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Use the factorization $A=P D P^{-1}$ to compute $A^{k},$ where $k$ represents an arbitrary positive integer.$\left[\begin{array}{rr}{-2} & {12} \\ {-1} & {5}\end{array}\right]=\left[\begin{array}{ll}{3} & {4} \\ {1} & {1}\end{array}\right]\left[\begin{array}{ll}{2} & {0} \\ {0} & {1}\end{array}\right]\left[\begin{array}{rr}{-1} & {4} \\ {1} & {-3}\end{array}\right]$

$\left[\begin{array}{ll}{3} & {4} \\ {1} & {1}\end{array}\right]\left[\begin{array}{cc}{2^{x}} & {0} \\ {0} & {1^{k}}\end{array}\right]\left[\begin{array}{cc}{-1} & {4} \\ {1} & {-3}\end{array}\right]$ = $\left[\begin{array}{ll}{3} & {4} \\ {1} & {1}\end{array}\right]\left[\begin{array}{cc}{-2^{k}} & {4 \cdot 2^{k}} \\ {1^{k}} & {-3 \cdot 1^{k}}\end{array}\right]$$\left[\begin{array}{cc}{3\left(-2^{k}\right)+4\left(1^{u}\right)} & {12\left(2^{k}\right)^{\prime}-12\left(1^{k}\right)} \\ {-2^{k}+1^{k}} & {4\left(2^{k}\right)-3\left(1^{k}\right)}\end{array}\right]$$\left[\begin{array}{cc}{-3(2^ k)+4} & {12\left(2^{k}\right)-12} \\ {-2^{k}+1} & {4\left(2^{k}\right)-3}\end{array}\right]$

00:40

Amy J.

Calculus 3

Chapter 5

Eigenvalues and Eigenvectors

Section 3

Diagonalization

Vectors

Missouri State University

Campbell University

University of Michigan - Ann Arbor

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So if we want to raise this to the K through power, let's just go ahead and do this really quickly. So first, remember, this is supposed to be a is a good P D. P in verse. So if we were to kind of raise this to the case of power on each side So Okay, okay. Remember, we can't just distribute the K Tau everything. We'd have to do this P D p Inverse times PDP inverse times PDP inverse Over and over again. So that tells us the PM PM versus will all cancel except for the ones on the end. So then we would have over here negative to 12. Negative. 15 raised to the Kate is going to be so we just have p first. So 3411 And then we have this middle one where it is raised to the k. So 201 race, the k and then we have our outside once a negative one for one negative three. And then remember, multiplying this one mental is a lot easier since all we do is take that K and then apply it to the diagonals. So be 3411 Then we get to to the K 001 to the K, which is just one. And now it just turns into multiply these matrices together. So let's do the first one over here. So we first would dio uh, three times two case that would be 32 k and then four times zero. So that's just going to be that. Then we would do 34 times, 01 So that just leaves us with four here. Then we would do 11 times two k zero. So that's just two K. And then we get zero there. Right? So that's the first, uh, zero. Because we still need to this last one. So 11 time 01 which gives us one, and then we still have our matrix over here. So negative one for one negative three. So now we do again. The first row, First column. So 32 k times night once. That would be negative. Three times to K and then four times one which is going to be plus four. Then we do same one, but for negative three. So that will be so. Four times three is going to be 12. So this would be 12 times two K and then four times three negative 12 and then we do two k times. Negative 11 So that would give us negative two K plus one and then forward last one, we'd get so two K times four so four times to UK and then one times three is negative three. And so this is what we're going to get if we were to x raise that original matrix to the case Power Assuming again K is just some positive integer.

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