Answer: (a) \(\begin{pmatrix}1\\1\\1\end{pmatrix}\) is an eigenvector because \(\mathbf{A}\begin{pmatrix}1\\1\\1\end{pmatrix}=\begin{pmatrix}-2\\-2\\-2\end{pmatrix}=-2\begin{pmatrix}1\\1\\1\end{pmatrix}\). The corresponding eigenvalue is \(-2\).

(b) The characteristic equation is \(\lambda^3-19\lambda-30=0\). Since \(\lambda=-2\) is a root, \(\lambda^3-19\lambda-30=(\lambda+2)(\lambda^2-2\lambda-15)=(\lambda+2)(\lambda-5)(\lambda+3)\). The other eigenvalues are \(5\) and \(-3\).

(c) \(\mathbf{A}^{-1}=\frac{1}{30}\begin{pmatrix}-26&-5&16\\0&-15&0\\-32&-5&22\end{pmatrix}.\)

(a) Multiply \(\mathbf{A}\) by \(\begin{pmatrix}1\\1\\1\end{pmatrix}\):

\(\mathbf{A}\begin{pmatrix}1\\1\\1\end{pmatrix}=\begin{pmatrix}-11&1&8\\0&-2&0\\-16&1&13\end{pmatrix}\begin{pmatrix}1\\1\\1\end{pmatrix}=\begin{pmatrix}-11+1+8\\0-2+0\\-16+1+13\end{pmatrix}=\begin{pmatrix}-2\\-2\\-2\end{pmatrix}.\)

Since this equals \(-2\begin{pmatrix}1\\1\\1\end{pmatrix}\), the vector is an eigenvector and the eigenvalue is \(\lambda=-2\).

(b) The characteristic equation is found from \(\det(\mathbf{A}-\lambda\mathbf{I})=0\):

\(\det\begin{pmatrix}-11-\lambda&1&8\\0&-2-\lambda&0\\-16&1&13-\lambda\end{pmatrix}=0.\)

Expanding along the second row,

\((-2-\lambda)\det\begin{pmatrix}-11-\lambda&8\\-16&13-\lambda\end{pmatrix}=0.\)

So

\((-2-\lambda)\big(( -11-\lambda)(13-\lambda)+128\big)=0.\)

Now

\(( -11-\lambda)(13-\lambda)+128=\lambda^2-2\lambda-143+128=\lambda^2-2\lambda-15.\)

Hence

\((-2-\lambda)(\lambda^2-2\lambda-15)=0.\)

This is equivalent to

\(\lambda^3-19\lambda-30=0.\)

Using the root \(\lambda=-2\),

\(\lambda^3-19\lambda-30=(\lambda+2)(\lambda^2-2\lambda-15)=(\lambda+2)(\lambda-5)(\lambda+3).\)

So the other eigenvalues are \(5\) and \(-3\).

(c) By Cayley-Hamilton, \(\mathbf{A}\) satisfies its characteristic equation:

\(\mathbf{A}^3-19\mathbf{A}-30\mathbf{I}=\mathbf{0}.\)

Multiply by \(\mathbf{A}^{-1}\):

\(\mathbf{A}^2-19\mathbf{I}-30\mathbf{A}^{-1}=\mathbf{0}.\)

So

\(30\mathbf{A}^{-1}=\mathbf{A}^2-19\mathbf{I}.\)

Now calculate

\(\mathbf{A}^2=\begin{pmatrix}-11&1&8\\0&-2&0\\-16&1&13\end{pmatrix}\begin{pmatrix}-11&1&8\\0&-2&0\\-16&1&13\end{pmatrix}=\begin{pmatrix}-7&-5&16\\0&4&0\\-32&-5&41\end{pmatrix}.\)

Therefore

\(\mathbf{A}^2-19\mathbf{I}=\begin{pmatrix}-7&-5&16\\0&4&0\\-32&-5&41\end{pmatrix}-\begin{pmatrix}19&0&0\\0&19&0\\0&0&19\end{pmatrix}=\begin{pmatrix}-26&-5&16\\0&-15&0\\-32&-5&22\end{pmatrix}.\)

Hence

\(\mathbf{A}^{-1}=\frac{1}{30}\begin{pmatrix}-26&-5&16\\0&-15&0\\-32&-5&22\end{pmatrix}.\)