Explanation

(b)
Since the product matrix is $3 \times 3$ matrix and the pre-multi-plier of A is a $3 \times 2$ matrix, therefore A is $2 \times 3$ matrix. Let,
$A = \begin{bmatrix}
l & m & n \\
x & y & \mathcal{z}
\end{bmatrix}$. Then the given equation becomes
$\begin{bmatrix}
2 & - 1 \\
1 & 0 \\
- 3 & 4
\end{bmatrix}\begin{bmatrix}
l & m & n \\
x & y & \mathcal{z}
\end{bmatrix} = \begin{bmatrix}
- 1 & - 8 & - 10 \\
1 & - 2 & - 5 \\
9 & 22 & 15
\end{bmatrix}$
$\Rightarrow \begin{bmatrix}
2l - x & 2m - y & 2n - \mathcal{z} \\
l & m & x \\
- 3l + 4x & - 3m + 4y & - 3n + 4\mathcal{z}
\end{bmatrix} = \begin{bmatrix}
- 1 & - 8 & - 10 \\
1 & - 2 & - 5 \\
9 & 22 & 15
\end{bmatrix}$
$\Rightarrow 2l - x = - 1,\ 2m - y = - 8,\ 2n - \mathcal{z} = - 10,\ l = 1,\ m = - 2,\ n = - 5$
$\Rightarrow x = 3,\ y = 4,\ \mathcal{z} = 0,\ l = 1,\ m = - 2,\ n = - 5$
$\Rightarrow A = \begin{bmatrix}
l & m & n \\
x & y & \mathcal{z}
\end{bmatrix} = \begin{bmatrix}
1 & - 2 & - 5 \\
3 & 4 & 0
\end{bmatrix}$