Laplace’s Formula

This section is optional. The only later sections that depends on this material is Five.III.

Determinants are a font of interesting and amusing formulas. Here is one that is often used to compute determinants by hand.

Laplace’s Expansion

In the permutation expansion of $| T |$ , every term has one and only one entry from each row and column of $T$ .

$| T | = \sum_{permutations ϕ} t_{1, ϕ (1)} t_{2, ϕ (2)} \dots t_{n, ϕ (n)} | P_{ϕ} |$

Here is the formula’s $3 \times 3$ instance.

$\begin{aligned} | \begin{array}{c} t_{1, 1} & t_{1, 2} & t_{1, 3} \\ t_{2, 1} & t_{2, 2} & t_{2, 3} \\ t_{3, 1} & t_{3, 2} & t_{3, 3} \end{array} | & = \begin{aligned} t_{1, 1} t_{2, 2} t_{3, 3} | P_{ϕ_{1}} | + t_{1, 1} t_{2, 3} t_{3, 2} | P_{ϕ_{2}} | + t_{1, 2} t_{2, 1} t_{3, 3} | P_{ϕ_{3}} | \\ + t_{1, 2} t_{2, 3} t_{3, 1} | P_{ϕ_{4}} | + t_{1, 3} t_{2, 1} t_{3, 2} | P_{ϕ_{5}} | + t_{1, 3} t_{2, 2} t_{3, 1} | P_{ϕ_{6}} | \end{aligned} \\ = \begin{aligned} t_{1, 1} t_{2, 2} t_{3, 3} - t_{1, 1} t_{2, 3} t_{3, 2} - t_{1, 2} t_{2, 1} t_{3, 3} \\ + t_{1, 2} t_{2, 3} t_{3, 1} + t_{1, 3} t_{2, 1} t_{3, 2} - t_{1, 3} t_{2, 2} t_{3, 1} \end{aligned} \end{aligned}$

We can focus on any row or column; here we bring out the entries from $T$ ’s first row.

$\begin{aligned} = t_{1, 1} \cdot [t_{2, 2} t_{3, 3} - t_{2, 3} t_{3, 2}] \\ - t_{1, 2} \cdot [t_{2, 1} t_{3, 3} - t_{2, 3} t_{3, 1}] \\ + t_{1, 3} \cdot [t_{2, 1} t_{3, 2} - t_{2, 2} t_{3, 1}] & (*) \end{aligned}$

Note that inside the first set of square brackets is the determinant

$\begin{matrix} | \begin{matrix} t_{2, 2} & t_{2, 3} \\ t_{3, 2} & t_{3, 3} \end{matrix} | & (* *) \end{matrix}$

and similarly, inside the other brackets are other determinants.

$\begin{matrix} | \begin{matrix} t_{2, 1} & t_{2, 3} \\ t_{3, 1} & t_{3, 3} \end{matrix} | | \begin{matrix} t_{2, 1} & t_{2, 2} \\ t_{3, 1} & t_{3, 2} \end{matrix} | & (* * *) \end{matrix}$

The explanation for these determinants lies with the permutations. If we fix some row and column—such as when we consider all terms containing $t_{1, 1}$ , or all terms containing $t_{1, 2}$ , or $t_{1, 3}$ —then what remains is for the permutation to pick one entry from each of the other rows and columns.

The matrices below illustrate. On the left, the row and column containing $t_{1, 1}$ are shaded and what remains is the matrix from ( $* *$ ) above. Similarly, in the other two what remains are the matrices from ( $* * *$ ).

$(\begin{array}{ccc} t_{1, 1} & t_{1, 2} & t_{1, 3} \\ t_{2, 1} & t_{2, 2} & t_{2, 3} \\ t_{3, 1} & t_{3, 2} & t_{3, 3} \end{array}) (\begin{array}{ccc} t_{1, 1} & t_{1, 2} & t_{1, 3} \\ t_{2, 1} & t_{2, 2} & t_{2, 3} \\ t_{3, 1} & t_{3, 2} & t_{3, 3} \end{array}) (\begin{array}{ccc} t_{1, 1} & t_{1, 2} & t_{1, 3} \\ t_{2, 1} & t_{2, 2} & t_{2, 3} \\ t_{3, 1} & t_{3, 2} & t_{3, 3} \end{array})$

So that accounts for the appearance in equation ( $*$ ) of the determinants of smaller-sized matrices. As to the minus sign on that equation’s second line, consider this version of $| T | = t_{1, 1} t_{2, 2} t_{3, 3} | P_{ϕ_{1}} | + \dots + t_{1, 3} t_{2, 2} t_{3, 1} | P_{ϕ_{6}} |$ .

