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Point) The vectors5322and w =11-1+k-3are linearly independent if and only if k #...

Question

Point) The vectors5322and w =11-1+k-3are linearly independent if and only if k #

point) The vectors 5 3 22 and w = 11 -1+k -3 are linearly independent if and only if k #



Answers

Determine all values of the constant $k$ for which the vectors $(1,1, k),(0,2, k)$ and $(1, k, 6)$ are linearly dependent in $\mathbb{R}^{3}$.

All right. I hope that this is going to work out well, because my system seems to have some lag right now, but maybe not a terrible amount. Um, I'm going to create of four by four matrix. And I would be writing faster if I did not have lag. So I am writing them transposed into the columns. So that's going to be if we follow along. Oh, my goodness. Follow along. This row right here, zero times a matrix. Um, that's row 1 to 3. Road three column to That's gonna be odd. So it's gonna be negative. The Matrix will be one to negative. One one one one negative. One one. Okay. Plus que times. Another three by three matrix. Thanks. Minus one times. Ah, 3rd 3 by three matrix. No, because of the lag. I'm not going to show you, actually calculating there's determinants. But you know how to calculate the determinants, The determinant of the 1st 1 where the first determinant is negative. K minus five. The determinant of the 2nd 1 is K squared, minus two k. Okay, I'm on minus three and the determinant of the last one three k minus one. Curry is. You simplify this. You get okay to the third power minus two. Okay. Supposed to be too. Okay. Squared. Minus five. Que. Oh, my goodness. That's definitely not OK. Less six. Try six again equals zero. Let me read that again. Que To the third power minus two K squared. Minus five. K plus six equals zero. Now, if you put that equation into a graphing calculator and you look for the zeros, you will see that there are zeros at 31 and negative, too. So what that means is that okay does not equal three one or negative to, and that's the solution to this problem.

Hello, guys. So now we got the following vectors. The vector you, which is minus three Thio 10 Then we got our vector V that these for seven mines three to on our vector W which is five minus 281 Thanks, Chief. Okay, so we got these three vectors on. We got the following operation here, so we got three you plus V minus two. Time is all of you is equal, Thio two times, all of you. Plus three x and, well, we want to know the value off these vectors x such that these equation here satisfied. So let's solve for X so x, it's going to be equals toe one third off three times the vector you plus V and minus four times the back Trudeau via So let's solve this inside part here, here, on the right part. Okay, so let's solve this inside the operation that is on the parenthesis. So three you plus V and minus four w So three you is just minus nine here. Six 30 The we got already. That information is just four seven minus three to on minus four times the of you is minus 20 eight, minus 32 on minus four. Okay, so let's sum all these vectors together on we obtain that. This is 25 minus 25. Here is 21 21 minus 32 on minus two. Okay, so this is what the This is the result off this year, operation on the rockets on the parenthesis. So then let's step. He's just multiplied by the one third that is here. So x at the end is going to be one third off. What off? Minus 25 21. Minus 32 on minus two on this is equals to minus 25. Divided three. Here is seven minus 32 divided three on finally minus two thirds. And that is the result.

Let s be the subspace of ah real three dimensional space span divided by the vectors V one, which is one comma, one common negative one be to which is to comma one comma three and V three, which is negative to calm, a negative to common to. So that ass is also spanned just by V one and V two. Well, if we multiply V three times negative 1/2 I guess I shouldn't do that right there. Let me do that somewhere else. Negative 1/2 times V three is negative. 1/2 times two, which is one one negative one. And that is V one. So V one and V three are in the exact same direction. So they're repetitive. Eso therefore v three does not provide you with any greater span. And therefore, um s is all

The question. If you use this killer, go ahead. Find the one better learn fight that has u e understand. Just on the that. You know it's the real estate. Be it is zero minus big wins. What's the solution? If you would not be the best What? Non zero weapons and free space. Then the volume off a parallel pipe that has you'll be interview. But that is given by these goals. You that dark three, which is a closed. Do you want you What? See people? Okay, One speak so way will be putting the values. And that's from like people get bacon. 264 has thio minus. For solving this, you will get minus the last week will be 16. Therefore, a volume being banner. That pipe be model is off minus 16. Practice 16. You You This is the question. Thank you.


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