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Let $A=P D P^{-1}$ and compute $A^{4}$$P=\left[\begin{array}{rr}{2} & {-3} \\ {-3} & {5}\end{array}\right], D=\left[\begin{array}{cc}{1} & {0} \\ {0} & {1 / 2}\end{array}\right]$

$A^{4}=\left[\begin{array}{cc}{2} & {-3} \\ {-3} & {5}\end{array}\right]\left[\begin{array}{ccc}{1} & {0} & {0} \\ {0} & {1 / 16}\end{array}\right]\left[\begin{array}{ll}{5} & {3} \\ {3} & {2}\end{array}\right]=\frac{1}{16}\left[\begin{array}{cc}{151} & {90} \\ {-225} & {-134}\end{array}\right]$

Calculus 3

Chapter 5

Eigenvalues and Eigenvectors

Section 3

Diagonalization

Vectors

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okay were given A is equal to two negative three a bit of 35 and the is equal to 1001 Have Let's get P Universe. Sorry, this is actually a P that's equal to 5332 and then these are for that's equal to 1001 over 16. Okay, so we have a four is equal to key, which is to linger three. Number 35 time seems to be par four, which is one's own. So one over 16 and P universe, which is 5332 It simplifies till when our sixth in 151 90. Bigger 2 to 5 and make of 134

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