5

1 P(A)= 1 1 1...

Question

1 P(A)= 1 1 1

1 P(A)= 1 1 1



Answers

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

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.

This video's gonna go through the answer to question number 11 from chapter 9.3. So ask to use real reduction to find the inverse off the matrix. That 11 one 121 Thio three. So So we conform the combination matrix with the identity and they tried refugees. Okay, so if we subtract to you off the top equation from the bomb equation, then we're gonna get zero one that to you, minus 20 Maybe it's gonna be minus 201 on the inside. And if we should bottle subtract one of the first question from the middle equation, that's gonna be zero That's gonna be one on that's going to zero months. Well, on zero on me, the top equation as it is, Savior zero. Okay, so now we get to be a stick in court because on left inside the bomb equation on the middle or after the bottom row of the majors in the middle of the matrix. All the same, which means that the ah, the row is off the matrix linearly dependence, which by their a born in the book, means that er the identity that's all right with me


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