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Determine whether these matrices are diagonalizable or not. If they are
diagonalizable, find $\mathrm{P}$ and $\mathrm{D}$ and compute $A^{6} .$

$A_{1}=\left(\begin{array}{ll}{13} & {-10} \\ {15} & {-12}\end{array}\right)$

$|\lambda I-A|=0$

$\left|\begin{array}{cc}{\lambda-13} & {10} \\ {15} & {\lambda+12}\end{array}\right|=0$

$(\lambda-13)(\lambda+12)-150=0$

$\lambda_{1}=-2, \lambda_{2}=3$

$A_{1}$ is Dia.

$D=\left(\begin{array}{ll}{\lambda_{1}} & {0} \\ {0} & {\lambda_{2}}\end{array}\right)$

$D=\left(\begin{array}{rr}{-2} & {0} \\ {0} & {3}\end{array}\right)$

For $\lambda_{1}=-2$

$(\lambda I-A) x=0$

$(-2 I-A) x=0$

$x_{1}=\left(\begin{array}{l}{2 / 3} \\ {1}\end{array}\right)$

For $\lambda_{2}=3$

$x_{2}=\left(\begin{array}{l} {1} \\ {1} \end{array}\right)$

$P=\left(\begin{array}{ll}{2 / 3} & {1} \\ {1} & {1}\end{array}\right)$

$A=P D P^{-1}$

$A^{6}=P D^{6} P^{-1}$

$(P | I) \Rightarrow\left(\begin{array}{cc|cc}{2 / 3} & {1} & {1} & {0} \\ {1} & {1} & {0} & {1}\end{array}\right)$

$\Rightarrow\left(I | P^{-1}\right) \Rightarrow P^{-1}=\left(\begin{array}{cc}{-3} & {3} \\ {3} & {-2}\end{array}\right)$

$A^{6}=\left(\begin{array}{cc}{2 / 3} & {1} \\ {1} & {1}\end{array}\right)\left(\begin{array}{cc}{(-2)^{6}} & {0} \\ {0} & {\left(3\right)^{6}}\end{array}\right)\left(\begin{array}{cc}{-3} & {3} \\ {3} & {-2}\end{array}\right)$

$=\left(\begin{array}{cc}{2059} & {-1330} \\ {1995} & {-1266}\end{array}\right)$

Determine whether these matrices are diagonalizable or not. If they are diagonaliz
find $\mathrm{P}$ and $\mathrm{D}$ and compute $A^{6} .$

$A_{2}=\left(\begin{array}{cc}{2} & {1} \\ {-1} & {4}\end{array}\right)$

$|\lambda I-A |=6$

$\left|\begin{array}{cc}{\lambda-2} & {-1} \\ {1} & {\lambda-4}\end{array}\right|=0$

$(\lambda-2)(\lambda-4)+1=0$

$\lambda_{1}=3=\lambda_{2}$

$\lambda_{1, 2}=3$

For $\lambda_{1, 2}=3$

$(3 I-A) x=0$

$X_{1,2}=\left(\begin{array}{l}{1} \\ {1}\end{array}\right)$

A is not Diag.

$D=\left(\begin{array}{ll}{3} & {0} \\ {0} & {3}\end{array}\right)$

$P=\left(\begin{array}{ll}{1} & {1} \\ {1} & {1}\end{array}\right) \Rightarrow \text {det}(P)=0$

P is not inv.