Question

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

Short answer

The required answers are

$$\begin{gathered} M^4=\left(\begin{array}{cc}136& 120\\ -120& 136\end{array}\right) \\ {M}^{10}=\left(\begin{array}{cc}524800& -523776\\ -523776& 524800\end{array}\right) \\ {e}^M=e^3\left(\begin{array}{cc}\cosh 1& -\sinh 1\\ -\sinh 1& \cosh 1\end{array}\right) \end{gathered}$$

Step-by-step solution

Given information

The given matrix

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

Eigen values

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

 Find eigen values

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 eigen vector

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

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

General function of M

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

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