$\begin{aligned} t_{1, 1} \cdot [t_{2, 2} t_{3, 3} | \begin{array}{c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} | + t_{2, 3} t_{3, 2} | \begin{array}{c} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{array} |] \\ + t_{1, 2} \cdot [t_{2, 1} t_{3, 3} | \begin{array}{c} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{array} | + t_{2, 3} t_{3, 1} | \begin{array}{c} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{array} |] \\ + t_{1, 3} \cdot [t_{2, 1} t_{3, 2} | \begin{array}{c} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{array} | + t_{2, 2} t_{3, 1} | \begin{array}{c} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{array} |] \end{aligned}$

The matrices $P_{ϕ_{1}}$ and $P_{ϕ_{2}}$ in the first line are all set up for the determinant ( $* *$ ), since their first row is as in the identity matrix and their other two rows look like the rows for that determinant. The matrices $P_{ϕ_{3}}$ and $P_{ϕ_{4}}$ in the second line will be set up in the same way with a single row swap, putting $(\begin{matrix} 1 & 0 & 0 \end{matrix})$ into the first row. Of course, that row swap puts a negative sign on that line. Finally, setting up the third line’s matrices in the same way takes two swaps, which keeps the sign positive.

In short, we have this.

$| T | = t_{1, 1} \cdot | \begin{matrix} t_{2, 2} & t_{2, 3} \\ t_{3, 2} & t_{3, 3} \end{matrix} | - t_{1, 2} \cdot | \begin{matrix} t_{2, 1} & t_{2, 3} \\ t_{3, 1} & t_{3, 3} \end{matrix} | + t_{1, 3} \cdot | \begin{matrix} t_{2, 1} & t_{2, 2} \\ t_{3, 1} & t_{3, 2} \end{matrix} |$

The formula given in Theorem 1.4, which generalizes this, is a recurrence since the determinant is expressed as a combination of determinants. This formula isn’t circular because it gives the $n \times n$ determinant in terms of smaller-sized matrices.

Definition 1.1 For any $n \times n$ matrix $T$ , the $(n - 1) \times (n - 1)$ matrix formed by deleting row $i$ and column $j$ of $T$ is the $i, j$ minor of $T$ . The $i, j$ cofactor $T_{i, j}$ of $T$ is $(- 1)^{i + j}$ times the determinant of the $i, j$ minor of $T$ .

Example 1.2 The $1, 2$ cofactor of the matrix from equation ( $*$ ) is the negative of the first determinant in ( $* * *$ ).

$T_{1, 2} = - 1 \cdot | \begin{matrix} t_{2, 1} & t_{2, 3} \\ t_{3, 1} & t_{3, 3} \end{matrix} |$

Example 1.3 Where

$T = (\begin{matrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{matrix})$

these are the $1, 2$ and $2, 2$ cofactors.

$T_{1, 2} = (- 1)^{1 + 2} \cdot | \begin{matrix} 4 & 6 \\ 7 & 9 \end{matrix} | = 6 T_{2, 2} = (- 1)^{2 + 2} \cdot | \begin{matrix} 1 & 3 \\ 7 & 9 \end{matrix} | = - 12$

Theorem 1.4 (Laplace Expansion of Determinants) Where $T$ is an $n \times n$ matrix, we can find the determinant by expanding by cofactors on any row $i$ or column $j$ .

$\begin{aligned} | T | & = t_{i, 1} \cdot T_{i, 1} + t_{i, 2} \cdot T_{i, 2} + \dots + t_{i, n} \cdot T_{i, n} \\ = t_{1, j} \cdot T_{1, j} + t_{2, j} \cdot T_{2, j} + \dots + t_{n, j} \cdot T_{n, j} \end{aligned}$

Proof Exercise 1.26.

QED

Example 1.5 We can compute the determinant

$| T | = | \begin{matrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{matrix} |$

by expanding along the first row.

$| T | = 1 \cdot (+ 1) | \begin{matrix} 5 & 6 \\ 8 & 9 \end{matrix} | + 2 \cdot (- 1) | \begin{matrix} 4 & 6 \\ 7 & 9 \end{matrix} | + 3 \cdot (+ 1) | \begin{matrix} 4 & 5 \\ 7 & 8 \end{matrix} | = - 3 + 12 - 9 = 0$

We could also expand down the second column.

$| T | = 2 \cdot (- 1) | \begin{matrix} 4 & 6 \\ 7 & 9 \end{matrix} | + 5 \cdot (+ 1) | \begin{matrix} 1 & 3 \\ 7 & 9 \end{matrix} | + 8 \cdot (- 1) | \begin{matrix} 1 & 3 \\ 4 & 6 \end{matrix} | = 12 - 60 + 48 = 0$

Example 1.6 A row or column with many zeroes makes a this expansion easier.

$| \begin{matrix} 1 & 5 & 0 \\ 2 & 1 & 1 \\ 3 & - 1 & 0 \end{matrix} | = 0 \cdot (+ 1) | \begin{matrix} 2 & 1 \\ 3 & - 1 \end{matrix} | + 1 \cdot (- 1) | \begin{matrix} 1 & 5 \\ 3 & - 1 \end{matrix} | + 0 \cdot (+ 1) | \begin{matrix} 1 & 5 \\ 2 & 1 \end{matrix} | = 16$

We finish by applying Laplace’s expansion to derive a new formula for the inverse of a matrix. With Theorem 1.4, we can calculate the determinant of a matrix by taking linear combinations of entries from a row with their associated cofactors.

