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Hww 7Ji 1Ii ] 1 1 V 1 1 1 J 1 L 1 1 | [ 1 11 (...

Question

Hww 7Ji 1Ii ] 1 1 V 1 1 1 J 1 L 1 1 | [ 1 11 (

Hww 7 Ji 1 Ii ] 1 1 V 1 1 1 J 1 L 1 1 | [ 1 1 1 (



Answers

$\mathbf{v}_{1}=\left(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}\right), \quad \mathbf{v}_{2}=\left(\frac{1}{2}, \frac{1}{2},-\frac{1}{2},-\frac{1}{2}\right)$ $\mathbf{v}_{3}=\left(\frac{1}{2},-\frac{1}{2}, \frac{1}{2},-\frac{1}{2}\right)$

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

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

They're So for this exercise we have this vector B. And subspace of our. For in this case Subspace of R four W. That is spanned by two vectors. These vectors corresponds to the one that is just a vector for once And the two that is 1 1- warming. Okay, so for this exercise we need to calculate your total projection of beyond this space of you. So just to remind you that this projection corresponds to the inner proud of be with with each of the generators of the subspace. So this case is the one the one over the the square of the norm of B. One plus the inner product of B would be to times two over the norm of B. Two. So just to give you a little bit of geometric intuition of this, let's suppose that this plane here corresponds to the subspace off. Clearly this is a case of our four. But still, this picture will help you a lot when when you try to solve this kind of projection problems, the calculations are quite easy. It's just like yeah algebra. But the important thing is you should keep in mind that that algebra tells you have a geometric intuition behind. And so let's suppose that this this this subspace W. And here is the Trevi So this factory B is generated by two vectors. We won. And this case to that is the span of these two factors. So when you, when the problem asked about the orthogonal projection of beyond our view, means that you pick from fee and you project a line that is orthogonal to this plane, let's say this point here. So this park here in response to the projection, but the actual vector that you will obtain is on the plane. So this vector here will be the projection of be on the subspace W. So when you're putting here this formula, this part here means a projection of be on one of the generators of this space, namely this factor here. So if you have this projection and this projection and you some these two projections of being on the vector V. Two and on the vector V. One, then if you sum them up, you will obtain the actual projection of beyond the subspace. So then you obtain the component of being on this uh subspace W. So that's the geometric intuition in this case. So, having that in mind the rest is just algebra. We we just need to make some calculations that are really straightforward. So let's calculate what the components that we need. So let's start by the inner problems. So the inner product B would be, one is equal two, one. The inner product of B would be to is equals two, five under norms for the square of the norms for this to remember that the square of the norms corresponds to the interpretation of the vector with itself. And in this case for B, one is equal to four. And actually in this case is the same for a bit too his equals. So with this in mind, we just need to replace the values on the formula for the projection of being on the subspace. Love you. And that means that the projection be at the subspace W is equal 2, 1/4 Times the victory we want. But there's 1, 1, 1 and one bless five over four of times one, one minus one. Right. And the result of this is 1/2 times the vector, 3, -2 -2. So there's a projection, the orthogonal projection of the vector beyond the subspace W.

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


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