Question

Note in Section 6 [see (6.15)] that, for the given matrix A, we found ${\mathit{A}}^{\mathbf{2}}\mathbf{=}\mathbf{-}\mathit{I}$, so it was easy to find all the powers of A. It is not usually this easy to find high powers of a matrix directly. Try it for the square matrix Min equation (11.1). Then use the method outlined in Problem 57 to find${\mathit{M}}^{\mathbf{4}}\mathbf{,}{\mathit{M}}^{\mathbf{10}}\mathbf{.}{\mathit{e}}^{\mathbf{M}}$.

Short answer

The required answers are

$$\begin{gathered} M^4=\left(\begin{array}{cc}1037& -518\\ -518& 260\end{array}\right) \\ {M}^{10}=\left(\begin{array}{cc}48372941& -24186470\\ -24186470& 12093236\end{array}\right) \\ {e}^M=\frac{e}{5}\left(\begin{array}{cc}1+4e^5& 2(1-e^5)\\ 2(1-e^5)& 4+e^5\end{array}\right) \end{gathered}$$

Step-by-step solution

Given information

The given matrix is$M=\left(\begin{array}{cc}5& -2\\ -2& 2\end{array}\right)$

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.

Calculate eigen value

Calculate the functions of this matrix, which requires the knowledge of all powers of M.

Calculate a couple of powers

$$\begin{gathered} M^2={\left(\begin{array}{cc}5& -2\\ -2& 2\end{array}\right)}^2 \\ =\left(\begin{array}{cc}29& 14\\ -14& 8\end{array}\right) \\ {\mathrm{M}}^3=\left(\begin{array}{cc}5& -2\\ -2& 2\end{array}\right){\left(\begin{array}{cc}5& -2\\ -2& 2\end{array}\right)}^2 \\ =\left(\begin{array}{cc}173& 86\\ -86& 44\end{array}\right) \end{gathered}$$

This is very complicated, so use the fact that

$M^n=C{D}^n{C}^{-1}$

Therefore, diagonalize the matrix M. Solve the characteristic equation

$$\begin{gathered} \left|\begin{array}{cc}5-\lambda & -2\\ -2& 2-\lambda \end{array}\right|=0 \\ \left(5-\lambda \right)\left(2-\lambda \right)-4=0 \\ {\lambda }^2-7\lambda +6=0 \end{gathered}$$

The eigenvalues are 1 and 6.

Find inverse of matrix

The eigenvector corresponding to $\lambda =1$is obtained as

role="math" localid="1658822190844" $$\begin{gathered} \left(\begin{array}{cc}5& -2\\ -2& 2\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{c}x\\ y\end{array}\right) \\ \left(\begin{array}{cc}4& -2\\ -2& 1\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)=0 \end{gathered}$$

This gives the equation

$-2x+y=0$

So, the eigenvector is

$v_1=\frac{1}{\sqrt{5}}\left(\frac{1}{2}\right)$

The eigenvector corresponding to$\lambda =6$is obtained as

$$\begin{gathered} \left(\begin{array}{cc}5& -2\\ -2& 2\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)=6\left(\begin{array}{c}x\\ y\end{array}\right) \\ \left(\begin{array}{cc}-1& -2\\ -2& -4\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)=0 \end{gathered}$$

This gives the equation

x+2y=0

So, the eigenvector is

$v_6=\frac{1}{\sqrt{5}}\left(\begin{array}{c}-2\\ 1\end{array}\right)$

Therefore, the matrix C is

$C=\frac{1}{\sqrt{5}}\left(\begin{array}{cc}1& -2\\ 2& 1\end{array}\right)$

and its inverse is easily obtained because it is a$2\times 2$ matrix as

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

Since the determinant is 1,

$C^{-1}=\frac{1}{\sqrt{5}}\left(\begin{array}{cc}1& 2\\ -2& 1\end{array}\right)$

General function of M

Obtain the general function of M as

$$\begin{gathered} f\left(M\right)=\underset{n}{\sum a_n{M}^n} \\ =C\left(\sum _n{a}_n{D}^n\right)C^{-1} \\ =\frac{1}{5}\left(\begin{array}{cc}1& -2\\ 2& 1\end{array}\right)\left(\begin{array}{cc}f\left(1\right)& 0\\ 0& f\left(6\right)\end{array}\right)\left(\begin{array}{cc}1& 2\\ -2& 1\end{array}\right) \\ =\frac{1}{5}\left(\begin{array}{cc}f\left(1\right)& -2f\left(6\right)\\ 2f\left(6\right)& f\left(6\right)\end{array}\right)\left(\begin{array}{cc}1& 2\\ -2& 1\end{array}\right) \end{gathered}$$

Solve further

$\frac{1}{5}\left(\begin{array}{cc}f\left(1\right)+4f\left(6\right)& -2f\left(1\right)-2f\left(6\right)\\ 2f\left(1\right)-2f\left(6\right)& 4f\left(1\right)+f\left(6\right)\end{array}\right)$

We now have

$$\begin{gathered} M^4=\frac{1}{5}\left(\begin{array}{cc}{1}^4+4.6^4& 2.1^4-2.6^4\\ 2.1^4-2.6^4& 4.1^4+6^4\end{array}\right) \\ =\left(\begin{array}{cc}1037& -518\\ -518& 260\end{array}\right), \\ {M}^{10}=\frac{1}{5}\left(\begin{array}{cc}{1}^4+4.6^4& 2.1^{10}-2.6^{10}\\ 2.1^{10}-2.6^{10}& 4.1^{10}+6^{10}\end{array}\right) \\ =\left(\begin{array}{cc}48372941& -24186470\\ -24186470& 12093236\end{array}\right) \end{gathered}$$

Solve further

$$\begin{gathered} e^M=\frac{1}{5}\left(\begin{array}{cc}e+4e^6& 2e-2e^6\\ 2e-2e^6& 4e+e^6\end{array}\right) \\ =\frac{e}{5}\left(\begin{array}{cc}1+4e^6& 2\left(1-2e^5\right)\\ 2\left(1-2e^6\right)& 4e+e^5\end{array}\right) \end{gathered}$$