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2g V 0 1 53 [ 1 2} F L L 7 [ F F 1 L 1 1 2: 1 1t 1 1 2 1 [ 6a: Ing L "1 2 [ F 2 L 1 4...

Question

2g V 0 1 53 [ 1 2} F L L 7 [ F F 1 L 1 1 2: 1 1t 1 1 2 1 [ 6a: Ing L "1 2 [ F 2 L 1 4

2g V 0 1 53 [ 1 2} F L L 7 [ F F 1 L 1 1 2: 1 1t 1 1 2 1 [ 6a: Ing L "1 2 [ F 2 L 1 4



Answers

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

In the problem we have 56. And this is yeah it is gen X get works. Get three X. Get your necks. It's two weeks is tricks plus if one X after works three X gen X G networks. GTX H one X. HdX history X plus. If you're next afterwards after the X. And the one x. Networks G three X H one X H two x. and three x. No further. We have fds a. And this equals two. If one yeah. Have to wear after a year. Do you want to G two A. Gayatri? H one A. It's too good. Is three. Okay. And these are reputed To two times more studies if one A. Have to A F three given a and geeta, Geeta A 20 H two A. It was today glass if one after where have today do you want to? They were the tree H one A H two A is three. Now this equals to zero plus zero plus zero which is equal to zero. Since F R A Is equal to zero. G R A is equal to H R A. Is I mean this equals to 04. All the times that is if R0 gr hr is also zero For Article 1, 2 and three. So we have this as the answer to the problem

In this video, we're gonna go through the answer to question number 17 from chapter 9.4. So we asked to find where these vectors x one x two x three Ah, linearly dependent on where they are linearly independence, which FYI is teeth independent much varsity that linearly dependent. So okay, so for them to be linearly dependent would need values off. See one c two c three it such that c one times x one c two times x two plus C three times Next three equal to zero association into values for these x one x two x three then we write This is a system of three equations. So the 1st 1 is just gonna be Ah, well, it's the the first element of each of the equations Time each of the vectors X one experience to text three times by the corresponding um constant C one C to C three. So we're gonna have Well, there's no, uh, ex threes are no component, Maxime, because the top of the next three is zero. So we're going off, uh, ex one next Tuesday, at both of which contain eats the to tease it. So take a common factor out. Get either to t C one plus C two. Ah equals there. Next up, we're gonna have eats the two tea. Close. It was C one that's actually to eat the TT plus C three he to the three tea equal Syria. And finally five. Easter to tee times five C one minus C two equals Sarah. Okay, so comparing the first of these equations on the last of these equations we find that C one don't see too must be equal to zero as you see that because, well, first look in the first equation e to the two tea for any tea can never be zero. So we basically just cancel by its beauty and saying with the bottom equation s So therefore, we have this bit is equal to zero this busy zero for those two to both equal to zero. Then C one and C t o. You can show that by rearranging one of those sub student in the other. Um and you're find this. You wanna see t birthday frequent zero. So therefore I see two is the rocks that this is zero. So then we have to see three times Three tier sequence There we can we can divide by eats and three team because that's never zero for any tea on DDE. All we're left with is C three sequences So far, all t ah the C one c to the T three. Must could only be for this for this equation. Thio be satisfied. Uh, this equation to be satisfied then Theo Dissolution of the Trivial Solution. Therefore the vectors x one x two x three are linearly independent for that tea in any value between minus infinity to infinity.

Is given to us in problem one here. And then we're going to estimate the value of the derivative at each point. So for part A, we're looking at we're X equals negative three. So we go to where X is equal to negative three on the graph, and then we take a look at our graft function. Now, we're sort of estimating the value of the derivatives. What is the slope look like here? Well, it's definitely negative, and it's not super negative, right? It's just a little bit. So we could maybe estimate that the derivative at negative three is equal to say negative 30.1. Okay, then. Part B wants us to look at negative where it's at his eagle. Native twos. We go to native to on the graph and take a look at it slope. Well, looks like the slope zero there to me. Um, well, estimate just from looking at the graph at the's slope at the dirt, that negative too, would be equal to zero. Okay, so then do the same thing for the point. Negative one. So you got a negative one and see what that looks like. And this time we have a positive slope and it looks to be, um maybe it is equal to one, so it's equally increasing as it is heading to the right. Uh, we're gonna take a look out where the derivative is equal to comes physical. Zero. In here, it looks it's even more positive than before. Eso will say half says equal to two. They were gonna take a look where our function is equal to one. So it's still positive. It's maybe a little less positive it was before. It's looking to be about the same slope, that negative one. So So that's equal to one. And in fact, in the whole ah, right size for seems to be a little bit of a mirror version said it to here. Looks like we have no slope again. And then at a positive three, we have just a very slight decrease. So I would guess that two it's gonna be equal to zero. And then for positive three, it will be equal to our negative three. Uh, which we'll just say is equal. Appoint one. Okay, so definitely not exact. But then when we draw a graph, uh, e derivative, you're always try use this chart. Even this guy. A little bit of race marks on it we're gonna see is at negative three. It is going to be just a little bit negative at two. It'll be that negative too. It'll be equal to zero at negative one. It will be equal to one at zero of equal to two. And then it's gonna mirror mirror the other side there in sewer graph of the derivative is gonna just look like a little humpback here.

Going on three times matrix F. So I've got matrix F right here, this is F. And we're just gonna modify it by three. So take each component and multiply it by three. I'll take one step in here. Make sure you put times three and all of these. Okay? If you want to do that step in your head, you can. But your result will be 99 negative three negative threes, just distributing that coefficient, distributing that number in front onto each element of the matrix.


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