Answer: (a) The system has a unique solution when the determinant of the coefficient matrix is non-zero. This gives \(-8a-24 \neq 0\), so \(a \neq -3\).

(b) The system is consistent for all values of \(a\).

If \(a \neq -3\), it has a unique solution, so it is consistent.

If \(a=-3\), the first two equations are

\(6x-3y=3\) and \(2x-y=1\),

and the first is just \(3\) times the second, so there are only two independent equations in three unknowns. Hence there are infinitely many solutions, so the system is still consistent.

(c) A suitable choice is

\(\mathbf{P}=\left(\begin{array}{ccc}14&0&0\\4&-1&0\\17&1&1\end{array}\right)\), \(\mathbf{D}=\left(\begin{array}{ccc}11664&0&0\\0&100&0\\0&0&6400\end{array}\right)\).

(d) The characteristic equation of \(\mathbf{A}\) is

\(\lambda^3-9\lambda^2+14\lambda+24=0\).

Hence \(\mathbf{A}^3-9\mathbf{A}^2+14\mathbf{A}+24\mathbf{I}=\mathbf{0}\), so

\(14\mathbf{A}+24\mathbf{I}=-\mathbf{A}^3+9\mathbf{A}^2\).

Therefore

\((14\mathbf{A}+24\mathbf{I})^2=\left(-\mathbf{A}^3+9\mathbf{A}^2\right)^2=\mathbf{A}^4(\mathbf{A}-9\mathbf{I})^2\),

so \(b=-9\).

(a) The coefficient matrix is

\(\left(\begin{array}{ccc}6&a&0\\2&-1&0\\1&5&4\end{array}\right)\).

A unique solution exists exactly when its determinant is non-zero.

Expanding along the third column,

\(\det\left(\begin{array}{ccc}6&a&0\\2&-1&0\\1&5&4\end{array}\right)=4\det\left(\begin{array}{cc}6&a\\2&-1\end{array}\right)=4(-6-2a)=-24-8a.\)

So we need

\(-24-8a\neq 0\),

which gives

\(a\neq -3\).

(b) If \(a\neq -3\), then from part (a) the system has a unique solution, so it is consistent.

Now consider \(a=-3\). The equations become

\(6x-3y=3\),

\(2x-y=1\),

\(x+5y+4z=2\).

The first equation is just \(3\) times the second, so there are only two distinct equations in three unknowns. Hence one variable is free, giving infinitely many solutions. Therefore the system is consistent when \(a=-3\) as well.

So the system is consistent for all values of \(a\).

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

\(6,-1,4\).

We find corresponding eigenvectors.

For \(\lambda=6\), solve \((\mathbf{A}-6\mathbf{I})\mathbf{v}=\mathbf{0}\):

\(\mathbf{A}-6\mathbf{I}=\left(\begin{array}{ccc}0&0&0\\2&-7&0\\1&5&-2\end{array}\right).\)

This gives

\(2x-7y=0\), \(x+5y-2z=0\).

Choose \(y=4\), then \(x=14\), and \(14+20-2z=0\Rightarrow z=17\).

So one eigenvector is

\(\left(\begin{array}{c}14\\4\\17\end{array}\right).\)

For \(\lambda=-1\), solve \((\mathbf{A}+\mathbf{I})\mathbf{v}=\mathbf{0}\):

\(\mathbf{A}+\mathbf{I}=\left(\begin{array}{ccc}7&0&0\\2&0&0\\1&5&5\end{array}\right).\)

This gives \(x=0\) and \(y=-z\). Taking \(z=1\), an eigenvector is

\(\left(\begin{array}{c}0\\-1\\1\end{array}\right).\)

For \(\lambda=4\), solve \((\mathbf{A}-4\mathbf{I})\mathbf{v}=\mathbf{0}\):

\(\mathbf{A}-4\mathbf{I}=\left(\begin{array}{ccc}2&0&0\\2&-5&0\\1&5&0\end{array}\right).\)

This gives \(x=0\), \(y=0\), and \(z\) free. Taking \(z=1\), an eigenvector is

\(\left(\begin{array}{c}0\\0\\1\end{array}\right).\)

Hence we may take

\(\mathbf{P}=\left(\begin{array}{ccc}14&0&0\\4&-1&0\\17&1&1\end{array}\right).\)

If \(\lambda\) is an eigenvalue of \(\mathbf{A}\), then the corresponding eigenvalue of \((14\mathbf{A}+24\mathbf{I})^2\) is \((14\lambda+24)^2\).

So the diagonal entries are

\((14\cdot 6+24)^2=108^2=11664\),

\((14\cdot (-1)+24)^2=10^2=100\),

\((14\cdot 4+24)^2=80^2=6400\).

Therefore

\(\mathbf{D}=\left(\begin{array}{ccc}11664&0&0\\0&100&0\\0&0&6400\end{array}\right),\)

and

\((14\mathbf{A}+24\mathbf{I})^2=\mathbf{PDP}^{-1}.\)

(d) The characteristic equation of \(\mathbf{A}\) is

\((\lambda-6)(\lambda+1)(\lambda-4)=\lambda^3-9\lambda^2+14\lambda+24=0.\)

By Cayley-Hamilton,

\(\mathbf{A}^3-9\mathbf{A}^2+14\mathbf{A}+24\mathbf{I}=\mathbf{0}.\)

Rearranging,

\(14\mathbf{A}+24\mathbf{I}=-\mathbf{A}^3+9\mathbf{A}^2.\)

Now square both sides:

\((14\mathbf{A}+24\mathbf{I})^2=\left(-\mathbf{A}^3+9\mathbf{A}^2\right)^2.\)

Factorising the right-hand side,

\((14\mathbf{A}+24\mathbf{I})^2=\mathbf{A}^4(\mathbf{A}-9\mathbf{I})^2.\)

This is of the form \(\mathbf{A}^4(\mathbf{A}+b\mathbf{I})^2\), so

\(b=-9\).