5

1 1 Lo erelta1...

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1 1 Lo erelta1

1 1 Lo erelta 1



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$\left[ \begin{array}{lll}{1} & {1} & {1} \\ {1} & {2} & {3} \\ {0} & {1} & {1}\end{array}\right]$

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

Okay, so I'm gonna be multiplying negative one times itself six times so negative one times negative. One time saying good one time single, one times negative, one times negative. One who feel like a broken record. So I'm going to pair each of these up and multiplying. What's nice is all All three of these pears are the exact same thing. So negative one times negative one is a positive one. Same with this. Same with this. Now, when I multiplied those together well, one times one is one times one again is one and they're all positive. So my answer stays positive if you also look, I'm multiplied. Well, 1/1 times. One times one. Right. It just stays one. But notice I had 123456 Because I had an even number of negative ones. The negatives cancel each other out. Right. This two pairs cancelled each other out. Thes two pairs. Cancel each other out. These two pairs cancel each other out. So if you have an even number of negative numbers, if they turn positive because each have a pair of another negative to cancel out

Came this question. I'm multiplying negative one times itself over and over and over. Seven times now, one times one times, one times, one times, one times one times one is just going to be one. Now, these two negatives multiply together, make a positive one. These two also make a positive one. These two also make a positive one. But because of this odd man out here, our answer is going to be negative. If we were to just have even numbers of negative one, we would have had a positive answer. But because of this odd right, we had on number of negative ones that we were multiplying together. My aunts, my final answer will be make divorce.

So this question tells us Matrix A is this minus one minus one one and one. And it wants the show that a squared is equal to zero. It was zero matrix. So, really, all you have to do is take minus one minus 11 and one, and multiply it by itself, minus one minus one, one and one. Figure out what you're gonna get. So in the first case, your take minus one minus one and multiply it by minus one and plus one. So that's gonna give you bus one minus one u zero. Then you're gonna take that and multiply it by this column. So again, minus one times minus one is one minus one times one is finest ones. You get zero if you do the same thing with the second row. Okay, First column you get minus one plus one, not zero. And then finally, you do it with one and one in minus one, minus one, and you also get zero. So that shows it. As long as you understand major small vacation, which I'm assuming you do. Um, then you can see that it gives you this Euro matrix


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