2.1 Determinants by Cofactor Expansion: Class Notes

Recall

$A = [\begin{matrix} a & b \\ c & d \end{matrix}]$ $\Rightarrow$ $A^{- 1} = \frac{1}{a d - b c} [\begin{matrix} d & - b \\ - c & a \end{matrix}]$

The number $a d - b c$ is the determinant of the $2 \times 2$ matrix $A$ . This section defines the determinant for larger square matrices.

Definition

If $A$ is a square matrix, then the minor of entry a_ij is denoted by $M_{i j}$ and is defined to be the determinant of the submatrix that remains after the $i$ th row and $j$ th column are deleted from $A$ . The number $(- 1)^{i + j} M_{i j}$ is denoted by $C_{i j}$ and is called the cofactor of entry a_ij.

In short, for a matrix $A = [a_{i j}]$ :

$C_{i j} = (- 1)^{i + j} M_{i j}$

Example

Find all the minors and cofactors of $A = [\begin{matrix} 2 & 3 & - 7 \\ 4 & - 2 & - 9 \\ 2 & 5 & 6 \end{matrix}]$ .

Solution

Notation: square brackets mean a matrix, as in $A = [\begin{matrix} 2 & 3 & - 7 \\ 4 & - 2 & - 9 \\ 2 & 5 & 6 \end{matrix}]$ . Vertical bars mean a determinant (a number):

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

The sign $(- 1)^{i + j}$ depends only on the position: $(- 1)^{1 + 1} = 1$ , $(- 1)^{1 + 2} = - 1$ , and so on, giving the pattern

$[\begin{matrix} + & - & + \\ - & + & - \\ + & - & + \end{matrix}]$

So in a + position, the cofactor equals the minor, and in a − position, the cofactor is the negative of the minor.

Minors and cofactors. Each minor is a $2 \times 2$ determinant. For example, deleting row 1 and column 1:

$M_{11} = | \begin{matrix} - 2 & - 9 \\ 5 & 6 \end{matrix} | = (- 2) (6) - (- 9) (5) = - 12 - (- 45) = 33$

The rest are found the same way.

$M_{11} = | \begin{matrix} - 2 & - 9 \\ 5 & 6 \end{matrix} | = 33, C_{11} = 33$

$M_{12} = | \begin{matrix} 4 & - 9 \\ 2 & 6 \end{matrix} | = 42, C_{12} = - 42$

$M_{13} = | \begin{matrix} 4 & - 2 \\ 2 & 5 \end{matrix} | = 24, C_{13} = 24$

$M_{21} = | \begin{matrix} 3 & - 7 \\ 5 & 6 \end{matrix} | = 53, C_{21} = - 53$

$M_{22} = | \begin{matrix} 2 & - 7 \\ 2 & 6 \end{matrix} | = 26, C_{22} = 26$

$M_{23} = | \begin{matrix} 2 & 3 \\ 2 & 5 \end{matrix} | = 4, C_{23} = - 4$

$M_{31} = | \begin{matrix} 3 & - 7 \\ - 2 & - 9 \end{matrix} | = - 41, C_{31} = - 41$

$M_{32} = | \begin{matrix} 2 & - 7 \\ 4 & - 9 \end{matrix} | = 10, C_{32} = - 10$

$M_{33} = | \begin{matrix} 2 & 3 \\ 4 & - 2 \end{matrix} | = - 16, C_{33} = - 16$

Definition

If $A$ is an $n \times n$ matrix, then the number obtained by multiplying the entries in any row or column of $A$ by the corresponding cofactors and adding the resulting products is called the determinant of A, and the sums themselves are called cofactor expansions of A. That is,

$\begin{matrix} det (A) & = a_{1 j} C_{1 j} + a_{2 j} C_{2 j} + \dots + a_{n j} C_{n j} (cofactor expansion along the j th column) \\ = a_{i 1} C_{i 1} + a_{i 2} C_{i 2} + \dots + a_{i n} C_{i n} (cofactor expansion along the i th row) \end{matrix}$

Example

Compute the determinant of the matrix $A$ above.

Solution

Tip: Use the right symbol: the matrix is $A = [\begin{matrix} 2 & 3 & - 7 \\ 4 & - 2 & - 9 \\ 2 & 5 & 6 \end{matrix}]$ , and its determinant is $det (A) = | A | = | \begin{matrix} 2 & 3 & - 7 \\ 4 & - 2 & - 9 \\ 2 & 5 & 6 \end{matrix} |$ . Writing brackets for a determinant (or bars for a matrix) is incorrect.