$t_{i, 1} \cdot T_{i, 1} + t_{i, 2} \cdot T_{i, 2} + \dots + t_{i, n} \cdot T_{i, n} = | T |$

Recall that a matrix with two identical rows has a zero determinant. Thus, weighting the cofactors by entries from row $k$ with $k \neq i$ gives zero

$t_{i, 1} \cdot T_{k, 1} + t_{i, 2} \cdot T_{k, 2} + \dots + t_{i, n} \cdot T_{k, n} = 0$

because it represents the expansion along the row $k$ of a matrix with row $i$ equal to row $k$ . This summarizes.

$(\begin{array}{cccc} t_{1, 1} & t_{1, 2} & \dots & t_{1, n} \\ t_{2, 1} & t_{2, 2} & \dots & t_{2, n} \\ ⋮ \\ t_{n, 1} & t_{n, 2} & \dots & t_{n, n} \end{array}) (\begin{matrix} T_{1, 1} & T_{2, 1} & \dots & T_{n, 1} \\ T_{1, 2} & T_{2, 2} & \dots & T_{n, 2} \\ ⋮ \\ T_{1, n} & T_{2, n} & \dots & T_{n, n} \end{matrix}) = (\begin{matrix} | T | & 0 & \dots & 0 \\ 0 & | T | & \dots & 0 \\ ⋮ \\ 0 & 0 & \dots & | T | \end{matrix})$

Note that the order of the subscripts in the matrix of cofactors is opposite to the order of subscripts in the other matrix; e.g., along the first row of the matrix of cofactors the subscripts are $1, 1$ then $2, 1$ , etc.

Definition 1.7 The matrix adjoint (or adjugate) to the square matrix $T$ (or the classical adjoint) is

$adj (T) = (\begin{matrix} T_{1, 1} & T_{2, 1} & \dots & T_{n, 1} \\ T_{1, 2} & T_{2, 2} & \dots & T_{n, 2} \\ ⋮ \\ T_{1, n} & T_{2, n} & \dots & T_{n, n} \end{matrix})$

where the row $i$ , column $j$ entry, $T_{j, i}$ , is the $j, i$ cofactor.

Theorem 1.8 Where $T$ is a square matrix, $T \cdot adj (T) = adj (T) \cdot T = | T | \cdot I$ . Thus if $T$ has an inverse, if $| T | \neq 0$ , then $T^{- 1} = (1 / | T |) \cdot adj (T)$ .

Proof The discussion before the theorem statement makes it clear.

QED

Example 1.9 If

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

then $adj (T)$ is

$(\begin{matrix} T_{1, 1} & T_{2, 1} & T_{3, 1} \\ T_{1, 2} & T_{2, 2} & T_{3, 2} \\ T_{1, 3} & T_{2, 3} & T_{3, 3} \end{matrix}) = (\begin{matrix} | \begin{matrix} 1 & - 1 \\ 0 & 1 \end{matrix} | & - | \begin{matrix} 0 & 4 \\ 0 & 1 \end{matrix} | & | \begin{matrix} 0 & 4 \\ 1 & - 1 \end{matrix} | \\ - | \begin{matrix} 2 & - 1 \\ 1 & 1 \end{matrix} | & | \begin{matrix} 1 & 4 \\ 1 & 1 \end{matrix} | & - | \begin{matrix} 1 & 4 \\ 2 & - 1 \end{matrix} | \\ | \begin{matrix} 2 & 1 \\ 1 & 0 \end{matrix} | & - | \begin{matrix} 1 & 0 \\ 1 & 0 \end{matrix} | & | \begin{matrix} 1 & 0 \\ 2 & 1 \end{matrix} | \end{matrix}) = (\begin{matrix} 1 & 0 & - 4 \\ - 3 & - 3 & 9 \\ - 1 & 0 & 1 \end{matrix})$

and taking the product with $T$ gives the diagonal matrix $| T | \cdot I$ .

$(\begin{matrix} 1 & 0 & 4 \\ 2 & 1 & - 1 \\ 1 & 0 & 1 \end{matrix}) (\begin{matrix} 1 & 0 & - 4 \\ - 3 & - 3 & 9 \\ - 1 & 0 & 1 \end{matrix}) = (\begin{matrix} - 3 & 0 & 0 \\ 0 & - 3 & 0 \\ 0 & 0 & - 3 \end{matrix})$

The inverse of $T$ is $(1 / - 3) \cdot adj (T)$ .

$T^{- 1} = (\begin{matrix} 1 / - 3 & 0 / - 3 & - 4 / - 3 \\ - 3 / - 3 & - 3 / - 3 & 9 / - 3 \\ - 1 / - 3 & 0 / - 3 & 1 / - 3 \end{matrix}) = (\begin{matrix} - 1 / 3 & 0 & 4 / 3 \\ 1 & 1 & - 3 \\ 1 / 3 & 0 & - 1 / 3 \end{matrix})$

The formulas from this subsection are often used for by-hand calculation and are sometimes useful with special types of matrices. However, for generic matrices they are not the best choice because they require more arithmetic than, for instance, the Gauss-Jordan method.

