If $A=\begin{bmatrix}2&3\\1&-4\end{bmatrix}$ and $B =\begin{bmatrix}1&-2\\-1&3\end{bmatrix}$, then $B^{-1} A^{-1}$ is equal to:

$\frac{1}{11}\begin{bmatrix}14&5\\5&1\end{bmatrix}$

The correct answer is Option (3) → $\frac{1}{11}=\begin{bmatrix}14&5\\5&1\end{bmatrix}$

$A = \begin{bmatrix}2 & 3 \\ 1 & -4\end{bmatrix}, \quad B = \begin{bmatrix}1 & -2 \\ -1 & 3\end{bmatrix}$

$B^{-1}A^{-1} = (AB)^{-1}$

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

$= \begin{bmatrix}2(1)+3(-1) & 2(-2)+3(3) \\ 1(1)+(-4)(-1) & 1(-2)+(-4)(3)\end{bmatrix}$

$= \begin{bmatrix}-1 & 5 \\ 5 & -14\end{bmatrix}$

$\det(AB) = (-1)(-14) - (5)(5) = 14 - 25 = -11$

$(AB)^{-1} = \frac{1}{-11}\begin{bmatrix}-14 & -5 \\ -5 & -1\end{bmatrix}$

$= \frac{1}{11}\begin{bmatrix}14 & 5 \\ 5 & 1\end{bmatrix}$

$B^{-1}A^{-1} = \frac{1}{11}\begin{bmatrix}14 & 5 \\ 5 & 1\end{bmatrix}$