Expand along the first row, using the cofactors found above:

$\begin{matrix} det (A) & = | \begin{matrix} 2 & 3 & - 7 \\ 4 & - 2 & - 9 \\ 2 & 5 & 6 \end{matrix} | \\ = 2 (33) + 3 (- 42) + (- 7) (24) \\ = 2 (33) - 3 (42) - 7 (24) \\ = 66 - 126 - 168 \end{matrix}$

Answer: $det (A) = - 228$

#11

Use the arrow technique of Figure 2.1.1 to evaluate the determinant.

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

Solution

Copy the first two columns to the right of the determinant. Multiply along the three down-right diagonals and add those products; multiply along the three up-right diagonals and subtract those products.

Arrow technique: the 3 by 3 entries with the first two columns repeated and diagonal arrows

The rows $- 2, 1, 4, - 2, 1$ ; $3, 5, - 7, 3, 5$ ; $1, 6, 2, 1, 6$ (the last two columns repeat the first two). Down-right diagonals: $(- 2) (5) (2) = - 20$ , $(1) (- 7) (1) = - 7$ , $(4) (3) (6) = 72$ . Up-right diagonals: $(1) (5) (4) = 20$ , $(6) (- 7) (- 2) = 84$ , $(2) (3) (1) = 6$ .

$(- 20 - 7 + 72) - (20 + 84 + 6) = 45 - 110$

Answer: $- 65$

Arrow method for 2 × 2 determinants

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

Multiply along the down-right diagonal and subtract the product along the up-right diagonal.

Tip: The arrow technique works only for $2 \times 2$ and $3 \times 3$ determinants; it does not work for $4 \times 4$ or larger.

Theorem 2.1.1

If $A$ is an $n \times n$ matrix, then regardless of which row or column of $A$ is chosen, the number obtained by multiplying the entries in that row or column by the corresponding cofactors and adding the resulting products is always the same.

#16

Find all values of $λ$ for which $det (A) = 0$ .

$A = [\begin{matrix} λ - 4 & 0 & 0 \\ 0 & λ & 2 \\ 0 & 3 & λ - 1 \end{matrix}]$

Solution

Expand across row 1: its last two entries are 0, so only the first term remains.

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

$\begin{matrix} (λ - 4) (λ^{2} - λ - 6) & = 0 \\ (λ - 4) (λ - 3) (λ + 2) & = 0 \end{matrix}$

Answer: $λ = - 2, 3, 4$

Theorem 2.1.2

If $A$ is an $n \times n$ triangular matrix (upper triangular, lower triangular, or diagonal), then $det (A)$ is the product of the entries on the main diagonal of the matrix; that is, $det (A) = a_{11} a_{22} \dots a_{n n}$ .

#31

Evaluate the determinant of the given matrix by inspection.

$[\begin{matrix} 1 & 2 & 7 & - 3 \\ 0 & 1 & - 4 & 1 \\ 0 & 0 & 2 & 7 \\ 0 & 0 & 0 & 3 \end{matrix}]$

Solution

The matrix is upper triangular, so by Theorem 2.1.2 its determinant is the product of the diagonal entries:

Answer: $det (A) = 1 (1) (2) (3) = 6$

Why this works. Expand along the first column again and again; each time, only the top entry is nonzero:

$\begin{matrix} | \begin{matrix} 1 & 2 & 7 & - 3 \\ 0 & 1 & - 4 & 1 \\ 0 & 0 & 2 & 7 \\ 0 & 0 & 0 & 3 \end{matrix} | & = 1 | \begin{matrix} 1 & - 4 & 1 \\ 0 & 2 & 7 \\ 0 & 0 & 3 \end{matrix} | \\ = 1 (1 | \begin{matrix} 2 & 7 \\ 0 & 3 \end{matrix} |) \\ = 1 (1) (2) (3) \end{matrix}$

Tip: There is no need to write all this out each time; it just shows why Theorem 2.1.2 is true.

Aside

Let $A = [\begin{matrix} 2 & 1 \\ - 3 & 5 \end{matrix}]$ and $B = [\begin{matrix} 4 & 8 & 7 \\ 5 & - 3 & 2 \\ 8 & 12 & 17 \end{matrix}]$ .

$A B = [\begin{matrix} 2 & 1 \\ - 3 & 5 \end{matrix}] [\begin{matrix} 4 & 8 & 7 \\ 5 & - 3 & 2 \\ 8 & 12 & 17 \end{matrix}]$

is not defined (the sizes don’t match), but

$det (A) det (B) = | \begin{matrix} 2 & 1 \\ - 3 & 5 \end{matrix} | | \begin{matrix} 4 & 8 & 7 \\ 5 & - 3 & 2 \\ 8 & 12 & 17 \end{matrix} |$

is defined, since each determinant is just a number.