Answer:

Either

The reduction formula is \((2n+1)I_n=2nI_{n-1}+2^n\).

\(I_0=\int_0^1 1\,dx=1\).

Now use the formula repeatedly:

For \(n=1\): \(3I_1=2I_0+2=4\), so \(I_1=\frac{4}{3}\).

For \(n=2\): \(5I_2=4I_1+4=\frac{16}{3}+4=\frac{28}{3}\), so \(I_2=\frac{28}{15}\).

For \(n=3\): \(7I_3=6I_2+8=\frac{56}{5}+8=\frac{96}{5}\), so \(I_3=\frac{96}{35}\).

To find \(I_{-3}\), write the formula as \(2nI_{n-1}=(2n+1)I_n-2^n\).

For \(n=-1\): \(-2I_{-2}=-I_{-1}-\frac12\), so \(I_{-2}=\frac{1}{2}\left(I_{-1}+\frac12\right)=\frac{\pi}{8}+\frac14\).

For \(n=-2\): \(-4I_{-3}=-3I_{-2}-\frac14\), so \(I_{-3}=\frac{3I_{-2}+\frac14}{4}=\frac{3\pi}{32}+\frac14\).

Final answers: \(I_0=1\), \(I_3=\frac{96}{35}\), \(I_{-3}=\frac{3\pi}{32}+\frac14\).

Or

If \(\mathbf{e}\) is an eigenvector of \(\mathbf{A}\) and \(\mathbf{B}\) with eigenvalues \(\lambda\) and \(\mu\), then \(\mathbf{Ae}=\lambda\mathbf{e}\) and \(\mathbf{Be}=\mu\mathbf{e}\). Hence

\((\mathbf{A}+\mathbf{B})\mathbf{e}=\mathbf{Ae}+\mathbf{Be}=(\lambda+\mu)\mathbf{e}\),

so \(\lambda+\mu\) is an eigenvalue of \(\mathbf{A}+\mathbf{B}\).

For \(\mathbf{A}\):

\(\mathbf{A}\begin{pmatrix}1\\1\\1\end{pmatrix}=\begin{pmatrix}-1\\-1\\-1\end{pmatrix}\), so the eigenvalue is \(-1\).

\(\mathbf{A}\begin{pmatrix}1\\-1\\1\end{pmatrix}=\begin{pmatrix}1\\-1\\1\end{pmatrix}\), so the eigenvalue is \(1\).

\(\mathbf{A}\begin{pmatrix}2\\0\\1\end{pmatrix}=\begin{pmatrix}6\\0\\3\end{pmatrix}=3\begin{pmatrix}2\\0\\1\end{pmatrix}\), so the eigenvalue is \(3\).

For \(\mathbf{M}=\mathbf{A}+\mathbf{B}-5\mathbf{I}\), the eigenvalues are obtained by adding the eigenvalues of \(\mathbf{A}\) and \(\mathbf{B}\), then subtracting 5. Thus they are \((-1)+(-2)-5=-8\), \(1+2-5=-2\), and \(3+4-5=2\).

Take the eigenvectors as columns of

\(\mathbf{R}=\begin{pmatrix}1&1&2\\1&-1&0\\1&1&1\end{pmatrix}\).

Then \(\mathbf{S}=\mathbf{R}^{-1}=\frac12\begin{pmatrix}-1&1&2\\-1&-1&2\\2&0&-2\end{pmatrix}\), and

\(\mathbf{D}=\begin{pmatrix}-32768&0&0\\0&-32&0\\0&0&32\end{pmatrix}\).

Therefore \(\mathbf{M}^5=\mathbf{RDS}\).

Either

Let \(I_n=\int_0^1 (1+x^2)^n\,dx\).

To derive a recurrence, integrate by parts with \(u=(1+x^2)^n\) and \(dv=dx\). Then \(du=2nx(1+x^2)^{n-1}\,dx\) and \(v=x\).

So

\(I_n=\left[x(1+x^2)^n\right]_0^1-\int_0^1 x\cdot 2nx(1+x^2)^{n-1}\,dx\).

This gives

\(I_n=2^n-2n\int_0^1 x^2(1+x^2)^{n-1}\,dx\).

Now write \(x^2=(1+x^2)-1\):

\(I_n=2^n-2n\int_0^1 \big((1+x^2)-1\big)(1+x^2)^{n-1}\,dx\).

Hence

\(I_n=2^n-2nI_n+2nI_{n-1}\).

