{matrix} s_{1} \\ ⋮ \\ s_{n} \end{matrix}) \vec{t} = (\begin{matrix} t_{1} \\ ⋮ \\ t_{n} \end{matrix})$

Also let $a_{i, 1} x_{1} + \dots + a_{i, n} x_{n} = 0$ be the $i$ -th equation in the homogeneous system.
1. The check is easy.
  
  $\begin{matrix} a_{i, 1} (s_{1} + t_{1}) + \dots + a_{i, n} (s_{n} + t_{n}) \\ = (a_{i, 1} s_{1} + \dots + a_{i, n} s_{n}) + (a_{i, 1} t_{1} + \dots + a_{i, n} t_{n}) = 0 + 0 \end{matrix}$
2. This is similar to the prior one.
  
  $a_{i, 1} (3 s_{1}) + \dots + a_{i, n} (3 s_{n}) = 3 (a_{i, 1} s_{1} + \dots + a_{i, n} s_{n}) = 3 \cdot 0 = 0$
3. This one is not much harder.
  
  $\begin{matrix} a_{i, 1} (k s_{1} + m t_{1}) + \dots + a_{i, n} (k s_{n} + m t_{n}) \\ = k (a_{i, 1} s_{1} + \dots + a_{i, n} s_{n}) + m (a_{i, 1} t_{1} + \dots + a_{i, n} t_{n}) = k \cdot 0 + m \cdot 0 \end{matrix}$
What is wrong with that argument is that any linear combination involving only the zero vector yields the zero vector.
Exercise 3.25 Worked answer

Prove that if a system with only rational coefficients and constants has a solution then it has at least one all-rational solution. Must it have infinitely many?

Back to Exercise 3.25

Answer. First the proof.

If the system has all rational coefficients then Gauss’s Method with back substitution will use only rationals (e.g., $- (m / n) ρ_{i} + ρ_{j}$ ). Thus we can express the solution set using only rational numbers as the components of each vector. Therefore, the particular solution is all rational.

As to the question about infinitely many, there are infinitely many rational vector solutions if and only if the associated homogeneous system has infinitely many real vector solutions. That’s because setting any parameters to be rationals will produce an all-rational solution.

References cited in this section

Onan

Michael Onan, Linear Algebra, Harcourt, 1990.

Anton

Howard Anton, Elementary Linear Algebra, John Wiley & Sons, 1987.

Con. Prob. 1955

The Contest Problem Book, 1955 number 38.

Am. Math. Mon., Jan. 1935

W. R. Ransom (proposer), Hansraj Gupta (solver), Elementary problem 105, American Mathematical Monthly, vol. 42 no. 1 (Jan. 1935), p. 47.

Wohascum no. 2

The Wohascum County Problem Book problem number 2.

joriki

Mathematics Stack Exchange user joriki, http://math.stackexchange.com/a/118914/205168, 2012.

Am. Math. Mon., Jan. 1963

Underwood Dudley, Arnold Lebow (proposers), David Rothman (solver), Elementary problem 1151, American Mathematical Monthly, vol. 70 no. 1 (Jan. 1963), p. 93.

Cleary

R. Cleary, private communication, Nov. 2011.

Shepelev

Anton Shepelev, private communication, Feb 19, 2011.

USSR Olympiad no. 174

The USSR Mathematics Olympiad, number 174.

Math. Mag., Sept. 1952

Dewey Duncan (proposer), W. H. Quelch (solver), Mathematics Magazine, volume 26 number 1 (Sept-Oct. 1952), p. 48.

More information on mathematical induction is in the appendix.↩︎
More information on set equality is in the appendix.↩︎

Aluminum	$2.7$	Alcohol	0.81
Copper	$8.9$	Benzene	$0.90$
Gold	$19.3$	Glycerin	$1.26$
Lead	$11.3$	Water	$1.00$
Silver	$10.8$

R:	$13$
G:	$15$
B:	$17$