Answer: (i) \(\mathbf{ABe}=\lambda\mu\mathbf e\).

(ii) The eigenvalues of \(\mathbf A\) are \(-1,2,3\), with corresponding eigenvectors \(\begin{pmatrix}1\\-1\\0\end{pmatrix}\), \(\begin{pmatrix}1\\-1\\1\end{pmatrix}\) and \(\begin{pmatrix}1\\0\\1\end{pmatrix}\).

(iii) The eigenvalues of \(\mathbf{AB}\) are \(3,-4,12\), with the same corresponding eigenvectors in that order.

(i) Since \(\mathbf{Ae}=\lambda\mathbf e\) and \(\mathbf{Be}=\mu\mathbf e\),

\(\mathbf{ABe}=\mathbf A(\mathbf{Be})=\mathbf A(\mu\mathbf e)=\mu\mathbf{Ae}=\mu\lambda\mathbf e=\lambda\mu\mathbf e.\)

So \(\mathbf e\) is an eigenvector of \(\mathbf{AB}\) with eigenvalue \(\lambda\mu\).

(ii) The characteristic equation is

\((\lambda+1)(\lambda^2-5\lambda+6)=0=(\lambda+1)(\lambda-2)(\lambda-3).\)

So \(\lambda=-1,2,3\). Solving \((\mathbf A-\lambda\mathbf I)\mathbf x=\mathbf0\) gives the corresponding eigenvectors

\(\lambda=-1:\begin{pmatrix}1\\-1\\0\end{pmatrix},\qquad \lambda=2:\begin{pmatrix}1\\-1\\1\end{pmatrix},\qquad \lambda=3:\begin{pmatrix}1\\0\\1\end{pmatrix}.\)

(iii) For the first eigenvector,

\(\mathbf B\begin{pmatrix}1\\-1\\0\end{pmatrix}=\begin{pmatrix}-3\\3\\0\end{pmatrix}=-3\begin{pmatrix}1\\-1\\0\end{pmatrix},\)

so the corresponding eigenvalue of \(\mathbf B\) is \(-3\). Similarly, the other two eigenvalues of \(\mathbf B\) are \(-2\) and \(4\). By part (i), the eigenvalues of \(\mathbf{AB}\) are the products of the matching eigenvalues:

\((-1)(-3)=3,\qquad 2(-2)=-4,\qquad 3(4)=12.\)

The corresponding eigenvectors are

\(\begin{pmatrix}1\\-1\\0\end{pmatrix},\qquad \begin{pmatrix}1\\-1\\1\end{pmatrix},\qquad \begin{pmatrix}1\\0\\1\end{pmatrix}.\)