Cramer’s Rule

A linear system is equivalent to a linear relationship among vectors.

$\begin{array}{rcrcr} x_{1} & + & 2 x_{2} & = & 6 \\ 3 x_{1} & + & x_{2} & = & 8 \end{array} ⟺ x_{1} \cdot (\begin{matrix} 1 \\ 3 \end{matrix}) + x_{2} \cdot (\begin{matrix} 2 \\ 1 \end{matrix}) = (\begin{matrix} 6 \\ 8 \end{matrix})$

In the picture below the small parallelogram is formed from the vectors $(\binom{1}{3})$ and $(\binom{2}{1})$ . It is nested inside a parallelogram with sides $x_{1} (\binom{1}{3})$ and $x_{2} (\binom{2}{1})$ . By the vector equation, the far corner of the larger parallelogram is $(\binom{6}{8})$ .

Original nested parallelograms. The small one is spanned by (1,3) and (2,1); the larger has sides x1(1,3) and x2(2,1), with opposite corner (6,8). Labels and arrows identify both directions and their dilations.

This drawing restates the algebraic question of finding the solution of a linear system into geometric terms: by what factors $x_{1}$ and $x_{2}$ must we dilate the sides of the starting parallelogram so that it will fill the other one?

We can use this picture, and our geometric understanding of determinants, to get a new formula for solving linear systems. Compare the sizes of these shaded boxes.

First shaded box: the small parallelogram spanned by (1,3) and (2,1), with the original larger parallelogram outlined behind it. Second shaded box: the parallelogram spanned by x1(1,3) and (2,1). Only the first of its two sides is dilated from the first shaded box. Third shaded box: the parallelogram spanned by (6,8) and (2,1). Its long side is the sum x1(1,3)+x2(2,1); the original larger outline remains visible for comparison. This is the shear of the second shaded box.

The second is defined by the vectors $x_{1} (\binom{1}{3})$ and $(\binom{2}{1})$ and one of the properties of the size function—the determinant—is that therefore the size of the second box is $x_{1}$ times the size of the first. The third box is derived from the second by shearing, adding $x_{2} (\binom{2}{1})$ to $x_{1} (\binom{1}{3})$ to get $x_{1} (\binom{1}{3}) + x_{2} (\binom{2}{1}) = (\binom{6}{8})$ , along with $(\binom{2}{1})$ . The determinant is not affected by shearing so the size of the third box equals that of the second.

Taken together we have this.

$x_{1} \cdot | \begin{matrix} 1 & 2 \\ 3 & 1 \end{matrix} | = | \begin{matrix} x_{1} \cdot 1 & 2 \\ x_{1} \cdot 3 & 1 \end{matrix} | = | \begin{matrix} x_{1} \cdot 1 + x_{2} \cdot 2 & 2 \\ x_{1} \cdot 3 + x_{2} \cdot 1 & 1 \end{matrix} | = | \begin{matrix} 6 & 2 \\ 8 & 1 \end{matrix} |$

Solving gives the value of one of the variables.

$x_{1} = \frac{| \begin{matrix} 6 & 2 \\ 8 & 1 \end{matrix} |}{| \begin{matrix} 1 & 2 \\ 3 & 1 \end{matrix} |} = \frac{- 10}{- 5} = 2$

The generalization of this example is Cramer’s Rule: if $| A | \neq 0$ then the system $A \vec{x} = \vec{b}$ has the unique solution $x_{i} = | B_{i} | / | A |$ where the matrix $B_{i}$ is formed from $A$ by replacing column $i$ with the vector $\vec{b}$ . The proof is Exercise 3.

For instance, to solve this system for $x_{2}$

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

we do this computation.

$x_{2} = \frac{| \begin{matrix} 1 & 2 & 4 \\ 2 & 1 & - 1 \\ 1 & - 1 & 1 \end{matrix} |}{| \begin{matrix} 1 & 0 & 4 \\ 2 & 1 & - 1 \\ 1 & 0 & 1 \end{matrix} |} = \frac{- 18}{- 3}$

Cramer’s Rule lets us by-eye solve systems that are small and simple. For example, we can solve systems with two equations and two unknowns, or three equations and three unknowns, where the numbers are small integers. Such cases appear often enough that many people find this formula handy.

But using it to solving large or complex systems is not practical, either by hand or by a computer. A Gauss’s Method-based approach is faster.

Exercises

Exercise 1 Supplied answer

Use Cramer’s Rule to solve each for each of the variables.
1. $\begin{array}{rcrcr} x & - & y & = & 4 \\ - x & + & 2 y & = & - 7 \end{array}$
2. $\begin{array}{rcrcr} - 2 x & + & y & = & - 2 \\ x & - & 2 y & = & - 2 \end{array}$
Back to Exercise 1

Answer.
1. Solve for the variables separately.
  
