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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.

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