Exercises

Exercise 1.10 Supplied answer

Recommended. Find the cofactor.

$T = (\begin{matrix} 1 & 0 & 2 \\ - 1 & 1 & 3 \\ 0 & 2 & - 1 \end{matrix})$
1. $T_{2, 3}$
2. $T_{3, 2}$
3. $T_{1, 3}$
Back to Exercise 1.10

Answer.
1. $(- 1)^{2 + 3} | \begin{matrix} 1 & 0 \\ 0 & 2 \end{matrix} | = - 2$
2. $(- 1)^{3 + 2} | \begin{matrix} 1 & 2 \\ - 1 & 3 \end{matrix} | = - 5$
3. $(- 1)^{4} | \begin{matrix} - 1 & 1 \\ 0 & 2 \end{matrix} | = - 2$
Exercise 1.11 Supplied answer

Recommended. Find the adjoint to this matrix.

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

Back to Exercise 1.11

Answer.

$adj (T) = (\begin{matrix} | \begin{matrix} 1 & 3 \\ 2 & - 1 \end{matrix} | & - | \begin{matrix} 0 & 2 \\ 2 & - 1 \end{matrix} | & | \begin{matrix} 0 & 2 \\ 1 & 3 \end{matrix} | \\ - | \begin{matrix} - 1 & 3 \\ 0 & - 1 \end{matrix} | & | \begin{matrix} 1 & 2 \\ 0 & - 1 \end{matrix} | & - | \begin{matrix} 1 & 2 \\ - 1 & 3 \end{matrix} | \\ | \begin{matrix} - 1 & 1 \\ 0 & 2 \end{matrix} | & - | \begin{matrix} 1 & 0 \\ 0 & 2 \end{matrix} | & | \begin{matrix} 1 & 0 \\ - 1 & 1 \end{matrix} | \end{matrix}) = (\begin{matrix} - 7 & 4 & - 2 \\ - 1 & - 1 & - 5 \\ - 2 & - 2 & 1 \end{matrix})$
Exercise 1.12 Supplied answer

This determinant is $0$ . Compute that by expanding on the first row.

$| \begin{matrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{matrix} |$

Back to Exercise 1.12

Answer.

$| \begin{matrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{matrix} | = 1 \cdot (+ 1) \cdot | \begin{matrix} 5 & 6 \\ 8 & 9 \end{matrix} | + 2 \cdot (- 1) \cdot | \begin{matrix} 4 & 6 \\ 7 & 9 \end{matrix} | + 3 \cdot (+ 1) \cdot | \begin{matrix} 4 & 5 \\ 7 & 8 \end{matrix} |$

The formula for $2 \times 2$ matrices gives $1 \cdot (- 3) - 2 \cdot (- 6) + 3 \cdot (- 3) = 0$ .
Exercise 1.13 Supplied answer

Recommended. Find the determinant by expanding

$| \begin{matrix} 3 & 0 & 1 \\ 1 & 2 & 2 \\ - 1 & 3 & 0 \end{matrix} |$
1. on the first row
2. on the second row
3. on the third column.
Back to Exercise 1.13

Answer.
1. $3 \cdot (+ 1) | \begin{matrix} 2 & 2 \\ 3 & 0 \end{matrix} | + 0 \cdot (- 1) | \begin{matrix} 1 & 2 \\ - 1 & 0 \end{matrix} | + 1 \cdot (+ 1) | \begin{matrix} 1 & 2 \\ - 1 & 3 \end{matrix} | = - 13$
2. $1 \cdot (- 1) | \begin{matrix} 0 & 1 \\ 3 & 0 \end{matrix} | + 2 \cdot (+ 1) | \begin{matrix} 3 & 1 \\ - 1 & 0 \end{matrix} | + 2 \cdot (- 1) | \begin{matrix} 3 & 0 \\ - 1 & 3 \end{matrix} | = - 13$
3. $1 \cdot (+ 1) | \begin{matrix} 1 & 2 \\ - 1 & 3 \end{matrix} | + 2 \cdot (- 1) | \begin{matrix} 3 & 0 \\ - 1 & 3 \end{matrix} | + 0 \cdot (+ 1) | \begin{matrix} 3 & 0 \\ 1 & 2 \end{matrix} | = - 13$
Exercise 1.14 Supplied answer

Find the adjoint of the matrix in Example 1.5.

Back to Exercise 1.14

Answer. This is $adj (T)$ .

