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Let $A=P D P^{-1}$ and compute $A^{4}$$P=\left[\begin{array}{ll}{5} & {7} \\ {2} & {3}\end{array}\right], D=\left[\begin{array}{ll}{2} & {0} \\ {0} & {1}\end{array}\right]$

$\left[\begin{array}{cc}{226} & {-525} \\ {90} & {-209}\end{array}\right]$

Calculus 3

Chapter 5

Eigenvalues and Eigenvectors

Section 3

Diagonalization

Vectors

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Okay, So in this problem, we have a which is K B P minus one. And what we want to know is A to the power. For so eight times, eight times a times we have P, which is a two by two matrix. So 57 to 3 and D is a diagonal me matrix to 001 We know that a four is equal to P D to the power for P minus one. Because it's a statement. Is it fear improvement? I am page 200 84. So all we have to do is compute d to the powerful. So d to the power for is simply two to the power for 001 to the power for because de is a diagonal matrix, which means that the four is 16 001 So if we want to know a four or we have to do is p times the four All right, the four times p minus one, which is ah fi seven 23 So p d to the power force of 16 001 and p minus one, which is either given or you have some compute in any case. It's a two by two matrix, so it's very easy to find the inverse matrix. So three minus seven minus two and five and then you can plug that in your calculator or your computer or you can computed by hand. It's not the purpose of the question. You should be able to do this. 226 minus 525 90 and minus 200. So to conclude we have a four, which is 200 26 500 minus 525 90 minus 200 I So there you have it pretty easy. When you have a diagonal ization of the Matrix, you can easily compute the power any power of that matrix using Ah, that's formula right here, which is presented on page 284 of the book

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