Answer: \(\mathbf{A}^{4}=10\mathbf{A}^{2}-9\mathbf{I}\).

Hence \(p=10\) and \(q=-9\).

Since \(\mathbf{A}\) is upper triangular, its eigenvalues are the diagonal entries \(-1\), \(1\) and \(3\).

So the characteristic equation is

\((\lambda+1)(\lambda-1)(\lambda-3)=0\).

Expanding,

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

Using the characteristic equation for \(\mathbf{A}\),

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

Hence

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

Multiplying the matrix equation \(\mathbf{A}^3-3\mathbf{A}^2-\mathbf{A}+3\mathbf{I}=\mathbf{0}\) by \(\mathbf{A}\),

\(\mathbf{A}^4-3\mathbf{A}^3-\mathbf{A}^2+3\mathbf{A}=\mathbf{0}\),

so

\(\mathbf{A}^4=3\mathbf{A}^3+\mathbf{A}^2-3\mathbf{A}\).

Now substitute for \(\mathbf{A}^3\):

\(\mathbf{A}^4=3(3\mathbf{A}^2+\mathbf{A}-3\mathbf{I})+\mathbf{A}^2-3\mathbf{A}\).

Simplifying,

\(\mathbf{A}^4=9\mathbf{A}^2+3\mathbf{A}-9\mathbf{I}+\mathbf{A}^2-3\mathbf{A}=10\mathbf{A}^2-9\mathbf{I}\).

Therefore \(\mathbf{A}^{4}=p\mathbf{A}^{2}+q\mathbf{I}\) with \(p=10\) and \(q=-9\).