$\begin{aligned} (\begin{array}{c} T_{1, 1} & T_{2, 1} & T_{3, 1} \\ T_{1, 2} & T_{2, 2} & T_{3, 2} \\ T_{1, 3} & T_{2, 3} & T_{3, 3} \end{array}) & = (\begin{array}{c} | \begin{matrix} 5 & 6 \\ 8 & 9 \end{matrix} | & - | \begin{matrix} 2 & 3 \\ 8 & 9 \end{matrix} | & + | \begin{matrix} 2 & 3 \\ 5 & 6 \end{matrix} | \\ - | \begin{matrix} 4 & 6 \\ 7 & 9 \end{matrix} | & + | \begin{matrix} 1 & 3 \\ 7 & 9 \end{matrix} | & - | \begin{matrix} 1 & 3 \\ 4 & 6 \end{matrix} | \\ + | \begin{matrix} 4 & 5 \\ 7 & 8 \end{matrix} | & - | \begin{matrix} 1 & 2 \\ 7 & 8 \end{matrix} | & + | \begin{matrix} 1 & 2 \\ 4 & 5 \end{matrix} | \end{array}) \\ = (\begin{array}{c} - 3 & 6 & - 3 \\ 6 & - 12 & 6 \\ - 3 & 6 & - 3 \end{array}) \end{aligned}$
Exercise 1.15 Supplied answer

Recommended. Find the matrix adjoint to each.
1. $(\begin{matrix} 2 & 1 & 4 \\ - 1 & 0 & 2 \\ 1 & 0 & 1 \end{matrix})$
2. $(\begin{matrix} 3 & - 1 \\ 2 & 4 \end{matrix})$
3. $(\begin{matrix} 1 & 1 \\ 5 & 0 \end{matrix})$
4. $(\begin{matrix} 1 & 4 & 3 \\ - 1 & 0 & 3 \\ 1 & 8 & 9 \end{matrix})$
Back to Exercise 1.15

Answer.
1. This is the adjoint.
  
  $\begin{aligned} (\begin{array}{c} T_{1, 1} & T_{2, 1} & T_{3, 1} \\ T_{1, 2} & T_{2, 2} & T_{3, 2} \\ T_{1, 3} & T_{2, 3} & T_{3, 3} \end{array}) & = (\begin{array}{c} | \begin{matrix} 0 & 2 \\ 0 & 1 \end{matrix} | & - | \begin{matrix} 1 & 4 \\ 0 & 1 \end{matrix} | & | \begin{matrix} 1 & 4 \\ 0 & 2 \end{matrix} | \\ - | \begin{matrix} - 1 & 2 \\ 1 & 1 \end{matrix} | & | \begin{matrix} 2 & 4 \\ 1 & 1 \end{matrix} | & - | \begin{matrix} 2 & 4 \\ - 1 & 2 \end{matrix} | \\ | \begin{matrix} - 1 & 0 \\ 1 & 0 \end{matrix} | & - | \begin{matrix} 2 & 1 \\ 1 & 0 \end{matrix} | & | \begin{matrix} 2 & 1 \\ - 1 & 0 \end{matrix} | \end{array}) \\ = (\begin{array}{c} 0 & - 1 & 2 \\ 3 & - 2 & - 8 \\ 0 & 1 & 1 \end{array}) \end{aligned}$
2. The minors are $1 \times 1$ . $(\begin{matrix} T_{1, 1} & T_{2, 1} \\ T_{1, 2} & T_{2, 2} \end{matrix}) = (\begin{matrix} | \begin{matrix} 4 \end{matrix} | & - | \begin{matrix} - 1 \end{matrix} | \\ - | \begin{matrix} 2 \end{matrix} | & | \begin{matrix} 3 \end{matrix} | \end{matrix}) = (\begin{matrix} 4 & 1 \\ - 2 & 3 \end{matrix})$
3. $(\begin{matrix} 0 & - 1 \\ - 5 & 1 \end{matrix})$
4. The minors are $2 \times 2$ .
  
  $\begin{aligned} (\begin{array}{c} T_{1, 1} & T_{2, 1} & T_{3, 1} \\ T_{1, 2} & T_{2, 2} & T_{3, 2} \\ T_{1, 3} & T_{2, 3} & T_{3, 3} \end{array}) & = (\begin{array}{c} | \begin{matrix} 0 & 3 \\ 8 & 9 \end{matrix} | & - | \begin{matrix} 4 & 3 \\ 8 & 9 \end{matrix} | & | \begin{matrix} 4 & 3 \\ 0 & 3 \end{matrix} | \\ - | \begin{matrix} - 1 & 3 \\ 1 & 9 \end{matrix} | & | \begin{matrix} 1 & 3 \\ 1 & 9 \end{matrix} | & - | \begin{matrix} 1 & 3 \\ - 1 & 3 \end{matrix} | \\ | \begin{matrix} - 1 & 0 \\ 1 & 8 \end{matrix} | & - | \begin{matrix} 1 & 4 \\ 1 & 8 \end{matrix} | & | \begin{matrix} 1 & 4 \\ - 1 & 0 \end{matrix} | \end{array}) \\ = (\begin{array}{c} - 24 & - 12 & 12 \\ 12 & 6 & - 6 \\ - 8 & - 4 & 4 \end{array}) \end{aligned}$
Exercise 1.16 Supplied answer

Recommended. Find the inverse of each matrix in the prior question with Theorem 1.8.
Back to Exercise 1.16

