Answer: (i) \(A^{-1}=\dfrac15\begin{pmatrix}4&-3\\-1&2\end{pmatrix}\). (ii) \(C=\dfrac15\begin{pmatrix}0&5\\-13&16\end{pmatrix}\). (iii) \(D=\begin{pmatrix}6&-20\\8&-20\end{pmatrix}\).

(i) For

\(A=\begin{pmatrix}2&3\\1&4\end{pmatrix},\)

the determinant is

\(2\cdot4-3\cdot1=5.\)

Hence

\(A^{-1} = \frac15 \begin{pmatrix}4&-3\\-1&2\end{pmatrix}.\)

(ii) Since \(CA=B\), post-multiply by \(A^{-1}\):

\(C=BA^{-1}.\)

So

\(C= \begin{pmatrix}1&4\\-2&5\end{pmatrix} \frac15 \begin{pmatrix}4&-3\\-1&2\end{pmatrix}.\)

Multiplying gives

\(C=\frac15 \begin{pmatrix} 4-4&-3+8\\ -8-5&6+10 \end{pmatrix} = \frac15 \begin{pmatrix}0&5\\-13&16\end{pmatrix}.\)

(iii) From

\(A^{-1}D+B=I,\)

we have

\(A^{-1}D=I-B.\)

Multiplying by \(A\) gives

\(D=A(I-B).\)

Now

\(I-B= \begin{pmatrix}1&0\\0&1\end{pmatrix} - \begin{pmatrix}1&4\\-2&5\end{pmatrix} = \begin{pmatrix}0&-4\\2&-4\end{pmatrix}.\)

Therefore

\(D= \begin{pmatrix}2&3\\1&4\end{pmatrix} \begin{pmatrix}0&-4\\2&-4\end{pmatrix} = \begin{pmatrix}6&-20\\8&-20\end{pmatrix}.\)