## 4 Answers

Characteristic Equation is $ |A -\mu I| = 0$

$ \begin{vmatrix} 1-\mu & 1\\1 &-1-\mu\end{vmatrix} = 0$

$\implies (1-\mu)(-1-\mu)-1=0$

$\implies -1-\mu+\mu+\mu^{2}-1=0$

$\implies \mu^{2}-2=0$

$\implies \mu=+\sqrt 2 $ and $-\sqrt 2$

According to properties of Eigen values,

eigen values of $A^{19}=$ (eigen value of A)$^{19}$

$=(\sqrt 2)^{19}$ and $(-\sqrt 2)^{19}$

Hence, Ans is **option (D).**

At the time of solving I just forgot this property & get two method, one method is working fine but other is not,but can't understand why the other method is not working.

**1st Method:**

By using Cayley-hamilton Theorem from the characteristic eqn I get $\lambda ^{2} - 2 = 0$ which is $A^{2} - 2 = 0$

$A^{2} = 2I \Rightarrow ({A^{2}})^9 = 2^{9}I$

i.e $A^{18}=512I$, Now post-multiplied by A on both side $A^{18}.A=512I.A$. which becomes $A^{19}=\begin{bmatrix} 512 &512 \\ 512&-512 \end{bmatrix}$

computing eigen values here we'll get $512\sqrt2$ & $-512\sqrt2$

This method is ok & it's working.

**2nd method:**

finding a pattern to compute A^19 which 1st compute A^2 by A*A , then A^3 by A^2 * A, then A^4 by A^3 * A

by this we'll get a pattern like,

$A^{n} =\begin{bmatrix} n &0 \\ 0 &n \end{bmatrix}$ when n is even

$A^{n} =\begin{bmatrix} n-1 &n-1 \\ n-1 &1-n \end{bmatrix}$ when n is odd

likewise $A^{19} =\begin{bmatrix} 18 &18 \\ 18 &-18 \end{bmatrix}$

but computing eigen values on $A^{19}$ will give $-18\sqrt2$ & $18\sqrt2$

don't know why this 2nd method is not working,Anyone please give a reason why this is not working.