Answer to Question #200602 in Linear Algebra for Mpopo

Question

Evaluate det(−A) and det(−A T ). Compare det(−A) and det(−A T ) for:

(2.1) A = -4 2

3 -3

(2.2) A = 3 1 -2

-5 3 -6

-1 0 -4

Answer 1

The determinant of the transpose of a square matrix is equal to the determinant of the matrix, that is,

$$\det(A^T)=\det (A)$$

2.1

$$A=\begin{bmatrix} -4 & 2 \\ 3 & -3 \end{bmatrix}$$

$$-A=\begin{bmatrix} 4 & -2 \\ -3 & 3 \end{bmatrix}$$

$$\det (-A)=\begin{vmatrix} 4 & -2 \\ -3 & 3 \end{vmatrix}=4(3)-(-2)(-3)=18$$

$$A^T=\begin{bmatrix} -4 & 2 \\ 3 & -3 \end{bmatrix}^T=\begin{bmatrix} -4 & 3 \\ 2 & -3 \end{bmatrix}$$

$$-A^T=\begin{bmatrix} 4 & -3 \\ -2 & 3 \end{bmatrix}$$

$$\det (-A^T)=\begin{vmatrix} 4 & -3 \\ -2 & 3 \end{vmatrix}=4(3)-(-2)(-3)=18$$

$$\det (-A)=\det (-A^T)$$

2.2

$$A=\begin{bmatrix} 3 & 1 & -2 \\ -5 & 3 & -6 \\ -1 & 0 & -4 \end{bmatrix}$$

$$-A=\begin{bmatrix} -3 & -1 & 2 \\ 5 & -3 & 6 \\ 1 & 0 & 4 \end{bmatrix}$$

$$\det (-A)=\begin{vmatrix} -3 & -1 & 2 \\ 5 & -3 & 6 \\ 1 & 0 & 4 \end{vmatrix}$$

$$=1\begin{vmatrix} -1& 2 \\ -3 & 6 \end{vmatrix}-0\begin{vmatrix} -3 & 2 \\ 5 & 6 \end{vmatrix}+4\begin{vmatrix} -3 & -1 \\ 5 & -3 \end{vmatrix}$$

$$=-6+6+4(9+5)=56$$

$$A^T=\begin{bmatrix} 3 & 1 & -2 \\ -5 & 3 & -6 \\ -1 & 0 & -4 \end{bmatrix}^T=\begin{bmatrix} 3 &- 5 & -1 \\ 1 & 3 & 0 \\ -2 & -6 & -4 \end{bmatrix}$$

$$-A^T=\begin{bmatrix} -3 & 5 & 1 \\ -1 & -3 & 0 \\ 2 & 6 & 4 \end{bmatrix}$$

$$\det (-A^T)=\begin{vmatrix} -3 & 5 & 1 \\ -1 & -3 & 0 \\ 2 & 6 & 4 \end{vmatrix}$$

$$=1\begin{vmatrix} -1& -3 \\ 2 & 6 \end{vmatrix}-0\begin{vmatrix} -3 & 5 \\ 2 & 6 \end{vmatrix}+4\begin{vmatrix} -3 &5 \\ -1 & -3 \end{vmatrix}$$

$$=-6+6-9+4(9+5)=56$$

$$\det (-A)=\det (-A^T)$$