Answer: (a)(i) \(\begin{pmatrix}-12&7\\11&7\end{pmatrix}\), (a)(ii) \(\begin{pmatrix}17&9\\14&18\\-3&-1\end{pmatrix}\), (b)(i) \(X^{-1}=\frac1{10}\begin{pmatrix}-2&3\\-4&1\end{pmatrix}\), (b)(ii) \(Y=\begin{pmatrix}3.5&8\\-0.5&6\end{pmatrix}\).
(a)(i) First find
\(3C=3\begin{pmatrix}-5&2\\3&1\end{pmatrix} = \begin{pmatrix}-15&6\\9&3\end{pmatrix}.\)
So
\(A+3C = \begin{pmatrix}3&1\\2&4\end{pmatrix} + \begin{pmatrix}-15&6\\9&3\end{pmatrix} = \begin{pmatrix}-12&7\\11&7\end{pmatrix}.\)
(a)(ii) Multiply \(B\) by \(A\):
\(BA= \begin{pmatrix}5&1\\2&4\\-1&0\end{pmatrix} \begin{pmatrix}3&1\\2&4\end{pmatrix}.\)
Row by column gives
\(BA= \begin{pmatrix} 5\cdot3+1\cdot2 & 5\cdot1+1\cdot4\\ 2\cdot3+4\cdot2 & 2\cdot1+4\cdot4\\ -1\cdot3+0\cdot2 & -1\cdot1+0\cdot4 \end{pmatrix} = \begin{pmatrix}17&9\\14&18\\-3&-1\end{pmatrix}.\)
(b)(i) For
\(X=\begin{pmatrix}1&-3\\4&-2\end{pmatrix},\)
the determinant is
\((1)(-2)-(-3)(4)=-2+12=10.\)
Therefore
\(X^{-1} = \frac1{10} \begin{pmatrix}-2&3\\-4&1\end{pmatrix}.\)
(b)(ii) Since \(XY=M\), where
\(M=\begin{pmatrix}5&-10\\15&20\end{pmatrix},\)
premultiply by \(X^{-1}\):
\(Y=X^{-1}M = \frac1{10} \begin{pmatrix}-2&3\\-4&1\end{pmatrix} \begin{pmatrix}5&-10\\15&20\end{pmatrix}.\)
Compute the product:
\(\begin{pmatrix}-2&3\\-4&1\end{pmatrix} \begin{pmatrix}5&-10\\15&20\end{pmatrix} = \begin{pmatrix}35&80\\-5&60\end{pmatrix}.\)
So
\(Y= \frac1{10} \begin{pmatrix}35&80\\-5&60\end{pmatrix} = \begin{pmatrix}3.5&8\\-0.5&6\end{pmatrix}.\)
