Answer: \(a=-6\) or \(a=2\); \(A^{-1}=\dfrac1{20}\begin{pmatrix}8&-3\\-4&4\end{pmatrix}\); \(B=\dfrac1{20}\begin{pmatrix}4&39\\8&-32\end{pmatrix}\).

The inverse does not exist when the determinant is zero:

\(\det A=a(a+4)-12.\)

So

\(a^2+4a-12=0.\)

Factorising,

\((a+6)(a-2)=0.\)

Hence \(A^{-1}\) does not exist when

\(a=-6\quad\text{or}\quad a=2.\)

When \(a=4\),

\(A=\begin{pmatrix}4&3\\4&8\end{pmatrix}.\)

The determinant is \(4(8)-3(4)=20\), so

\(A^{-1}=\frac1{20}\begin{pmatrix}8&-3\\-4&4\end{pmatrix}.\)

Since \(AB=\begin{pmatrix}2&3\\4&-5\end{pmatrix}\),

\(B=A^{-1}\begin{pmatrix}2&3\\4&-5\end{pmatrix}.\)

Therefore

\(B=\frac1{20} \begin{pmatrix}8&-3\\-4&4\end{pmatrix} \begin{pmatrix}2&3\\4&-5\end{pmatrix} =\frac1{20}\begin{pmatrix}4&39\\8&-32\end{pmatrix}.\)