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E2 u/8 e/1 ( #k 71 8 1 23...

Question

E2 u/8 e/1 ( #k 71 8 1 23

e2 u/8 e/ 1 ( #k 7 1 8 1 2 3



Answers

$\mathbf{A}=\left[ \begin{array}{ccc}{-1} & {-8} & {1} \\ {-1} & {-3} & {2} \\ {-4} & {-16} & {7}\end{array}\right]$

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.

They were multiplying matrices, E and F. Together here. Eight times F. We know this will give us a two by two matrix as a result. So. Top right element row one, column 11 times three plus three times negative one. Get zero there. Okay. Top right. Row one column to one times three was three times negative one. This is another zero there. Okay. Moving our way through. It just takes a bit of practice. These guys working through all the elements, say it to yourself in your head or out loud. If you want. Row one column, try road to column one. Two times three plus six times negative one. Okay, It's gonna give us another zero there. I guess. I think we're getting zero matrix out here, but we'll check with the last element Road to college too. Two times 36 times negative one. Another zero out there. Okay, so our result here is again zero metrics.

Matrix. So this is negative one times the two by two matrix of all the ones. Okay, pretty simple here, we just distribute this negative one in until all the elements of this matrix. With modification. If you want to skip that step and do it in your head, you can. It's basically going to switch to sign on every single element in our matrix, a very natural calculation to do here.

Okay, this question we have to do two times Matrix A. So I've got Matrix A sitting right here. Okay, I'm just gonna drag it over here and we're gonna multiply it in front by the two. Ok, so essentially it's two is just going to kind of distribute onto every little element in. Our Matrix gets kind of complicated with arrows, but basically we just multiplying each piece by two, so two times one, two times two, two times four, and then two times three is the last element. Okay, you can probably do some of this in your hand. These simple calculations should give you a final matrix of 2486


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