Answer: The eigenvalues of \(\mathbf{A}\) are \(2\), \(1\) and \(-1\).

For \(\mathbf{B}\), the vector \(\mathbf{e}=\begin{pmatrix}1\\1\\1\end{pmatrix}\) is an eigenvector with eigenvalue \(3\).

Hence an eigenvector of \(\mathbf{A B}\) is \(\mathbf{e}=\begin{pmatrix}1\\1\\1\end{pmatrix}\), with corresponding eigenvalue \(6\).

First, for \(\mathbf{A}\) and \(\mathbf{e}=\begin{pmatrix}1\\1\\1\end{pmatrix}\),

\(\mathbf{A e}=\begin{pmatrix}4 & -5 & 3\\3 & -4 & 3\\1 & -1 & 2\end{pmatrix}\begin{pmatrix}1\\1\\1\end{pmatrix}=\begin{pmatrix}4-5+3\\3-4+3\\1-1+2\end{pmatrix}=\begin{pmatrix}2\\2\\2\end{pmatrix}=2\mathbf{e}.\)

So \(\mathbf{e}\) is an eigenvector of \(\mathbf{A}\), with eigenvalue \(2\).

To find the other eigenvalues, use the characteristic equation:

\(\det(\mathbf{A}-\lambda\mathbf{I})=0.\)

This gives

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

Factorising,

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

Hence the other two eigenvalues are \(1\) and \(-1\).

Now for \(\mathbf{B}\),

\(\mathbf{B e}=\begin{pmatrix}-1 & 4 & 0\\-1 & 3 & 1\\1 & -1 & 3\end{pmatrix}\begin{pmatrix}1\\1\\1\end{pmatrix}=\begin{pmatrix}-1+4+0\\-1+3+1\\1-1+3\end{pmatrix}=\begin{pmatrix}3\\3\\3\end{pmatrix}=3\mathbf{e}.\)

So \(\mathbf{e}\) is an eigenvector of \(\mathbf{B}\), with eigenvalue \(3\).

Therefore

\(\mathbf{A B e}=\mathbf{A}(\mathbf{B e})=\mathbf{A}(3\mathbf{e})=3\mathbf{A e}=3(2\mathbf{e})=6\mathbf{e}.\)

So \(\mathbf{e}=\begin{pmatrix}1\\1\\1\end{pmatrix}\) is an eigenvector of \(\mathbf{A B}\), with corresponding eigenvalue \(6\).