  $x = \frac{| \begin{matrix} 4 & - 1 \\ - 7 & 2 \end{matrix} |}{| \begin{matrix} 1 & - 1 \\ - 1 & 2 \end{matrix} |} = \frac{1}{1} = 1 y = \frac{| \begin{matrix} 1 & 4 \\ - 1 & - 7 \end{matrix} |}{| \begin{matrix} 1 & - 1 \\ - 1 & 2 \end{matrix} |} = \frac{- 3}{1} = - 3$
2. $x = 2$ , $y = 2$
Exercise 2 Supplied answer

Use Cramer’s Rule to solve this system for $z$ .

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

Back to Exercise 2

Answer. $z = 1$
Exercise 3 Supplied answer

Prove Cramer’s Rule.

Back to Exercise 3

Answer. Determinants are unchanged by combinations, including column combinations, so $det (B_{i}) = det ({\vec{a}}_{1}, \dots, x_{1} {\vec{a}}_{1} + \dots + x_{i} {\vec{a}}_{i} + \dots + x_{n} {\vec{a}}_{n}, \dots, {\vec{a}}_{n})$ . Use the operation of taking $- x_{1}$ times the first column and adding it to the $i$ -th column, etc., to see this is equal to $det ({\vec{a}}_{1}, \dots, x_{i} {\vec{a}}_{i}, \dots, {\vec{a}}_{n})$ . In turn, that is equal to $x_{i} \cdot det ({\vec{a}}_{1}, \dots, {\vec{a}}_{i}, \dots, {\vec{a}}_{n}) = x_{i} \cdot det (A)$ , as required.
Exercise 4 Supplied answer

Here is an alternative proof of Cramer’s Rule that doesn’t overtly contain any geometry. Write $X_{i}$ for the identity matrix with column $i$ replaced by the vector $\vec{x}$ of unknowns $x_{1}$ , …, $x_{n}$ .
1. Observe that $A X_{i} = B_{i}$ .
2. Take the determinant of both sides.
Back to Exercise 4

Answer.
1. Here is the case of a $2 \times 2$ system with $i = 2$ .
  
  $\begin{array}{rcrcr} a_{1, 1} x_{1} & + & a_{1, 2} x_{2} & = & b_{1} \\ a_{2, 1} x_{1} & + & a_{2, 2} x_{2} & = & b_{2} \end{array} ⟺ (\begin{matrix} a_{1, 1} & a_{1, 2} \\ a_{2, 1} & a_{2, 2} \end{matrix}) (\begin{matrix} 1 & x_{1} \\ 0 & x_{2} \end{matrix}) = (\begin{matrix} a_{1, 1} & b_{1} \\ a_{2, 1} & b_{2} \end{matrix})$
2. The determinant function is multiplicative $det (B_{i}) = det (A X_{i}) = det (A) \cdot det (X_{i})$ . The Laplace expansion shows that $det (X_{i}) = x_{i}$ , and solving for $x_{i}$ gives Cramer’s Rule.
Exercise 5 Supplied answer

Suppose that a linear system has as many equations as unknowns, that all of its coefficients and constants are integers, and that its matrix of coefficients has determinant $1$ . Prove that the entries in the solution are all integers. (Remark. This is often used to invent linear systems for exercises.)

Back to Exercise 5

Answer. Because the determinant of $A$ is nonzero, Cramer’s Rule applies and shows that $x_{i} = | B_{i} | / 1$ . Since $B_{i}$ is a matrix of integers, its determinant is an integer.
Exercise 6 Supplied answer

Use Cramer’s Rule to give a formula for the solution of a two equations/two unknowns linear system.

Back to Exercise 6

Answer. The solution of

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

is

$x = \frac{e d - f b}{a d - b c} y = \frac{a f - e c}{a d - b c}$

provided of course that the denominators are not zero.
Exercise 7 Supplied answer

Can Cramer’s Rule tell the difference between a system with no solutions and one with infinitely many?

Back to Exercise 7

Answer. Of course, singular systems have $| A |$ equal to zero, but we can characterize the infinitely many solutions case is by the fact that all of the $| B_{i} |$ are zero as well.
Exercise 8 Supplied answer

The first picture in this Topic (the one that doesn’t use determinants) shows a unique solution case. Produce a similar picture for the case of infinitely many solutions, and the case of no solutions.

Back to Exercise 8

Answer. We can consider the two nonsingular cases together with this system

$\begin{array}{rcrcr} x_{1} & + & 2 x_{2} & = & 6 \\ x_{1} & + & 2 x_{2} & = & c \end{array}$

where $c = 6$ of course yields infinitely many solutions, and any other value for $c$ yields no solutions. The corresponding vector equation

$x_{1} \cdot (\begin{matrix} 1 \\ 1 \end{matrix}) + x_{2} \cdot (\begin{matrix} 2 \\ 2 \end{matrix}) = (\begin{matrix} 6 \\ c \end{matrix})$

gives a picture of two overlapping vectors. Both lie on the line $y = x$ . In the $c = 6$ case the vector on the right side also lies on the line $y = x$ but in any other case it does not.