Answer: If \(\mathbf e\) is a common eigenvector of \(\mathbf A\) and \(\mathbf B\), then it is an eigenvector of \(\mathbf A+\mathbf B\) with eigenvalue \(\lambda+\mu\).

For the given common eigenvectors, one possible diagonalisation is

\(\mathbf P=\begin{pmatrix}1&1&1\\1&2&-1\\-2&-3&1\end{pmatrix},\qquad \mathbf D=\begin{pmatrix}81&0&0\\0&256&0\\0&0&1296\end{pmatrix},\)

so \((\mathbf A+\mathbf B)^4=\mathbf P\mathbf D\mathbf P^{-1}\).

Suppose

\(\mathbf A\mathbf e=\lambda\mathbf e,\qquad \mathbf B\mathbf e=\mu\mathbf e.\)

Then

\((\mathbf A+\mathbf B)\mathbf e=\mathbf A\mathbf e+\mathbf B\mathbf e=\lambda\mathbf e+\mu\mathbf e=(\lambda+\mu)\mathbf e.\)

Thus \(\mathbf e\) is an eigenvector of \(\mathbf A+\mathbf B\) with eigenvalue \(\lambda+\mu\).

For each given common eigenvector, add the corresponding eigenvalues of \(\mathbf A\) and \(\mathbf B\). The resulting eigenvalues of \(\mathbf A+\mathbf B\) are \(3,4,6\).

Therefore the eigenvalues of \((\mathbf A+\mathbf B)^4\) are

\(3^4=81,\qquad 4^4=256,\qquad 6^4=1296.\)

Taking the common eigenvectors as columns, one possible choice is

\(\mathbf P=\begin{pmatrix}1&1&1\\1&2&-1\\-2&-3&1\end{pmatrix}.\)

Then

\((\mathbf A+\mathbf B)^4=\mathbf P\begin{pmatrix}81&0&0\\0&256&0\\0&0&1296\end{pmatrix}\mathbf P^{-1}.\)

The order of the columns of \(\mathbf P\) may be changed, provided the diagonal entries are changed in the same order.