Rearranging gives

\((2n+1)I_n=2nI_{n-1}+2^n\).

Now evaluate \(I_0\):

\(I_0=\int_0^1 1\,dx=1\).

Use the recurrence for \(n=1,2,3\):

For \(n=1\), \(3I_1=2I_0+2=4\), so \(I_1=\frac43\).

For \(n=2\), \(5I_2=4I_1+4=\frac{16}{3}+4=\frac{28}{3}\), so \(I_2=\frac{28}{15}\).

For \(n=3\), \(7I_3=6I_2+8=6\cdot\frac{28}{15}+8=\frac{56}{5}+8=\frac{96}{5}\), so

\(I_3=\frac{96}{35}\).

To find \(I_{-3}\), first rewrite the recurrence as

\(2nI_{n-1}=(2n+1)I_n-2^n\).

For \(n=-1\):

\(-2I_{-2}=-I_{-1}-\frac12\).

Given \(I_{-1}=\frac\pi4\), this gives

\(I_{-2}=\frac{1}{2}\left(\frac\pi4+\frac12\right)=\frac{\pi}{8}+\frac14\).

For \(n=-2\):

\(-4I_{-3}=-3I_{-2}-\frac14\).

So

\(I_{-3}=\frac{3I_{-2}+\frac14}{4}=\frac{3\left(\frac{\pi}{8}+\frac14\right)+\frac14}{4}=\frac{3\pi}{32}+\frac14\).

Therefore \(I_0=1\), \(I_3=\frac{96}{35}\), and \(I_{-3}=\frac{3\pi}{32}+\frac14\).

Or

If \(\mathbf{e}\) is an eigenvector of \(\mathbf{A}\) with eigenvalue \(\lambda\), and of \(\mathbf{B}\) with eigenvalue \(\mu\), then

\(\mathbf{Ae}=\lambda\mathbf{e}\) and \(\mathbf{Be}=\mu\mathbf{e}\).

Adding these gives

\((\mathbf{A}+\mathbf{B})\mathbf{e}=\mathbf{Ae}+\mathbf{Be}=(\lambda+\mu)\mathbf{e}\).

Since \(\mathbf{e}\neq\mathbf{0}\), \(\lambda+\mu\) is an eigenvalue of \(\mathbf{A}+\mathbf{B}\).

For \(\mathbf{A}\), use each given eigenvector.

\(\mathbf{A}\begin{pmatrix}1\\1\\1\end{pmatrix}=\begin{pmatrix}-1\\-1\\-1\end{pmatrix}=-1\begin{pmatrix}1\\1\\1\end{pmatrix}\), so one eigenvalue is \(-1\).

\(\mathbf{A}\begin{pmatrix}1\\-1\\1\end{pmatrix}=\begin{pmatrix}1\\-1\\1\end{pmatrix}=1\begin{pmatrix}1\\-1\\1\end{pmatrix}\), so one eigenvalue is \(1\).

\(\mathbf{A}\begin{pmatrix}2\\0\\1\end{pmatrix}=\begin{pmatrix}6\\0\\3\end{pmatrix}=3\begin{pmatrix}2\\0\\1\end{pmatrix}\), so one eigenvalue is \(3\).

For \(\mathbf{M}=\mathbf{A}+\mathbf{B}-5\mathbf{I}\), the eigenvectors are the same, and the eigenvalues are shifted by \(-5\).

Since the eigenvalues of \(\mathbf{B}\) are \(-2,2,4\), the eigenvalues of \(\mathbf{M}\) are

\((-1)+(-2)-5=-8,\quad 1+2-5=-2,\quad 3+4-5=2.\)

Take the eigenvectors in the order corresponding to these eigenvalues. Then

\(\mathbf{R}=\begin{pmatrix}1&1&2\\1&-1&0\\1&1&1\end{pmatrix}\).

Its inverse is

\(\mathbf{S}=\mathbf{R}^{-1}=\frac12\begin{pmatrix}-1&1&2\\-1&-1&2\\2&0&-2\end{pmatrix}\).

The diagonal matrix of fifth powers of the eigenvalues is

\(\mathbf{D}=\begin{pmatrix}(-8)^5&0&0\\0&(-2)^5&0\\0&0&2^5\end{pmatrix}=\begin{pmatrix}-32768&0&0\\0&-32&0\\0&0&32\end{pmatrix}\).

Hence \(\mathbf{M}^5=\mathbf{RDS}\).