Repeat the last part of Problem for the matrix $\mathit{M}\mathbf{=}\left(\begin{array}{cc}3& -1\\ -1& 3\end{array}\right)$

The given matrix

$M=\left(\begin{array}{cc}3& -1\\ -1& 3\end{array}\right)$

Eigen values are the special set of scalar values that is associated with the set of linear equations most probably in the matrix equations.

Calculate functions of the matrix

$M=\left(\begin{array}{cc}3& -1\\ -1& 3\end{array}\right)$

Use the formula

${\mathrm{M}}^{\mathrm{n}}={\mathrm{CD}}^{\mathrm{n}}{\mathrm{C}}^{-1}$

First diagonalize the matrix, then solve the equation

$$\begin{gathered} \left|\begin{array}{cc}3-\lambda & -1\\ -1& 3-\lambda \end{array}\right|=0 \\ {(3-\lambda )}^2-1=0 \\ 3-\lambda =\underline{+}1 \\ \lambda =3\underline{+}1 \end{gathered}$$

We see that the eigenvalues are $2$and $4$.

The eigenvector corresponding to $\lambda =2$is obtained as

$$\begin{gathered} \left(\begin{array}{cc}3& -1\\ -1& 3\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)=2\left(\begin{array}{c}x\\ y\end{array}\right) \\ \left(\begin{array}{cc}1& -1\\ -1& 1\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)=0 \end{gathered}$$

which gives the equation

$x-y=0$

The eigenvector is therefore

$v_2=\frac{1}{\sqrt{2}}\left(\begin{array}{c}1\\ 1\end{array}\right)$

The eigenvector corresponding to$\lambda =4$ is obtained as

$$\begin{gathered} \left(\begin{array}{cc}3& -1\\ -1& 3\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)=4\left(\begin{array}{c}x\\ y\end{array}\right) \\ \left(\begin{array}{cc}1& -1\\ -1& 1\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)=0 \end{gathered}$$

which gives the equation

$x+y=0$

The eigenvector is therefore

$v_2=\frac{1}{\sqrt{2}}\left(\begin{array}{c}1\\ -1\end{array}\right)$

The matrix is therefore

$c=\frac{1}{\sqrt{2}}\left(\begin{array}{cc}1& 1\\ 1& -1\end{array}\right)$

and its inverse is obtained as

$$\begin{gathered} C^{-1}=\frac{1}{detC}\left(\begin{array}{cc}{c}_{22}& -c_{21}\\ -c_{12}& c_{11}\end{array}\right) \\ =\frac{1}{\sqrt{2}}\left(\begin{array}{cc}1& 1\\ 1& -1\end{array}\right) \end{gathered}$$

A general function of $M$is obtained as

$$\begin{gathered} f\left(M\right)=\sum _n{a}_n{M}^n \\ =C\left(\sum _n{a}_n{D}^n\right)C^{-1} \\ =\frac{1}{2}\left(\begin{array}{cc}1& 1\\ 1& -1\end{array}\right)\left(\begin{array}{cc}f\left(2\right)& 0\\ 0& -f\left(4\right)\end{array}\right)\left(\begin{array}{cc}1& 1\\ 1& -1\end{array}\right) \\ =\frac{1}{2}\left(\begin{array}{cc}f\left(2\right)+f\left(4\right)& f\left(2\right)-f\left(4\right)\\ f\left(2\right)-f\left(4\right)& f\left(2\right)+f\left(4\right)\end{array}\right) \end{gathered}$$

Therefore,

$$\begin{gathered} M^4=\frac{1}{2}\left(\begin{array}{cc}{2}^4+2^4& 2^4-2^4\\ 2^4-4^4& 2^4+2^4\end{array}\right) \\ =\left(\begin{array}{cc}136& -120\\ -120& 136\end{array}\right) \\ {M}^{10}=\frac{1}{2}\left(\begin{array}{cc}{2}^{10}+4^{10}& 2^{10}-4^{10}\\ 2^{10}-4^{10}& 2^{10}+4^{10}\end{array}\right) \end{gathered}$$

Solve further

$$\begin{gathered} e^M=\frac{1}{2}\left(\begin{array}{cc}{e}^2+e^4& e^2-e^4\\ e^2-e^4& e^2+e^4\end{array}\right) \\ =\frac{e^3}{2}\left(\begin{array}{cc}{e}^{-1}+e^1& e^{-1}-e^1\\ e^{-1}-e^1& e^{-1}+e^1\end{array}\right) \\ =e^3\left(\begin{array}{cc}\cosh 1& -\sinh 1\\ -\sinh 1& \cosh 1\end{array}\right) \end{gathered}$$