Answer: First alternative

(i) \(\alpha+\beta+\gamma+\delta=-4\)

(ii) \(\alpha^2+\beta^2+\gamma^2+\delta^2=12\)

(iii) \(\dfrac1\alpha+\dfrac1\beta+\dfrac1\gamma+\dfrac1\delta=4\)

(iv) \(\dfrac{\alpha}{\beta\gamma\delta}+\dfrac{\beta}{\alpha\gamma\delta}+\dfrac{\gamma}{\alpha\beta\delta}+\dfrac{\delta}{\alpha\beta\gamma}=12\)

With \(y=x+1\), the equation becomes \(y^4-4y^2+4=0\), so \((y^2-2)^2=0\). Hence \(y=\pm\sqrt2\), each repeated twice, and therefore the roots are \(x=\sqrt2-1\) (twice) and \(x=-\sqrt2-1\) (twice).

Second alternative

(i) If \(\mathbf{A}\mathbf{e}=\lambda\mathbf{e}\) and \(\mathbf{A}\) is non-singular, then \(\mathbf{A}\mathbf{e}\neq\mathbf{0}\), so \(\lambda\neq 0\).

(ii) Multiplying \(\mathbf{A}\mathbf{e}=\lambda\mathbf{e}\) by \(\mathbf{A}^{-1}\) gives \(\mathbf{e}=\lambda\mathbf{A}^{-1}\mathbf{e}\), hence \(\mathbf{A}^{-1}\mathbf{e}=\lambda^{-1}\mathbf{e}\).

For \(\mathbf{B}=(\mathbf{A}+3\mathbf{I})^{-1}\), one suitable diagonalisation is

\(\mathbf{B}=\mathbf{PDP}^{-1}\), with

\(\mathbf{P}=\begin{pmatrix}1&2&-6\\0&1&25\\0&0&20\end{pmatrix}\), \(\mathbf{D}=\begin{pmatrix}1&0&0\\0&\tfrac12&0\\0&0&\tfrac16\end{pmatrix}\).

First alternative

For \(x^4+4x^3+2x^2-4x+1=0\) with roots \(\alpha,\beta,\gamma,\delta\), use the relations between roots and coefficients.

Since the coefficient of \(x^3\) is \(4\),

\(\alpha+\beta+\gamma+\delta=-4\).

Also

\(\alpha\beta+\alpha\gamma+\alpha\delta+\beta\gamma+\beta\delta+\gamma\delta=2\).

Therefore

\(\alpha^2+\beta^2+\gamma^2+\delta^2=(\alpha+\beta+\gamma+\delta)^2-2(\alpha\beta+\alpha\gamma+\alpha\delta+\beta\gamma+\beta\delta+\gamma\delta) = (-4)^2-2\cdot2=12\).

Also

\(\alpha\beta\gamma\delta=1\).

So

\(\dfrac1\alpha+\dfrac1\beta+\dfrac1\gamma+\dfrac1\delta =\dfrac{\beta\gamma\delta+\alpha\gamma\delta+\alpha\beta\delta+\alpha\beta\gamma}{\alpha\beta\gamma\delta} =\dfrac{-(-4)}1=4\).

For part (iv),

\(\dfrac{\alpha}{\beta\gamma\delta}+\dfrac{\beta}{\alpha\gamma\delta}+\dfrac{\gamma}{\alpha\beta\delta}+\dfrac{\delta}{\alpha\beta\gamma} =\dfrac{\alpha^2+\beta^2+\gamma^2+\delta^2}{\alpha\beta\gamma\delta} =\dfrac{12}{1}=12\).

Now use \(y=x+1\), so \(x=y-1\).

Substitute into the quartic:

\((y-1)^4+4(y-1)^3+2(y-1)^2-4(y-1)+1=0\).

Expand in parts:

\((y-1)^4+4(y-1)^3=y^4-6y^2+8y-3\),

\(2(y-1)^2-4(y-1)+1=2y^2-8y+7\).

So the equation becomes

\(y^4-4y^2+4=0\).

Hence

\((y^2-2)^2=0\).

Therefore \(y=\pm\sqrt2\), each repeated twice.

Since \(x=y-1\), the roots are

\(x=\sqrt2-1\) (twice), \(x=-\sqrt2-1\) (twice).

Second alternative

If \(\mathbf{A}\mathbf{e}=\lambda\mathbf{e}\) and \(\mathbf{A}\) is non-singular, then \(\mathbf{e}\neq\mathbf{0}\) and \(\mathbf{A}\mathbf{e}\neq\mathbf{0}\). So \(\lambda\mathbf{e}\neq\mathbf{0}\), which gives \(\lambda\neq0\).

Now multiply \(\mathbf{A}\mathbf{e}=\lambda\mathbf{e}\) on the left by \(\mathbf{A}^{-1}\):

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

Thus

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

so

\(\mathbf{A}^{-1}\mathbf{e}=\dfrac1\lambda\mathbf{e}\).

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

For

\(\mathbf{A}=\begin{pmatrix}-2&2&-4\\0&-1&5\\0&0&3\end{pmatrix}\),

the eigenvalues are the diagonal entries, since \(\mathbf{A}\) is upper triangular:

\(-2,-1,3\).

Corresponding eigenvectors are:

for \(-2\): \(\begin{pmatrix}1\\0\\0\end{pmatrix}\),

for \(-1\): \(\begin{pmatrix}2\\1\\0\end{pmatrix}\),

for \(3\): \(\begin{pmatrix}-6\\25\\20\end{pmatrix}\).

So one suitable matrix is

\(\mathbf{P}=\begin{pmatrix}1&2&-6\\0&1&25\\0&0&20\end{pmatrix}\).

Now \(\mathbf{A}+3\mathbf{I}\) has eigenvalues \(1,2,6\), so \(\mathbf{B}=(\mathbf{A}+3\mathbf{I})^{-1}\) has eigenvalues \(1,\dfrac12,\dfrac16\), with the same eigenvectors.

Therefore

\(\mathbf{D}=\begin{pmatrix}1&0&0\\0&\dfrac12&0\\0&0&\dfrac16\end{pmatrix}\),

and

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