The matrix \(\mathbf{M}\) is upper triangular, so its eigenvalues are the diagonal entries \(2\), \(m\) and \(1\). Since \(m\neq 0,1,2\), these eigenvalues are distinct, so \(\mathbf{M}\) is diagonalisable.

We find eigenvectors for each eigenvalue.

For \(\lambda=2\):

\[ \mathbf{M}-2\mathbf{I} = \begin{pmatrix} 0 & m & 1\\ 0 & m-2 & 7\\ 0 & 0 & -1 \end{pmatrix}. \]

Let \(\mathbf{x}=(x,y,z)^T\). Then \((\mathbf{M}-2\mathbf{I})\mathbf{x}=\mathbf{0}\) gives \(z=0\), then \((m-2)y+7z=0\), so \(y=0\). Hence \(x\) is free, and an eigenvector is

\[ \mathbf{v}_1= \begin{pmatrix} 1\\ 0\\ 0 \end{pmatrix}. \]

For \(\lambda=m\):

\[ \mathbf{M}-m\mathbf{I} = \begin{pmatrix} 2-m & m & 1\\ 0 & 0 & 7\\ 0 & 0 & 1-m \end{pmatrix}. \]

From the last row, \((1-m)z=0\). Since \(m\neq 1\), we get \(z=0\). The first row becomes

\[ (2-m)x+my=0. \]

A convenient choice is \(x=m\), \(y=m-2\). Thus an eigenvector is

\[ \mathbf{v}_2= \begin{pmatrix} m\\ m-2\\ 0 \end{pmatrix}. \]

For \(\lambda=1\):

\[ \mathbf{M}-\mathbf{I} = \begin{pmatrix} 1 & m & 1\\ 0 & m-1 & 7\\ 0 & 0 & 0 \end{pmatrix}. \]

From \((m-1)y+7z=0\), choose \(z=m-1\). Then \(y=-7\). The first row gives

\[ x+m(-7)+(m-1)=0, \]

so

\[ x=6m+1. \]

Hence an eigenvector is

\[ \mathbf{v}_3= \begin{pmatrix} 6m+1\\ -7\\ m-1 \end{pmatrix}. \]

Therefore, we may take

\[ \mathbf{P} = \begin{pmatrix} 1 & m & 6m+1\\ 0 & m-2 & -7\\ 0 & 0 & m-1 \end{pmatrix}, \qquad \mathbf{D} = \begin{pmatrix} 2 & 0 & 0\\ 0 & m & 0\\ 0 & 0 & 1 \end{pmatrix}. \]

Then

\[ \mathbf{M}=\mathbf{P}\mathbf{D}\mathbf{P}^{-1}. \]

Now

\[ \mathbf{M}^7 = \left(\mathbf{P}\mathbf{D}\mathbf{P}^{-1}\right)^7 = \mathbf{P}\mathbf{D}^7\mathbf{P}^{-1}. \]

Multiplying by \(\mathbf{P}\) on the right gives

\[ \mathbf{M}^7\mathbf{P} = \mathbf{P}\mathbf{D}^7. \]

Since

\[ \mathbf{D}^7 = \operatorname{diag}(2^7,m^7,1^7) = \operatorname{diag}(128,m^7,1), \]

we get

\[ \mathbf{M}^7\mathbf{P} = \mathbf{P} \begin{pmatrix} 128 & 0 & 0\\ 0 & m^7 & 0\\ 0 & 0 & 1 \end{pmatrix}. \]

This means that the first column of \(\mathbf{P}\) is multiplied by \(128\), the second column by \(m^7\), and the third column is unchanged.

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