let-ap-d-p-1-and-compute-a4-pleftbeginarrayrr2-3-3-5endarrayright-dleftbeginarraycc1-0-0-1-2endarray

Question

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

Amy Jiang · Accepted Answer

okay were given A is equal to two negative three a bit of 35 and the is equal to 1001 Have Let&#x27;s get P Universe. Sorry, this is actually a P that&#x27;s equal to 5332 and then these are for that&#x27;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&#x27;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

Samuel Goyette · Answer

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&#x27;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&#x27;s a two by two matrix, so it&#x27;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&#x27;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&#x27;s formula right here, which is presented on page 284 of the book

Heather Zimmers · Answer

here were given p of X and R of X and in this problem were asked to find four times P of a. So to find Pia, they were going to substitute a in for X in both of those positions in the P function. So we&#x27;re going to have four times three a squared minus to a plus fat and then we&#x27;ll go ahead and distribute the four, and that gives us 12 a squared minus a plus 20.

James Kiss · Answer

we have the function. P of X equals two. Expert was three over X Square. We&#x27;re going to evaluate P A five and then p of three House, followed by p of three any and finally p of a minus one. So let&#x27;s get started with p of five. We find p of five by playing five in for X, and the function was two times five squared, plus 3/5 squared. That&#x27;s two times 25. That&#x27;s five squared two times 25 is 50. Post three is 53 25th. P three has would be two times 3/2 squared, plus three over three has squared or two times 9/4. That&#x27;s 300 square plus 3/9 4th which is 18 force plus 3/9 4th in 4th 3 is 12 4th 18 plus 12 is 30. It&#x27;s to be 30 fourths over 9/4 which would be 34th times for a 19 month by by the reciprocal four isn&#x27;t cancelled. We have 30/9. Three goes into both of those that&#x27;s 13th. So 13th is our most simplified answer. Under three a B two times three a squared plus 3/3 a squared B two times nine a squared plus 3/9 A squared or 18 a squared plus 3/9 a squared in P V minus one B, two times a minus one squared, plus three over a minus one squared B two times a A squared minus two. A plus one that&#x27;s a minus one squared lost three over a minus, one squared again. It&#x27;s a minus. One times a minus one. A squared minus two, a plus one or special right below distribute the to to a squared minus four. A plus two plus three is plus five over a squared mass to a plus one.

James Kiss · Answer

I say we have the function. P of X equals three X squared minus five over X squared. We want to evaluate p of five p of three halfs if you have three A then p of a minus one. So let&#x27;s start with your five if I m. P of five by plugging five in for X three times five squared is three times 25 minus 5/5. Spurred 25 three times 25 is 75 minus five. That&#x27;s 70/25 and reduce 70/25 to 14 5th You can divide both by five p. Three. House would be three times three have squared, minus 5/3 halves squared. That would be three times 9/4 minus 5/9 4th three times nine force would be 27 4th minus five overnight fourths. You don&#x27;t really have to do this by hand, but in fourths we&#x27;ve got 27 minus. I would be 24th 27 minutes. 20 would be 7/4 over nine force. You divide by a fraction by multiplying by the reciprocal with 7/4 times four nights and the fourth canceled. You get seven nights. Next we API of three ace would be three times three a squared minus five over three. A square three a squared is nine a square, three times nine a squared is 27 a squared, minus 5/9, a squared and last but not least, P of a minus one B three times any money This one squared minus five over a minus, one squared a minus. One squared is a squared minus to a plus one. And again, a minus one squared is a squared minus to a plus one. We can distribute the three in the numerator yet to get three a squared minus six a plus three minus five would be minus two over a squared minus to a plus one.

Heather Zimmers · Answer

here we have polynomial function p of X, and we&#x27;re going to find Pew three and p of negative one. So to find p of three, we need to substitute three and the function for X. And that gives us three to the fourth minus three times three cubed plus two times three squared minus five times three plus one and then we&#x27;ll simplify it. Three to the fourth is 81 minus three times three. Cubed is 27 plus two times three squared is nine minus five times 3 15 plus one. So we have 81 minus 81 plus 18 minus 15 plus one, 81 minus 81 a zero and 18 minus 15 plus one is four. Soapy of three is four for p of negative one. We do the same sort of thing. We substitute negative one in for X for all the exits. So we have negative one to the fourth, minus three times negative one cubed plus two times negative. One squared minus five times negative one plus one negative. One to the fourth is one minus three times negative. One cubed is negative. One plus two times negative. One squared is positive one minus five times negative one which is plus five and then at the end, plus one. So we have one plus three plus two plus five plus one, and that adds all together to give you 12 soapy of negative one is 12.

Problem

Use the factorization $A=P D P^{-1}$ to compute $…

Problem 2 Medium Difficulty

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

Answer

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

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

Linear Algebra and Its Applications

Heather Z.

Samuel H.

Michael J.

Joseph L.

Vectors Intro

Vector Basics - Example 1

David C. Lay, Steven R. Lay, Judi J. McDonaldLinear Algebra and Its Applications

Heather Z.

Samuel H.

Michael J.

Joseph L.

David C. Lay, Steven R. Lay, Judi J. McDonald

Linear Algebra and Its Applications