Answer: Points matrix \(\begin{pmatrix}10\\10\\6\\2\\12\end{pmatrix}\), so team E; \(A^{-1}=\frac16\begin{pmatrix}4&1\\-2&1\end{pmatrix}\); \(C=\frac16\begin{pmatrix}21&-2\\-9&-2\end{pmatrix}\).

(a)(i) The results matrix can be multiplied by the points matrix:

\(\begin{pmatrix} 2&1&1\\ 1&3&0\\ 1&1&2\\ 0&1&3\\ 3&0&1 \end{pmatrix} \begin{pmatrix}4\\2\\0\end{pmatrix}.\)

(a)(ii) Multiplying gives

\(\begin{pmatrix} 2&1&1\\ 1&3&0\\ 1&1&2\\ 0&1&3\\ 3&0&1 \end{pmatrix} \begin{pmatrix}4\\2\\0\end{pmatrix} = \begin{pmatrix}10\\10\\6\\2\\12\end{pmatrix}.\)

The largest total is \(12\), so team \(E\) was awarded the most points.

(b)(i) For

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

the determinant is

\(1(4)-(-1)(2)=6.\)

Therefore

\(A^{-1}=\frac16\begin{pmatrix}4&1\\-2&1\end{pmatrix}.\)

(b)(ii) Since \(AC=B\), multiply both sides by \(A^{-1}\):

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

So

\(C=\frac16\begin{pmatrix}4&1\\-2&1\end{pmatrix} \begin{pmatrix}5&0\\1&-2\end{pmatrix}.\)

This gives

\(C=\frac16\begin{pmatrix}21&-2\\-9&-2\end{pmatrix}.\)