Answer: (a(i)) Stretch parallel to the \(x\)-axis, scale factor \(k\).

(a(ii)) Enlargement, centre the origin, scale factor \(k\).

(b) \(ABC\) is singular because \(\det(ABC)=0\).

(a(i))

When \(m=1\),

\(B=\begin{pmatrix}k&0\\0&1\end{pmatrix}.\)

This maps \((x,y)\) to \((kx,y)\), so it is a stretch parallel to the \(x\)-axis with scale factor \(k\).

(a(ii))

When \(m=k\),

\(B=\begin{pmatrix}k&0\\0&k\end{pmatrix}=kI.\)

This maps \((x,y)\) to \((kx,ky)\), so it is an enlargement with centre the origin and scale factor \(k\).

(b)

First multiply \(A\) and \(B\):

\(AB=\begin{pmatrix}0&m\\-k&m\\k&m\end{pmatrix}.\)

Then

\(ABC=\begin{pmatrix}m&m&2m\\-2k+m&k+m&-k+2m\\2k+m&-k+m&k+2m\end{pmatrix}.\)

Expanding the determinant along the first row gives

\(\det(ABC)=m\begin{vmatrix}k+m&-k+2m\\-k+m&k+2m\end{vmatrix}-m\begin{vmatrix}-2k+m&-k+2m\\2k+m&k+2m\end{vmatrix}+2m\begin{vmatrix}-2k+m&k+m\\2k+m&-k+m\end{vmatrix}.\)

Expanding and simplifying gives

\(\det(ABC)=0.\)

Therefore \(ABC\) is singular.

Mark scheme