Let $A = [a_{ij}]$ is given by $A = \begin{bmatrix} 1 & -1 & 2 \\ 3 & 4 & -5 \\ 2 & -1 & 3 \end{bmatrix}$. Then the matrix $B = [b_{ij}]$, where $b_{ij}$ = Minor of $a_{ij}$ is:

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

The minor of an element $a_{ij}$ is the determinant of the matrix obtained by deleting row $i$ and column $j$.

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For $b_{11}$, delete row 1 and column 1:

$\begin{bmatrix} 4 & -5 \\ -1 & 3 \end{bmatrix}$

$b_{11} = (4)(3) - (-5)(-1)$

$b_{11} = 12 - 5$

$b_{11} = 7$

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For $b_{12}$, delete row 1 and column 2:

$\begin{bmatrix} 3 & -5 \\ 2 & 3 \end{bmatrix}$

$b_{12} = (3)(3) - (-5)(2)$

$b_{12} = 9 + 10$

$b_{12} = 19$

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For $b_{13}$, delete row 1 and column 3:

$\begin{bmatrix} 3 & 4 \\ 2 & -1 \end{bmatrix}$

$b_{13} = (3)(-1) - (4)(2)$

$b_{13} = -3 - 8$

$b_{13} = -11$

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For $b_{21}$, delete row 2 and column 1:

$\begin{bmatrix} -1 & 2 \\ -1 & 3 \end{bmatrix}$

$b_{21} = (-1)(3) - (2)(-1)$

$b_{21} = -3 + 2$

$b_{21} = -1$

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For $b_{22}$, delete row 2 and column 2:

$\begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix}$

$b_{22} = (1)(3) - (2)(2)$

$b_{22} = 3 - 4$

$b_{22} = -1$

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For $b_{23}$, delete row 2 and column 3:

$\begin{bmatrix} 1 & -1 \\ 2 & -1 \end{bmatrix}$

$b_{23} = (1)(-1) - (-1)(2)$

$b_{23} = -1 + 2$

$b_{23} = 1$

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For $b_{31}$, delete row 3 and column 1:

$\begin{bmatrix} -1 & 2 \\ 4 & -5 \end{bmatrix}$

$b_{31} = (-1)(-5) - (2)(4)$

$b_{31} = 5 - 8$

$b_{31} = -3$

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For $b_{32}$, delete row 3 and column 2:

$\begin{bmatrix} 1 & 2 \\ 3 & -5 \end{bmatrix}$

$b_{32} = (1)(-5) - (2)(3)$

$b_{32} = -5 - 6$

$b_{32} = -11$

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For $b_{33}$, delete row 3 and column 3:

$\begin{bmatrix} 1 & -1 \\ 3 & 4 \end{bmatrix}$

$b_{33} = (1)(4) - (-1)(3)$

$b_{33} = 4 + 3$

$b_{33} = 7$

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Therefore, the matrix of minors is:

$B = \begin{bmatrix} 7 & 19 & -11 \\ -1 & -1 & 1 \\ -3 & -11 & 7 \end{bmatrix}$