Answer: (i) The corresponding eigenvalue is \(-2\).

(ii) A corresponding eigenvector is \(\begin{pmatrix}1\\0\\-1\end{pmatrix}\).

(iii) The third eigenvalue is \(5\), and a corresponding eigenvector is \(\begin{pmatrix}1\\1\\1\end{pmatrix}\).

(iv) One suitable choice is \(\mathbf{P}=\begin{pmatrix}1&1&1\\1&0&1\\0&-1&1\end{pmatrix}\) and \(\mathbf{D}=\begin{pmatrix}-2&0&0\\0&-1&0\\0&0&5\end{pmatrix}\). Hence \(\mathbf{A}^n=\mathbf{P}\mathbf{D}^n\mathbf{P}^{-1}\), so

\(\mathbf{A}^{n}=\begin{pmatrix}5^{n}+(-1)^{n}-(-2)^{n} & 2(-2)^{n}+(-1)^{n+1}-5^{n} & 5^{n}-(-2)^{n}\\5^{n}-(-2)^{n} & 2(-2)^{n}-5^{n} & 5^{n}-(-2)^{n}\\5^{n}-(-1)^{n} & (-1)^{n}-5^{n} & 5^{n}\end{pmatrix}.\)

(i) Since \(\begin{pmatrix}1\\1\\0\end{pmatrix}\) is an eigenvector, we compute \(\mathbf{A}\begin{pmatrix}1\\1\\0\end{pmatrix}\):

\(\mathbf{A}\begin{pmatrix}1\\1\\0\end{pmatrix}=\begin{pmatrix}6-8\\7-9\\6-6\end{pmatrix}=\begin{pmatrix}-2\\-2\\0\end{pmatrix}= -2\begin{pmatrix}1\\1\\0\end{pmatrix}.\)

So the corresponding eigenvalue is \(\lambda=-2\).

(ii) To find an eigenvector for \(\lambda=-1\), solve \((\mathbf{A}+\mathbf{I})\mathbf{x}=\mathbf{0}\).

\(\mathbf{A}+\mathbf{I}=\begin{pmatrix}7&-8&7\\7&-8&7\\6&-6&6\end{pmatrix}.\)

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

\(7x-8y+7z=0\) and \(6x-6y+6z=0\), so from the second equation \(x-y+z=0\), i.e. \(x=y-z\).

Substitute into the first:

\(7(y-z)-8y+7z=0 \Rightarrow -y=0 \Rightarrow y=0.\)

Then \(x=-z\). Taking \(z=-1\) gives \(x=1\), so one eigenvector is

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

(iii) The determinant is the product of the eigenvalues. First calculate \(\det\mathbf{A}\):

Expand along the first row:

\(\det\mathbf{A}=6\begin{vmatrix}-9&7\\-6&5\end{vmatrix}-(-8)\begin{vmatrix}7&7\\6&5\end{vmatrix}+7\begin{vmatrix}7&-9\\6&-6\end{vmatrix}.\)

So

\(\det\mathbf{A}=6(({-9})(5)-7({-6})) + 8(7\cdot 5-7\cdot 6) + 7(7({-6})-({-9})6).\)

\(=6(-45+42)+8(35-42)+7(-42+54)=6(-3)+8(-7)+7(12).\)

\(=-18-56+84=10.\)

Thus the product of the three eigenvalues is \(10\). Since two eigenvalues are \(-2\) and \(-1\), the third eigenvalue \(\lambda_3\) satisfies

\(({-2})(-1)\lambda_3=10 \Rightarrow 2\lambda_3=10 \Rightarrow \lambda_3=5.\)

For \(\lambda=5\), solve \((\mathbf{A}-5\mathbf{I})\mathbf{x}=\mathbf{0}\):

\(\mathbf{A}-5\mathbf{I}=\begin{pmatrix}1&-8&7\\7&-14&7\\6&-6&0\end{pmatrix}.\)

From the third row, \(6x-6y=0\), so \(x=y\). Then the first row gives

\(x-8y+7z=0.\)

With \(x=y\), this becomes \(-7y+7z=0\), so \(y=z\). Hence \(x=y=z\), and a corresponding eigenvector is

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

(iv) Use the eigenvectors as the columns of \(\mathbf{P}\), in the same order as the eigenvalues in \(\mathbf{D}\):

\(\mathbf{P}=\begin{pmatrix}1&1&1\\1&0&1\\0&-1&1\end{pmatrix},\qquad \mathbf{D}=\begin{pmatrix}-2&0&0\\0&-1&0\\0&0&5\end{pmatrix}.\)

Then \(\mathbf{P}^{-1}\mathbf{A}\mathbf{P}=\mathbf{D}\), so \(\mathbf{A}=\mathbf{P}\mathbf{D}\mathbf{P}^{-1}\), and therefore

\(\mathbf{A}^n=\mathbf{P}\mathbf{D}^n\mathbf{P}^{-1}.\)

Now \(\mathbf{D}^n=\begin{pmatrix}(-2)^n&0&0\\0&(-1)^n&0\\0&0&5^n\end{pmatrix}.\)

Also \(\det\mathbf{P}=-1\), so

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

Multiplying out gives

\(\mathbf{A}^{n}=\begin{pmatrix}5^{n}+(-1)^{n}-(-2)^{n} & 2(-2)^{n}+(-1)^{n+1}-5^{n} & 5^{n}-(-2)^{n}\\5^{n}-(-2)^{n} & 2(-2)^{n}-5^{n} & 5^{n}-(-2)^{n}\\5^{n}-(-1)^{n} & (-1)^{n}-5^{n} & 5^{n}\end{pmatrix}.\)