Answer.
1. $(1 / 3) \cdot (\begin{matrix} 0 & - 1 & 2 \\ 3 & - 2 & - 8 \\ 0 & 1 & 1 \end{matrix}) = (\begin{matrix} 0 & - 1 / 3 & 2 / 3 \\ 1 & - 2 / 3 & - 8 / 3 \\ 0 & 1 / 3 & 1 / 3 \end{matrix})$
2. $(1 / 14) \cdot (\begin{matrix} 4 & 1 \\ - 2 & 3 \end{matrix}) = (\begin{matrix} 2 / 7 & 1 / 14 \\ - 1 / 7 & 3 / 14 \end{matrix})$
3. $(1 / - 5) \cdot (\begin{matrix} 0 & - 1 \\ - 5 & 1 \end{matrix}) = (\begin{matrix} 0 & 1 / 5 \\ 1 & - 1 / 5 \end{matrix})$
4. The matrix has a zero determinant, and so has no inverse.
Exercise 1.17 Supplied answer

Find the matrix adjoint to this one.

$(\begin{matrix} 2 & 1 & 0 & 0 \\ 1 & 2 & 1 & 0 \\ 0 & 1 & 2 & 1 \\ 0 & 0 & 1 & 2 \end{matrix})$

Back to Exercise 1.17

Answer. $(\begin{matrix} T_{1, 1} & T_{2, 1} & T_{3, 1} & T_{4, 1} \\ T_{1, 2} & T_{2, 2} & T_{3, 2} & T_{4, 2} \\ T_{1, 3} & T_{2, 3} & T_{3, 3} & T_{4, 3} \\ T_{1, 4} & T_{2, 4} & T_{3, 4} & T_{4, 4} \end{matrix}) = (\begin{matrix} 4 & - 3 & 2 & - 1 \\ - 3 & 6 & - 4 & 2 \\ 2 & - 4 & 6 & - 3 \\ - 1 & 2 & - 3 & 4 \end{matrix})$
Exercise 1.18 Supplied answer

Recommended. Expand across the first row to derive the formula for the determinant of a $2 \times 2$ matrix.

Back to Exercise 1.18

Answer. The determinant

$| \begin{matrix} a & b \\ c & d \end{matrix} |$

expanded on the first row gives $a \cdot (+ 1) | d | + b \cdot (- 1) | c | = a d - b c$ (note the two $1 \times 1$ minors).
Exercise 1.19 Supplied answer

Recommended. Expand across the first row to derive the formula for the determinant of a $3 \times 3$ matrix.

Back to Exercise 1.19

Answer. The determinant of

$(\begin{matrix} a & b & c \\ d & e & f \\ g & h & i \end{matrix})$

is this.

$a \cdot | \begin{matrix} e & f \\ h & i \end{matrix} | - b \cdot | \begin{matrix} d & f \\ g & i \end{matrix} | + c \cdot | \begin{matrix} d & e \\ g & h \end{matrix} | = a (e i - f h) - b (d i - f g) + c (d h - e g)$
Exercise 1.20 Supplied answer

Recommended.
1. Give a formula for the adjoint of a $2 \times 2$ matrix.
2. Use it to derive the formula for the inverse.
Back to Exercise 1.20

Answer.
1. $(\begin{matrix} T_{1, 1} & T_{2, 1} \\ T_{1, 2} & T_{2, 2} \end{matrix}) = (\begin{matrix} | \begin{matrix} t_{2, 2} \end{matrix} | & - | \begin{matrix} t_{1, 2} \end{matrix} | \\ - | \begin{matrix} t_{2, 1} \end{matrix} | & | \begin{matrix} t_{1, 1} \end{matrix} | \end{matrix}) = (\begin{matrix} t_{2, 2} & - t_{1, 2} \\ - t_{2, 1} & t_{1, 1} \end{matrix})$
2. $(1 / t_{1, 1} t_{2, 2} - t_{1, 2} t_{2, 1}) \cdot (\begin{matrix} t_{2, 2} & - t_{1, 2} \\ - t_{2, 1} & t_{1, 1} \end{matrix})$
Exercise 1.21 Supplied answer

Recommended. Can we compute a determinant by expanding down the diagonal?

Back to Exercise 1.21

Answer. No. Here is a determinant whose value

$| \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} | = 1$

doesn’t equal the result of expanding down the diagonal.

$1 \cdot (+ 1) | \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} | + 1 \cdot (+ 1) | \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} | + 1 \cdot (+ 1) | \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} | = 3$
Exercise 1.22 Supplied answer

Give a formula for the adjoint of a diagonal matrix.

Back to Exercise 1.22

Answer. Consider this diagonal matrix.

$D = (\begin{matrix} d_{1} & 0 & 0 & \dots \\ 0 & d_{2} & 0 \\ 0 & 0 & d_{3} \\ ⋱ \\ d_{n} \end{matrix})$

If $i \neq j$ then the $i, j$ minor is an $(n - 1) \times (n - 1)$ matrix with only $n - 2$ nonzero entries, because we have deleted both $d_{i}$ and $d_{j}$ . Thus, at least one row or column of the minor is all zeroes, and so the cofactor $D_{i, j}$ is zero. If $i = j$ then the minor is the diagonal matrix with entries $d_{1}$ , …, $d_{i - 1}$ , $d_{i + 1}$ , …, $d_{n}$ . Its determinant is obviously $(- 1)^{i + j} = (- 1)^{2 i} = 1$ times the product of those.

