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:

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

The correct answer is Option (1) → $\begin{bmatrix}7&19&-11\\-1&-1&1\\-3&-11&7\end{bmatrix}$

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

Compute minors $b_{ij}$:

$b_{11} = \begin{vmatrix} 4 & -5 \\ -1 & 3 \end{vmatrix} = 4 \cdot 3 - (-5) \cdot (-1) = 12 - 5 = 7$
$b_{12} = \begin{vmatrix} 3 & -5 \\ 2 & 3 \end{vmatrix} = 3 \cdot 3 - (-5) \cdot 2 = 9 + 10 = 19$
$b_{13} = \begin{vmatrix} 3 & 4 \\ 2 & -1 \end{vmatrix} = 3 \cdot (-1) - 4 \cdot 2 = -3 - 8 = -11$

$b_{21} = \begin{vmatrix} -1 & 2 \\ -1 & 3 \end{vmatrix} = (-1) \cdot 3 - 2 \cdot (-1) = -3 + 2 = -1$
$b_{22} = \begin{vmatrix} 1 & 2 \\ 2 & 3 \end{vmatrix} = 1 \cdot 3 - 2 \cdot 2 = 3 - 4 = -1$
$b_{23} = \begin{vmatrix} 1 & -1 \\ 2 & -1 \end{vmatrix} = 1 \cdot (-1) - (-1) \cdot 2 = -1 + 2 = 1$

$b_{31} = \begin{vmatrix} -1 & 2 \\ 4 & -5 \end{vmatrix} = (-1) \cdot (-5) - 2 \cdot 4 = 5 - 8 = -3$
$b_{32} = \begin{vmatrix} 1 & 2 \\ 3 & -5 \end{vmatrix} = 1 \cdot (-5) - 2 \cdot 3 = -5 - 6 = -11$
$b_{33} = \begin{vmatrix} 1 & -1 \\ 3 & 4 \end{vmatrix} = 1 \cdot 4 - (-1) \cdot 3 = 4 + 3 = 7$

So, matrix of minors is:
$B = \begin{bmatrix} 7 & 19 & -11 \\ -1 & -1 & 1 \\ -3 & -11 & 7 \end{bmatrix}$