Answer: Either

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

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

(iii) \(\dfrac{1}{\alpha}+\dfrac{1}{\beta}+\dfrac{1}{\gamma}+\dfrac{1}{\delta}=4\)

(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 \(y=\pm\sqrt{2}\), each repeated, and therefore \(x=-1\pm\sqrt{2}\), each repeated twice.

Or

If \(\mathbf{A}\mathbf{e}=\lambda\mathbf{e}\) and \(\mathbf{A}\) is non-singular, then \(\lambda\neq 0\), and \(\mathbf{A}^{-1}\mathbf{e}=\lambda^{-1}\mathbf{e}\).

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

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

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

Either

For \(x^{4}+4x^{3}+2x^{2}-4x+1=0\) with roots \(\alpha,\beta,\gamma,\delta\), Vieta's formulas give

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

Also

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

So

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

The product of the roots is

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

and the sum of the triple products is

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

Hence

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

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 let \(y=x+1\), so \(x=y-1\). Substitute into the quartic:

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

Expanding,

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

and

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

Therefore

\(y^{4}-6y^{2}+8y-3+2y^{2}-8y+7=0\),

so

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

Thus

\((y^{2}-2)^{2}=0\),

so \(y=\pm\sqrt{2}\), each a repeated root.

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

\(x=-1+\sqrt{2}\) and \(x=-1-\sqrt{2}\), each repeated twice.

Or

If \(\mathbf{A}\mathbf{e}=\lambda\mathbf{e}\) and \(\mathbf{A}\) is non-singular, then \(\mathbf{e}\neq \mathbf{0}\). If \(\lambda=0\), then \(\mathbf{A}\mathbf{e}=\mathbf{0}\) for a non-zero vector \(\mathbf{e}\), which would make \(\mathbf{A}\) singular. Therefore \(\lambda\neq 0\).

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

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

So

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

hence

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

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

Now

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

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

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

Corresponding eigenvectors may be taken as

\(\begin{pmatrix}1\\0\\0\end{pmatrix},\;\begin{pmatrix}2\\1\\0\end{pmatrix},\;\begin{pmatrix}-6\\25\\20\end{pmatrix}.\)

So we can choose

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

The matrix \(\mathbf{A}+3\mathbf{I}\) has eigenvalues \(1,2,6\), so \(\mathbf{B}=(\mathbf{A}+3\mathbf{I})^{-1}\) has eigenvalues

\(1,\dfrac12,\dfrac16\).

Therefore

\(\mathbf{D}=\begin{pmatrix}1&0&0\\0&\tfrac12&0\\0&0&\tfrac16\end{pmatrix},\)

and hence

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