Answer: (i) \(A=\dfrac{1}{b+1}\begin{pmatrix}1+2b&1&1\\-b&3b+2&-1\\2b&-2b&b+3\end{pmatrix}\).

(ii) \(A^{-1}\begin{pmatrix}0\\2\\-2\end{pmatrix}=\dfrac{2}{3}\begin{pmatrix}0\\1\\-1\end{pmatrix}=\begin{pmatrix}0\\\frac23\\-\frac23\end{pmatrix}\).

(iii) \(n=2\) and \(b=1\).

(i) Put the three eigenvectors as the columns of a matrix \(P\), in the same order as their eigenvalues:

\(P=\begin{pmatrix}1&-1&0\\1&0&1\\0&b&-1\end{pmatrix}\), and \(D=\begin{pmatrix}2&0&0\\0&1&0\\0&0&3\end{pmatrix}\).

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

Since \(b\) is positive, \(b+1\neq0\), so \(P\) is invertible. Its inverse is

\(P^{-1}=\dfrac{1}{b+1}\begin{pmatrix}b&1&1\\-1&1&1\\-b&b&-1\end{pmatrix}\).

Now

\(PD=\begin{pmatrix}2&-1&0\\2&0&3\\0&b&-3\end{pmatrix}\).

Therefore

\(A=PDP^{-1}=\dfrac{1}{b+1}\begin{pmatrix}2&-1&0\\2&0&3\\0&b&-3\end{pmatrix}\begin{pmatrix}b&1&1\\-1&1&1\\-b&b&-1\end{pmatrix}\).

Multiplying gives

\(A=\dfrac{1}{b+1}\begin{pmatrix}1+2b&1&1\\-b&3b+2&-1\\2b&-2b&b+3\end{pmatrix}\).

(ii) Notice that

\(\begin{pmatrix}0\\2\\-2\end{pmatrix}=2\begin{pmatrix}0\\1\\-1\end{pmatrix}\).

The vector \(\begin{pmatrix}0\\1\\-1\end{pmatrix}\) is an eigenvector of \(A\) with eigenvalue \(3\). Hence applying \(A^{-1}\) to it divides it by \(3\):

\(A^{-1}\begin{pmatrix}0\\1\\-1\end{pmatrix}=\dfrac13\begin{pmatrix}0\\1\\-1\end{pmatrix}\).

So

\(A^{-1}\begin{pmatrix}0\\2\\-2\end{pmatrix}=\dfrac23\begin{pmatrix}0\\1\\-1\end{pmatrix}=\begin{pmatrix}0\\\frac23\\-\frac23\end{pmatrix}\).

(iii) Since \(\begin{pmatrix}1\\1\\0\end{pmatrix}\) is an eigenvector with eigenvalue \(2\),

\(A^n\begin{pmatrix}1\\1\\0\end{pmatrix}=2^n\begin{pmatrix}1\\1\\0\end{pmatrix}\).

But this is given as \(\begin{pmatrix}4\\4\\0\end{pmatrix}\), so \(2^n=4\). Therefore \(n=2\).

The vector \(\begin{pmatrix}-1\\0\\b\end{pmatrix}\) is an eigenvector with eigenvalue \(1\), so

\(A^n\begin{pmatrix}-1\\0\\b\end{pmatrix}=1^n\begin{pmatrix}-1\\0\\b\end{pmatrix}=\begin{pmatrix}-1\\0\\b\end{pmatrix}\).

This is given as \(\begin{pmatrix}-1\\0\\b^{-1}\end{pmatrix}\), so \(b=b^{-1}\). Since \(b\) is positive, \(b^2=1\) gives \(b=1\).