Answer: (a) Since \(\mathbf{A}\mathbf{e}=\lambda \mathbf{e}\) and \(\mathbf{A}\) is non-singular, multiply on the left by \(\mathbf{A}^{-1}\):

\(\mathbf{e}=\lambda \mathbf{A}^{-1}\mathbf{e}\), so \(\mathbf{A}^{-1}\mathbf{e}=\lambda^{-1}\mathbf{e}\).

Hence \(\lambda^{-1}\) is an eigenvalue of \(\mathbf{A}^{-1}\) with corresponding eigenvector \(\mathbf{e}\).

(b) A corresponding eigenvector for eigenvalue \(-1\) is \(\begin{pmatrix}1\\4\\-1\end{pmatrix}\).

(c) The corresponding eigenvalues are:

• for \(\begin{pmatrix}0\\1\\0\end{pmatrix}\): \(-4\)
• for \(\begin{pmatrix}1\\2\\1\end{pmatrix}\): \(5\)

(d) One suitable choice is

\(\mathbf{P}=\begin{pmatrix}1&0&1\\4&1&2\\-1&0&1\end{pmatrix}\), \(\mathbf{D}=\begin{pmatrix}-1&0&0\\0&-\frac14&0\\0&0&\frac15\end{pmatrix}\),

so that \(\mathbf{A}^{-1}=\mathbf{PDP}^{-1}\).

(e) \((\lambda+1)(\lambda+4)(\lambda-5)=\lambda^3-21\lambda-20=0\), so

\(\mathbf{A}^3-21\mathbf{A}-20\mathbf{I}=\mathbf{0}.\)

Multiplying by \(\mathbf{A}^{-1}\) gives

\(20\mathbf{A}^{-1}=\mathbf{A}^2-21\mathbf{I}\).

Hence \(\mathbf{A}^{-1}=\frac1{20}\mathbf{A}^2-\frac{21}{20}\mathbf{I}\), so \(p=\frac1{20}\), \(q=-\frac{21}{20}\).

(a) We are given

\(\mathbf{A}\mathbf{e}=\lambda\mathbf{e}.\)

Since \(\mathbf{A}\) is non-singular, \(\mathbf{A}^{-1}\) exists. Multiply on the left by \(\mathbf{A}^{-1}\):

\(\mathbf{A}^{-1}\mathbf{A}\mathbf{e}=\mathbf{A}^{-1}(\lambda\mathbf{e}).\)

So

\(\mathbf{e}=\lambda\mathbf{A}^{-1}\mathbf{e}.\)

Hence

\(\mathbf{A}^{-1}\mathbf{e}=\lambda^{-1}\mathbf{e}.\)

Therefore \(\lambda^{-1}\) is an eigenvalue of \(\mathbf{A}^{-1}\) and \(\mathbf{e}\) is a corresponding eigenvector.

(b) For eigenvalue \(-1\), solve

\((\mathbf{A}+\mathbf{I})\mathbf{x}=\mathbf{0}.\)

Now

\(\mathbf{A}+\mathbf{I}=\begin{pmatrix}3&0&3\\15&-3&3\\3&0&3\end{pmatrix}.\)

Let \(\mathbf{x}=\begin{pmatrix}x\\y\\z\end{pmatrix}\). Then

\(3x+3z=0\),

\(15x-3y+3z=0\).

From the first equation, \(z=-x\).

Substitute into the second:

\(15x-3y-3x=0\Rightarrow 12x-3y=0\Rightarrow y=4x.\)

Taking \(x=1\), we get \(y=4\), \(z=-1\).

So a corresponding eigenvector is

\(\begin{pmatrix}1\\4\\-1\end{pmatrix}.\)

(c) For \(\begin{pmatrix}0\\1\\0\end{pmatrix}\),

\(\mathbf{A}\begin{pmatrix}0\\1\\0\end{pmatrix}=\begin{pmatrix}0\\-4\\0\end{pmatrix}=-4\begin{pmatrix}0\\1\\0\end{pmatrix},\)

so the eigenvalue is \(-4\).

For \(\begin{pmatrix}1\\2\\1\end{pmatrix}\),

\(\mathbf{A}\begin{pmatrix}1\\2\\1\end{pmatrix}=\begin{pmatrix}5\\10\\5\end{pmatrix}=5\begin{pmatrix}1\\2\\1\end{pmatrix},\)

so the eigenvalue is \(5\).

(d) The eigenvectors of \(\mathbf{A}\) found are

\(\begin{pmatrix}1\\4\\-1\end{pmatrix},\quad \begin{pmatrix}0\\1\\0\end{pmatrix},\quad \begin{pmatrix}1\\2\\1\end{pmatrix},\)

with eigenvalues \(-1\), \(-4\), \(5\) respectively.

Therefore the eigenvalues of \(\mathbf{A}^{-1}\) are

\(-1,\quad -\frac14,\quad \frac15.\)

Using the same eigenvectors, take

\(\mathbf{P}=\begin{pmatrix}1&0&1\\4&1&2\\-1&0&1\end{pmatrix}\),

and

\(\mathbf{D}=\begin{pmatrix}-1&0&0\\0&-\frac14&0\\0&0&\frac15\end{pmatrix}.\)

Then

\(\mathbf{A}^{-1}=\mathbf{PDP}^{-1}.\)

(e) Since the eigenvalues are \(-1\), \(-4\), and \(5\), the characteristic equation is

\((\lambda+1)(\lambda+4)(\lambda-5)=0.\)

Expanding,

\((\lambda+1)(\lambda+4)(\lambda-5)=\lambda^3-21\lambda-20,\)

so the characteristic equation is

\(\lambda^3-21\lambda-20=0.\)

By Cayley-Hamilton,

\(\mathbf{A}^3-21\mathbf{A}-20\mathbf{I}=\mathbf{0}.\)

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

\(\mathbf{A}^2-21\mathbf{I}-20\mathbf{A}^{-1}=\mathbf{0}.\)

Hence

\(20\mathbf{A}^{-1}=\mathbf{A}^2-21\mathbf{I},\)

so

\(\mathbf{A}^{-1}=\frac1{20}\mathbf{A}^2-\frac{21}{20}\mathbf{I}.\)

Therefore \(p=\frac1{20}\) and \(q=-\frac{21}{20}\).