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L L 1 L L J 1 1 1 1 1 7 2 0 : T2 8 2 8 1 1 1 H 1 7 1 1 12] 1 1 1 22 { J 0...

Question

L L 1 L L J 1 1 1 1 1 7 2 0 : T2 8 2 8 1 1 1 H 1 7 1 1 12] 1 1 1 22 { J 0

L L 1 L L J 1 1 1 1 1 7 2 0 : T2 8 2 8 1 1 1 H 1 7 1 1 12] 1 1 1 22 { J 0



Answers

$$ \left[\begin{array}{rrr} 1 & 1 & 1 \\ 1 & 0 & 2 \\ 1 & -1 & 1 \end{array}\right] $$

We want to use a calculator to find the inverse of a matrix. I'm going to illustrate this with the decimals matrix calculator. So we're going to hit new matrix, identify this as a three by three matrix and then put in the elements 120 which is optional, type -1, -1 2 -10. Want us to find, then you just type Matrix A. And then the inverse. And you can then identify the inverse as 0.20 point 4.40 negative point to 1.4 negative one negative 1.2. Yeah.

We're given this magic A We're universe first. Me from the convertible meeting room. Without it, the determinant they You could have zero in a but not in veritable works. Check it. A convertible. What do you say? Actually, I read it down here. Let's find a determining a check of the convertible or not. 111 First, I'm gonna high road to buy world one by negative one and had it wrote to So I guess one might one minus one zero. Making one plus 21 No one here. Next I'm gonna multiply growth three by negative one times wrote to I get 111 negative. Q one is negative one. You know, the determining the mortification old the numbers in the pivot in the diagonal interment is clearly not so. Therefore, we can find a neighbor nullifying chambers through the over inside the right chambers on this side. Very eight in the side. You're a here you have the identity matrix for three by three One here is you here alone? Now we're gonna really do until this side here. It looks like this. Once we do that, we will get a members on this side Look for reduced. Well, we already thought before we're finding the determinant. Do it again. First they can about this here. 11 now weaken Can't hold this position here. Negative one. They won negative times. Negative. 101 Negative. 101 Here one. Now weaken. We can scale the throw here. We can divide the group by negative one. We get negative here. Also here. Positive here. Now I can scale road three by minus a few. Added row to cancel this The negative too. Times road Here. That positive you minus one. That's one. Make a few plus one minus one. Two zeros too. You get one Next. We just need get rid of this. We can. He gave the period. Rowing added to the first would be a bit of this one. Here. You never get here from zero and in one one to negative one plus 01 and 101 Now you get a second road out of the first road in negative. One plus two. Just one. You have one plus minus +10 You have minus to plus one. You hear? This is the identity matrix implies on this side. He had a members say in verse, should be one bureau minus one one, minus 12 and minus 11 minus 11

Okay for this one. We have a is equal to 12 negative. One 011 and zero. Negative 11 So the characteristic equation for this problem is given by negative Lambda Cubed plus three Lambda squared minus four. Lambda plus two is equal to zero. And when we solve this equation, it's a cubic equation. So you will get the three argon values as 11 plus I and one minus. I noticed that these talking visor complex congregants. So now if we have the Lambda equals one, then upon solving the system, eh? Minus I times u equals zero. We will get that use equal t times 100 and for Lambda equals one. Plus I, using a similar process, we end up getting that you sequel to a T times I'm minus two negative I and one no, for Lambda Contra Kit equals one minus I, which is given here. Then that implies that you is equal to it. Turns out that the Egan vectors are also conflicts congregates of each other. So you was simply gonna be equal to a T times negative. I'm minus two guy and one

In this question we have to use row reduction to find the inverse is of the given batteries if they exist. And check it by multiplication. No, let us consider the metrics. Yeah. 123, 4 01, 2, 3 0012 0001. And on the right side identity metrics. Or for the four 1000 0100 0010 0001. Now we will row reduce the all metrics. We will apply the operation are funny those two our than minus two. Our two stores too, uh minus Artie And our three stores too. Our 3 -R4. On applying these operations, we get the metrics 1111. Yeah 01 11 00 11 little little 01. And on the right hand side we get 1 -10 needle 01 minus one deedle needle needle one minus one 0001. Again. We will apply the operation are one stores too, Urban -R2. Our two stores too, Our 2- Artery. And our three stores too. Our three minus are full on applying these operations to get the metrics 1000 0100 0010 0001. And on the right hand side one minus two, one needle 0 1 -2, one digital hell one minus two 0001. So in investment taxes, one minus two, one hero 01 minus 21 001 -2 0001. Yeah. Now we will check it by multiplication. We will multiply A. And N. Was matrix. So we can write a Multiplied by a invest metrics equals two 1234 0123 beetle beetle 12 0001. Multiplied by in west metrics 1 -210 0 1 -2 one. You know the middle one by understood deedle deedle? They're all one. No. Yeah all multiplication. We will first multiplied by stroke with first column. So one multiplied by one plus two multiplied by zero Plus three multiplied by zero Plus four, multiplied by zero On simplifying it we get one similarly. Now we will multiply first row by second column one multiplied by -2 plus two, multiplied by one, three multiplied by zero Plus four, multiplied by zero and simplifying it. Be good feel Now we will write these values in the desire my tricks, €1. By following a similar method we will find the other elements of the metrics. So you know beetle 0100 0010 0001. Hence hey multiplied by and was metrics. It were to identity matrix. Thank you


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