Explanation

(d)
Let, $A = \begin{bmatrix}
1 & - 4 \\
3 & - 2
\end{bmatrix}$ and $B = \begin{bmatrix}
- 16 & - 6 \\
7 & 2
\end{bmatrix}$
Then the matrix equation is $AX = B.$
$\therefore|A| = \begin{bmatrix}
1 & - 4 \\
3 & - 2
\end{bmatrix} = - 2 + 12 \neq 0$
So A is and invertible matrix. Also,
adj $A = \begin{bmatrix}
- 2 & - 3 \\
4 & 1
\end{bmatrix}^{T} = \begin{bmatrix}
- 2 & 4 \\
- 3 & 1
\end{bmatrix}$
So,
$A^{- 1} = \frac{1}{|A|}$adj $A = \frac{1}{10}\begin{bmatrix}
- 2 & 4 \\
- 3 & 1
\end{bmatrix}$
Now,
$AX = B$
$\Rightarrow A^{- 1}(AX) = a^{- 1}B$
$\Rightarrow \left( A^{- 1}A \right)X = A^{- 1}B$
$\Rightarrow IX = A^{- 1}B$
$\Rightarrow X = A^{1}B$
$\Rightarrow X = \frac{1}{10}\begin{bmatrix}
- 2 & 4 \\
- 3 & 1
\end{bmatrix}\begin{bmatrix}
- 16 & - 6 \\
7 & 2
\end{bmatrix} = \begin{bmatrix}
6 & 2 \\
\frac{11}{2} & 2
\end{bmatrix}$