$adj (D) = (\begin{matrix} d_{2} \dots d_{n} & 0 & 0 \\ 0 & d_{1} d_{3} \dots d_{n} & 0 \\ ⋱ \\ d_{1} \dots d_{n - 1} \end{matrix})$

By the way, Theorem 1.8 provides a slicker way to derive this conclusion.
Exercise 1.23 Supplied answer

Recommended. Prove that the transpose of the adjoint is the adjoint of the transpose.

Back to Exercise 1.23

Answer. Just note that if $S = T^{𝖳}$ then the cofactor $S_{j, i}$ equals the cofactor $T_{i, j}$ because $(- 1)^{j + i} = (- 1)^{i + j}$ and because the minors are the transposes of each other (and the determinant of a transpose equals the determinant of the matrix).
Exercise 1.24 Supplied answer

Prove or disprove: $adj (adj (T)) = T$ .

Back to Exercise 1.24

Answer. It is false; here is an example.

$T = (\begin{matrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{matrix}) adj (T) = (\begin{matrix} - 3 & 6 & - 3 \\ 6 & - 12 & 6 \\ - 3 & 6 & - 3 \end{matrix}) adj (adj (T)) = (\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix})$
Exercise 1.25 Supplied answer

A square matrix is upper triangular if each $i, j$ entry is zero in the part above the diagonal, that is, when $i > j$ .
1. Must the adjoint of an upper triangular matrix be upper triangular? Lower triangular?
2. Prove that the inverse of a upper triangular matrix is upper triangular, if an inverse exists.
Back to Exercise 1.25

Answer.
1. An example
  
  $M = (\begin{matrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{matrix})$
  
  suggests the right answer.
  
  $\begin{aligned} adj (M) = (\begin{array}{c} M_{1, 1} & M_{2, 1} & M_{3, 1} \\ M_{1, 2} & M_{2, 2} & M_{3, 2} \\ M_{1, 3} & M_{2, 3} & M_{3, 3} \end{array}) & = (\begin{array}{c} | \begin{matrix} 4 & 5 \\ 0 & 6 \end{matrix} | & - | \begin{matrix} 2 & 3 \\ 0 & 6 \end{matrix} | & | \begin{matrix} 2 & 3 \\ 4 & 5 \end{matrix} | \\ - | \begin{matrix} 0 & 5 \\ 0 & 6 \end{matrix} | & | \begin{matrix} 1 & 3 \\ 0 & 6 \end{matrix} | & - | \begin{matrix} 1 & 3 \\ 0 & 5 \end{matrix} | \\ | \begin{matrix} 0 & 4 \\ 0 & 0 \end{matrix} | & - | \begin{matrix} 1 & 2 \\ 0 & 0 \end{matrix} | & | \begin{matrix} 1 & 2 \\ 0 & 4 \end{matrix} | \end{array}) \\ = (\begin{array}{c} 24 & - 12 & - 2 \\ 0 & 6 & - 5 \\ 0 & 0 & 4 \end{array}) \end{aligned}$
  
  The result is indeed upper triangular.
  
  This check is detailed but not hard. The entries in the upper triangle of the adjoint are $M_{a, b}$ where $a > b$ . We need to verify that the cofactor $M_{a, b}$ is zero if $a > b$ . With $a > b$ , row $a$ and column $b$ of $M$ ,
  
  $(\begin{matrix} m_{1, 1} & \dots & m_{1, b} & \dots \\ m_{2, 1} & \dots & m_{2, b} \\ ⋮ & ⋮ \\ m_{a, 1} & \dots & m_{a, b} & \dots & m_{a, n} \\ ⋮ & ⋮ \\ m_{n, b} \end{matrix})$
  
  when deleted, leave an upper triangular minor, because entry $i, j$ of the minor is either entry $i, j$ of $M$ (this happens if $a > i$ and $b > j$ ; in this case $i < j$ implies that the entry is zero) or it is entry $i, j + 1$ of $M$ (this happens if $i < a$ and $j > b$ ; in this case, $i < j$ implies that $i < j + 1$ , which implies that the entry is zero), or it is entry $i + 1, j + 1$ of $M$ (this last case happens when $i > a$ and $j > b$ ; obviously here $i < j$ implies that $i + 1 < j + 1$ and so the entry is zero). Thus the determinant of the minor is the product down the diagonal. Observe that the $a - 1, a$ entry of $M$ is the $a - 1, a - 1$ entry of the minor (it doesn’t get deleted because the relation $a > b$ is strict). But this entry is zero because $M$ is upper triangular and $a - 1 < a$ . Therefore the cofactor is zero, and the adjoint is upper triangular. (The lower triangular case is similar.)
2. This is immediate from the prior part, by Theorem 1.8.
Exercise 1.26 Supplied answer

This question requires material from the optional Determinants Exist subsection. Prove Theorem 1.4 by using the permutation expansion.

