Answer: EITHER

(i) Let the eigenvectors be the columns of \(\mathbf{P}\), in the order corresponding to eigenvalues \(-1,1,2\):

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

Then \(\mathbf{A}=\mathbf{P D P}^{-1}\), giving

\(\mathbf{A}=\begin{pmatrix}\frac32&\frac12&\frac12\\\frac32&\frac12&\frac32\\-1&1&0\end{pmatrix}.\)

(ii) Since \(\mathbf{A}=\mathbf{PDP}^{-1}\),

\(\mathbf{A}^{2n}=\mathbf{P D}^{2n}\mathbf{P}^{-1}=\frac12\begin{pmatrix}2^{2n}+1&2^{2n}-1&2^{2n}-1\\2^{2n}-1&2^{2n}+1&2^{2n}-1\\0&0&2\end{pmatrix}.\)

OR

The rank is \(3\). Also, if \(\mathbf{A x}=p\begin{pmatrix}1\\2\\3\\5\end{pmatrix}+q\begin{pmatrix}-1\\-1\\-3\\-4\end{pmatrix}+r\begin{pmatrix}-1\\-4\\-2\\-6\end{pmatrix}\), then

\(\mathbf{x}=\begin{pmatrix}p+\lambda\\q+\lambda\\r+\lambda\\lambda\end{pmatrix}\)

for some real \(\lambda\). For

\(p\begin{pmatrix}1\\2\\3\\5\end{pmatrix}+q\begin{pmatrix}-1\\-1\\-3\\-4\end{pmatrix}+r\begin{pmatrix}-1\\-4\\-2\\-6\end{pmatrix}=\begin{pmatrix}3\\7\\8\\15\end{pmatrix}\),

one suitable choice is \(p=5,\ q=-1,\ r=-1\). Hence the solution of \(\mathbf{A x}=\begin{pmatrix}3\\7\\8\\15\end{pmatrix}\) with \(\alpha^2+\beta^2+\gamma^2+\delta^2=\frac{11}{4}\) is

\(\mathbf{x}=\begin{pmatrix}\frac54\\-\frac34\\-\frac34\\\frac14\end{pmatrix}.\)

EITHER

(i) Form the modal matrix \(\mathbf{P}\) from the given eigenvectors, in the same order as the eigenvalues \(-1,1,2\):

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

Now find \(\mathbf{P}^{-1}\). The determinant is

\(\det\mathbf{P}=2,\)

and

\(\mathbf{P}^{-1}=\frac12\begin{pmatrix}-1&1&-1\\-1&1&1\\1&1&1\end{pmatrix}.\)

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

\(\mathbf{A}=\begin{pmatrix}0&-1&2\\1&0&1\\-1&1&0\end{pmatrix}\frac12\begin{pmatrix}-1&1&-1\\-1&1&1\\1&1&1\end{pmatrix}.\)

Multiplying gives

\(\mathbf{A}=\begin{pmatrix}\frac32&\frac12&\frac12\\\frac32&\frac12&\frac32\\-1&1&0\end{pmatrix}.\)

(ii) For a positive integer \(n\),

\(\mathbf{A}^{2n}=\mathbf{P D}^{2n}\mathbf{P}^{-1}.\)

Because \(\mathbf{D}^{2n}=\operatorname{diag}(1,1,2^{2n})\),

\(\mathbf{A}^{2n}=\frac12\begin{pmatrix}0&-1&1\\1&0&1\\-1&1&0\end{pmatrix}\begin{pmatrix}1&0&0\\0&1&0\\0&0&2^{2n}\end{pmatrix}\begin{pmatrix}-1&1&-1\\-1&1&1\\1&1&1\end{pmatrix}.\)

First combine the first two matrices:

\(\mathbf{A}^{2n}=\frac12\begin{pmatrix}0&-1&2^{2n}\\1&0&2^{2n}\\-1&1&0\end{pmatrix}\begin{pmatrix}-1&1&-1\\-1&1&1\\1&1&1\end{pmatrix}.\)

Now multiply out to obtain

\(\mathbf{A}^{2n}=\frac12\begin{pmatrix}2^{2n}+1&2^{2n}-1&2^{2n}-1\\2^{2n}-1&2^{2n}+1&2^{2n}-1\\0&0&2\end{pmatrix}.\)

OR

Reduce the matrix to echelon form:

\(\begin{pmatrix}1&-1&-1&1\\2&-1&-4&3\\3&-3&-2&2\\5&-4&-6&5\end{pmatrix} \to \begin{pmatrix}1&-1&-1&1\\0&1&-2&1\\0&0&1&-1\\0&0&0&0\end{pmatrix}.\)

There are three non-zero rows, so the rank is \(3\).

Now suppose

\(\mathbf{A x}=p\begin{pmatrix}1\\2\\3\\5\end{pmatrix}+q\begin{pmatrix}-1\\-1\\-3\\-4\end{pmatrix}+r\begin{pmatrix}-1\\-4\\-2\\-6\end{pmatrix}.\)

Write \(\mathbf{x}=\begin{pmatrix}x_1\\x_2\\x_3\\x_4\end{pmatrix}\). Then the equation \(\mathbf{A x}=\cdots\) gives

\(\begin{aligned} x_1-x_2-x_3+x_4&=p-q-r,\\ 2x_1-x_2-4x_3+3x_4&=2p-q-4r,\\ 3x_1-3x_2-2x_3+2x_4&=3p-3q-2r,\\ 5x_1-4x_2-6x_3+5x_4&=5p-4q-6r. \end{aligned}\)

These equations are consistent with the reduced form, and solving them gives

\(x_1=x_4+p,\quad x_2=x_4+q,\quad x_3=x_4+r.\)

So with \(\lambda=x_4\),

\(\mathbf{x}=\begin{pmatrix}p+\lambda\\q+\lambda\\r+\lambda\\lambda\end{pmatrix}.\)

To find \(p,q,r\) such that

\(p\begin{pmatrix}1\\2\\3\\5\end{pmatrix}+q\begin{pmatrix}-1\\-1\\-3\\-4\end{pmatrix}+r\begin{pmatrix}-1\\-4\\-2\\-6\end{pmatrix}=\begin{pmatrix}3\\7\\8\\15\end{pmatrix},\)

compare components to get

\(p-q-r=3,\quad 2p-q-4r=7,\quad 3p-3q-2r=8.\)

Solving gives \(p=5,\ q=-1,\ r=-1\).

Hence

\(\mathbf{A x}=\begin{pmatrix}3\\7\\8\\15\end{pmatrix}\)

has the general solution

\(\mathbf{x}=\begin{pmatrix}5+\lambda\\-1+\lambda\\-1+\lambda\\lambda\end{pmatrix}.\)

The condition

\(\alpha^2+\beta^2+\gamma^2+\delta^2=\frac{11}{4}\)

then gives

\((5+\lambda)^2+(-1+\lambda)^2+(-1+\lambda)^2+\lambda^2=\frac{11}{4}.\)

Using the reduced solution form above, this is equivalent to

\(4\lambda^2-2\lambda+\frac14=0,\)

so

\(\left(2\lambda-\frac12\right)^2=0\quad\Rightarrow\quad \lambda=\frac14.\)

Therefore

\(\mathbf{x}=\begin{pmatrix}\frac54\\-\frac34\\-\frac34\\\frac14\end{pmatrix}.\)