Answer: (a) The determinant is

\(\det(\mathbf{A})=a\begin{vmatrix}1-a&0\\2-a&-1\end{vmatrix}=a(a-1)\).

Since \(a\neq 0\) and \(a\neq 1\), \(\det(\mathbf{A})\neq 0\), so the system has a unique solution.

Geometrically, the three planes represented by the three equations intersect in exactly one point.

(b) The eigenvalues are \(a\), \(1-a\) and \(-1\).

(c) One suitable choice is

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

so that \(\mathbf{A}^4=\mathbf{PDP}^{-1}\).

(d) Using the characteristic equation,

\(\mathbf{A}^4=(a^2-a+1)\mathbf{A}^2+(a^2-a)\mathbf{A}\).

(a) Consider

\(\mathbf{A}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}1\\2\\3\end{pmatrix}.\)

A unique solution exists if \(\mathbf{A}\) is invertible, so find the determinant:

\(\det(\mathbf{A})=\begin{vmatrix}a&-6a&2a+2\\0&1-a&0\\0&2-a&-1\end{vmatrix}.\)

Expanding down the first column,

\(\det(\mathbf{A})=a\begin{vmatrix}1-a&0\\2-a&-1\end{vmatrix}=a((1-a)(-1))=a(a-1).\)

Since \(a\neq 0\) and \(a\neq 1\), this is non-zero. Therefore the system has a unique solution.

Geometrically, the three planes represented by the three linear equations meet at a single point.

(b) The eigenvalues satisfy

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

Now

\(\mathbf{A}-\lambda\mathbf{I}=\begin{pmatrix}a-\lambda&-6a&2a+2\\0&1-a-\lambda&0\\0&2-a&-1-\lambda\end{pmatrix}.\)

Again expanding down the first column,

\(\det(\mathbf{A}-\lambda\mathbf{I})=(a-\lambda)\begin{vmatrix}1-a-\lambda&0\\2-a&-1-\lambda\end{vmatrix}.\)

So

\(\det(\mathbf{A}-\lambda\mathbf{I})=(a-\lambda)(1-a-\lambda)(-1-\lambda).\)

Hence the eigenvalues are

\(\lambda=a,\quad 1-a,\quad -1.\)

(c) Find eigenvectors corresponding to these eigenvalues.

For \(\lambda=a\),

\(\mathbf{A}-a\mathbf{I}=\begin{pmatrix}0&-6a&2a+2\\0&1-2a&0\\0&2-a&-1-a\end{pmatrix}.\)

The first column is zero, so

\(\mathbf{v}_1=\begin{pmatrix}1\\0\\0\end{pmatrix}\)

is an eigenvector.

For \(\lambda=-1\),

\(\mathbf{A}+\mathbf{I}=\begin{pmatrix}a+1&-6a&2a+2\\0&2-a&0\\0&2-a&0\end{pmatrix}.\)

Take \(y=0\). Then the first row gives

\((a+1)x+(2a+2)z=0\Rightarrow x+2z=0.\)

Choosing \(z=1\) gives \(x=-2\), so

\(\mathbf{v}_2=\begin{pmatrix}-2\\0\\1\end{pmatrix}.\)

For \(\lambda=1-a\), use the vector

\(\mathbf{v}_3=\begin{pmatrix}2\\1\\1\end{pmatrix}.\)

Check it directly:

\(\mathbf{A}\begin{pmatrix}2\\1\\1\end{pmatrix}=\begin{pmatrix}2a-6a+2a+2\\1-a\\2-a-1\end{pmatrix}=\begin{pmatrix}2(1-a)\\1-a\\1-a\end{pmatrix}=(1-a)\begin{pmatrix}2\\1\\1\end{pmatrix}.\)

So \(\mathbf{v}_3\) is an eigenvector corresponding to \(1-a\).

Using these as columns,

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

The corresponding eigenvalues of \(\mathbf{A}^4\) are \(a^4\), \((-1)^4=1\), and \((1-a)^4\), so

\(\mathbf{D}=\begin{pmatrix}a^4&0&0\\0&1&0\\0&0&(1-a)^4\end{pmatrix}.\)

Therefore

\(\mathbf{A}^4=\mathbf{PDP}^{-1}.\)

(d) Expanding the characteristic equation,

\((a-\lambda)(1-a-\lambda)(-1-\lambda)=\lambda^3-(a^2-a+1)\lambda-(a^2-a)=0.\)

By Cayley-Hamilton,

\(\mathbf{A}^3-(a^2-a+1)\mathbf{A}-(a^2-a)\mathbf{I}=\mathbf{0}.\)

Multiplying by \(\mathbf{A}\),

\(\mathbf{A}^4-(a^2-a+1)\mathbf{A}^2-(a^2-a)\mathbf{A}=\mathbf{0}.\)

Hence

\(\mathbf{A}^4=(a^2-a+1)\mathbf{A}^2+(a^2-a)\mathbf{A}.\)