If \(A=\left[\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right], B=\left[\begin{array}{ll}1 & 0 \\ 3 & 1\end{array}\right]\), then \(\mathrm{B}^{-1} \mathrm{~A}^{-1}=\)

(A)
We have \(A=\left[\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right] \Rightarrow|A|=4-3=1 \Rightarrow A^{-1}\) exists.
\(B=\left[\begin{array}{ll}
1 & 0 \\
3 & 1
\end{array}\right] \Rightarrow|B|=1 \Rightarrow B^{-1} \text { exists }\)
\(\begin{aligned} A^{-1} &=\left[\begin{array}{cc}2 & -3 \\ -1 & 2\end{array}\right] \text { and } B^{-1}=\left[\begin{array}{cc}1 & 0 \\ -3 & 1\end{array}\right] \\ \therefore B^{-1} A^{-1} &=\left[\begin{array}{cc}1 & 0 \\ -3 & 1\end{array}\right]\left[\begin{array}{cc}2 & -3 \\ -1 & 2\end{array}\right] \\ &=\left[\begin{array}{cc}2+0 & -3+0 \\ -6-1 & 9+2\end{array}\right]=\left[\begin{array}{cc}2 & -3 \\ -7 & 11\end{array}\right] \end{aligned}\)