Back to Exercise 1.26

Answer. We will show that each determinant can be expanded along row $i$ . The argument for column $j$ is similar.

Each term in the permutation expansion contains one and only one entry from each row. As in Example [unresolved original reference: ex:ExpThreeFirstRow], factor out each row $i$ entry to get $| T | = t_{i, 1} \cdot {\hat{T}}_{i, 1} + \dots + t_{i, n} \cdot {\hat{T}}_{i, n}$ , where each ${\hat{T}}_{i, j}$ is a sum of terms not containing any elements of row $i$ . We will show that ${\hat{T}}_{i, j}$ is the $i, j$ cofactor.

Consider the $i, j = n, n$ case first:

$t_{n, n} \cdot {\hat{T}}_{n, n} = t_{n, n} \cdot \sum_{ϕ} t_{1, ϕ (1)} t_{2, ϕ (2)} \dots t_{n - 1, ϕ (n - 1)} sgn (ϕ)$

where the sum is over all $n$ -permutations $ϕ$ such that $ϕ (n) = n$ . To show that ${\hat{T}}_{i, j}$ is the minor $T_{i, j}$ , we need only show that if $ϕ$ is an $n$ -permutation such that $ϕ (n) = n$ and $σ$ is an $n - 1$ -permutation with $σ (1) = ϕ (1)$ , …, $σ (n - 1) = ϕ (n - 1)$ then $sgn (σ) = sgn (ϕ)$ . But that’s true because $ϕ$ and $σ$ have the same number of inversions.

Back to the general $i, j$ case. Swap adjacent rows until the $i$ -th is last and swap adjacent columns until the $j$ -th is last. Observe that the determinant of the $i, j$ -th minor is not affected by these adjacent swaps because inversions are preserved (since the minor has the $i$ -th row and $j$ -th column omitted). On the other hand, the sign of $| T |$ and ${\hat{T}}_{i, j}$ changes $n - i$ plus $n - j$ times. Thus ${\hat{T}}_{i, j} = (- 1)^{n - i + n - j} | T_{i, j} | = (- 1)^{i + j} | T_{i, j} |$ .
Exercise 1.27 Supplied answer

Prove that the determinant of a matrix equals the determinant of its transpose using Laplace’s expansion and induction on the size of the matrix.

Back to Exercise 1.27

Answer. This is obvious for the $1 \times 1$ base case.

For the inductive case, assume that the determinant of a matrix equals the determinant of its transpose for all $1 \times 1$ , …, $(n - 1) \times (n - 1)$ matrices. Expanding on row $i$ gives $| T | = t_{i, 1} T_{i, 1} + \dots + t_{i, n} T_{i, n}$ and expanding on column $i$ gives $| T^{𝖳} | = t_{1, i} (T^{𝖳})_{1, i} + \dots + t_{n, i} (T^{𝖳})_{n, i}$ Since $(- 1)^{i + j} = (- 1)^{j + i}$ the signs are the same in the two summations. Since the $j, i$ minor of $T^{𝖳}$ is the transpose of the $i, j$ minor of $T$ , the inductive hypothesis gives $| (T^{𝖳})_{i, j} | = | T_{i, j} |$ .
Exercise 1.28 Supplied answer

Puzzle. Show that

$F_{n} = | \begin{matrix} 1 & - 1 & 1 & - 1 & 1 & - 1 & \dots \\ 1 & 1 & 0 & 1 & 0 & 1 & \dots \\ 0 & 1 & 1 & 0 & 1 & 0 & \dots \\ 0 & 0 & 1 & 1 & 0 & 1 & \dots \\ . & . & . & . & . & . & \dots \end{matrix} |$

where $F_{n}$ is the $n$ -th term of $1, 1, 2, 3, 5, \dots, x, y, x + y, \dots$ , the Fibonacci sequence, and the determinant is of order $n - 1$ . [Am. Math. Mon., Jun. 1949]

Back to Exercise 1.28

Answer. This is how the answer was given in the cited source. Denoting the above determinant by $D_{n}$ , it is seen that $D_{2} = 1$ , $D_{3} = 2$ . It remains to show that $D_{n} = D_{n - 1} + D_{n - 2}, n \geq 4$ . In $D_{n}$ subtract the $(n - 3)$ -th column from the $(n - 1)$ -th, the $(n - 4)$ -th from the $(n - 2)$ -th, …, the first from the third, obtaining

$F_{n} = | \begin{matrix} 1 & - 1 & 0 & 0 & 0 & 0 & \dots \\ 1 & 1 & - 1 & 0 & 0 & 0 & \dots \\ 0 & 1 & 1 & - 1 & 0 & 0 & \dots \\ 0 & 0 & 1 & 1 & - 1 & 0 & \dots \\ . & . & . & . & . & . & \dots \end{matrix} | .$

By expanding this determinant with reference to the first row, there results the desired relation.

References cited in this section

Am. Math. Mon., Jun. 1949

Don Walter (proposer), Alex Tytun (solver), Elementary problem 834, American Mathematical Monthly, vol. 56 no. 6 (June-July 1949), p. 409.