We are given a 3x3 matrix $A$ and asked to find all the minors $|A_{ij}|$ of the matrix. The given matrix $A$ is $A = \begin{bmatrix} 3 & 1 & 0 \\ 2 & 4 & -2 \\ -3 & 5 & 2 \end{bmatrix}$.

1. Problem Description

We are given a 3x3 matrix $A$ and asked to find all the minors $|A_{ij}|$ of the matrix. The given matrix $A$ is

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

2. Solution Steps

The minor $|A_{ij}|$ is the determinant of the matrix obtained by deleting the $i$-th row and $j$-th column of $A$.

We need to compute all nine minors of $A$:

$|A_{11}|$, $|A_{12}|$, $|A_{13}|$, $|A_{21}|$, $|A_{22}|$, $|A_{23}|$, $|A_{31}|$, $|A_{32}|$, $|A_{33}|$.

$|A_{11}| = \begin{vmatrix} 4 & -2 \\ 5 & 2 \end{vmatrix} = (4)(2) - (-2)(5) = 8 + 10 = 18$

$|A_{12}| = \begin{vmatrix} 2 & -2 \\ -3 & 2 \end{vmatrix} = (2)(2) - (-2)(-3) = 4 - 6 = -2$

$|A_{13}| = \begin{vmatrix} 2 & 4 \\ -3 & 5 \end{vmatrix} = (2)(5) - (4)(-3) = 10 + 12 = 22$

$|A_{21}| = \begin{vmatrix} 1 & 0 \\ 5 & 2 \end{vmatrix} = (1)(2) - (0)(5) = 2 - 0 = 2$

$|A_{22}| = \begin{vmatrix} 3 & 0 \\ -3 & 2 \end{vmatrix} = (3)(2) - (0)(-3) = 6 - 0 = 6$

$|A_{23}| = \begin{vmatrix} 3 & 1 \\ -3 & 5 \end{vmatrix} = (3)(5) - (1)(-3) = 15 + 3 = 18$

$|A_{31}| = \begin{vmatrix} 1 & 0 \\ 4 & -2 \end{vmatrix} = (1)(-2) - (0)(4) = -2 - 0 = -2$

$|A_{32}| = \begin{vmatrix} 3 & 0 \\ 2 & -2 \end{vmatrix} = (3)(-2) - (0)(2) = -6 - 0 = -6$

$|A_{33}| = \begin{vmatrix} 3 & 1 \\ 2 & 4 \end{vmatrix} = (3)(4) - (1)(2) = 12 - 2 = 10$

3. Final Answer

$|A_{11}| = 18$

$|A_{12}| = -2$

$|A_{13}| = 22$

$|A_{21}| = 2$

$|A_{22}| = 6$

$|A_{23}| = 18$

$|A_{31}| = -2$

$|A_{32}| = -6$

$|A_{33}| = 10$