Answer: (a) \((BA)^{-1}=\dfrac1{-30}\begin{pmatrix}12&-6\\-1&-2\end{pmatrix}\). (b)(i) \(2\times3\). (b)(ii) \(X=\begin{pmatrix}2&-\frac12\end{pmatrix}\).

(a) First calculate \(BA\):

\(BA= \begin{pmatrix}0&-2\\3&-5\end{pmatrix} \begin{pmatrix}2&-1\\1&-3\end{pmatrix}.\)

This gives

\(BA= \begin{pmatrix} -2&6\\ 1&12 \end{pmatrix}.\)

The determinant is

\((-2)(12)-6(1)=-30.\)

Therefore

\((BA)^{-1} =\frac1{-30} \begin{pmatrix} 12&-6\\ -1&-2 \end{pmatrix}.\)

(b)(i) The matrix \(C\) has \(2\) rows and \(3\) columns, so its order is

\(2\times3.\)

(b)(ii) Since \(XC=D\), and \(C\) is \(2\times3\) while \(D\) is \(1\times3\), the matrix \(X\) must be \(1\times2\).

Let

\(X=\begin{pmatrix}a&b\end{pmatrix}.\)

Then

\(\begin{pmatrix}a&b\end{pmatrix} \begin{pmatrix}-2&5&3\\0&10&4\end{pmatrix} = \begin{pmatrix}-2a&5a+10b&3a+4b\end{pmatrix}.\)

Equating this with \(\begin{pmatrix}-4&5&4\end{pmatrix}\),

\(-2a=-4, \qquad 5a+10b=5, \qquad 3a+4b=4.\)

From \(-2a=-4\), \(a=2\). Then

\(5(2)+10b=5, \qquad 10b=-5, \qquad b=-\frac12.\)

This also satisfies \(3a+4b=4\), since \(6-2=4\). Hence

\(X=\begin{pmatrix}2&-\frac12\end{pmatrix}.\)