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Points) Justificution required 146J-U) Etermine !hether the vectors {V.VV-Va} an linearly dependent . lincarly independent. Hint computation should = ryuircd Thinkp...

Question

Points) Justificution required 146J-U) Etermine !hether the vectors {V.VV-Va} an linearly dependent . lincarly independent. Hint computation should = ryuircd Thinkpaints) Justilcation mquired:(J-UH-U} Letemmine whether the vectors {V.%V-Va} an lincurl dependent = linearly independent. Hing No computation should mquired. Think:

points) Justificution required 14 6J-U) Etermine !hether the vectors {V.VV-Va} an linearly dependent . lincarly independent. Hint computation should = ryuircd Think paints) Justilcation mquired: (J-UH-U} Letemmine whether the vectors {V.%V-Va} an lincurl dependent = linearly independent. Hing No computation should mquired. Think:



Answers

In Exercises 24–26, use determinants to decide if the set of vectors is linearly independent.
$$
\left[\begin{array}{r}{3} \\ {5} \\ {-6} \\ {4}\end{array}\right],\left[\begin{array}{r}{2} \\ {-6} \\ {0} \\ {7}\end{array}\right],\left[\begin{array}{r}{-2} \\ {-1} \\ {3} \\ {0}\end{array}\right],\left[\begin{array}{r}{0} \\ {0} \\ {0} \\ {-2}\end{array}\right]
$$

Now we have a set of vectors and we want to see if they are linearly independent. If the set is linearly independent or if the set is linearly dependent. Now I noticed that there are Ford four vectors and it's only three dimensional space. So therefore they are dependent. The set is linearly dependent. Now, I don't think it's obvious to me how these go together, how they're depended on each other. So I'm going to create a matrix notice that I'm writing, um, each vector in column forms. I'm writing the transpose of each vector. Whoops. That should have been to, um 312 negative one negative 11 And I'm going to write an augmented matrix because when you add a constant times all of these vectors together, um, you're going to get zero. That's the definition of linearly dependent. I'm gonna verify that. I wrote everything down correctly because once a few problems ago made a writing mistake, and that takes some time to figure out what was going wrong. All right, well, I'm gonna take negative one times the first row. That's gonna give me 10 Negative. 31 that, uh ah. Zero Okay, Now I'm gonna add the first road in the second row zero to four 00 No, myself. Some more room here. No, I didn't really need more room. Um, now I'm gonna take the first road times, too, and add it to the third row. Zero negative one three times to its six plus two is eight negative. One times two is negative. Two plus one is negative. One. All right, I'm gonna take the second row and divided by two. Now, I'm gonna add that to the third row. Um, two plus eight is 10 zero. All right, now I'm going. Teoh, divide the third row by 10. Now I'm going to take negative too times this and add it to that. It's gonna give me 01 zero negative too. Times this. So that's going to be negative. Two times negative. 1/10 Which is positive. Too tense, which is positive 1/5. Now, I'm going to take, um, three times the third row, the new third row, and I'm gonna add ah, Teoh whips to that. That's going to give me 10 zero, because one times three plus negative. 30 Okay, negative. 3/10 plus one. So that would be 10 tens minus three times. That's 7/10 zero. Now I can write some equations. C one plus 7/10 C four is zero see to plus 1/5 C four is zero C three minus 1/10 c four zero and I'm gonna solve these for C one, C two and C three. C one is 7/10 C four si two is It's negative. 7/10 negative 1/5 c four c three is 1/10 C four. Now I'm going to write that. See one times the first vector. And you know what? I'm just going to write Vector one plus C two times Vector two plus C three times Vector three plus C four times Vector four is zero because they are linearly dependent. But see one is negative. 7/10 C four si two is negative. 1/5 c four c three is 1/10 c four. Not doing anything with this one Since see Ford is not equal zero I can write negative 7/10 times V one minus 1/5 times V two plus 1/10 times V three plus V four equals zero and this should be the zero vector. Um, so let's right the vectors in here now. Negative 7/10 times The first vector, which was negative. One 12 minus 1/5 time's the second vector zero to negative one plus 1/10 times. The third vector plus the last vector. All right, and that's the answer for the dependency, However, let's check in for if it's correct. Um, I'm going to write just for the ah the X coordinate, which has to be zero negative. Seven tense times Negative one is 7/10 plus 3/10 minus one, which is indeed zero. Let's let's try all that. Let's check all the coordinates. Negative 7/10 minus two tents, which I mean two fists, which is 4/10 plus 1/10 minus one zero. That doesn't look right. Negative. 7/10 minus 4/10 plus 1/10 minus one equals 00 that is negative. 11 10th plus one is is negative one, but negative one minus one is is negative. Two. Makes me wonder if I wrote that last vector incorrectly, and that's going to be very annoying. I did not write it incorrectly. So now when I added, Oh, no. When I added row one and row to negative one plus neg. Plus one is zero. Zero plus two is two, um, three plus one is four negative. One plus negative. One is negative. Two. Oh, my goodness. Well, when you make a mistake, you just go back and you fix it. Okay? Now, let's see how this problem affected everything else. All I did here is I divided by to So that's going to give me negative one right there. Okay, Now, I added two rows together. Such that one plus negative one is 08 plus two is 10 and negative too. Plus negative one is negative. Three. Okay, Now, I divided that last row by 10 which would give me negative 3/10. Then that's going to effect this and this. So what I did is I mall applied by three, and I added it to the first row. So negative. Three tense times three is negative. 9/10. And, um, that's not right, either. Negative. Three tense times three is negative. 9/10. But then I add that to the first row, which is just 1 10 10 10 tens minus 9/10 is 1/10. All right, now we're on the right track now. I also multiplied that last row by two and added it to the second row. So negative 3/10. No, I'm all the planet by negative till let's just think of this. Think this through last road times negative Dio would be negative. Two plus two is zero k last row times negative to would be 6/10 which is the same street fifths and 3/5 minus one B 3/5 minus 5/5. It's gonna give us negative too Fifth. Okay, that, of course, changes everything over here. Okay, so let's fix this. 1/10 minus 2/5. Um, minus three tense. So let's fix over here. Negative one times. Positive. 2/5. Positive. 3/10. Okay, now we've got to fix it down here. I have it in three different places, so I'm just going to go ahead and erase and the rays and the rays. Now we can put the new numbers in. The 1st 1 is negative. 1/10. The 2nd 1 is 2/5. The third wine is the re tempts. Let's do our check again. So at least one thing we're learning is that it's it's very important to check your answers. Um X is 1/10 plus 9/10 minus one, which is indeed zero. Why is negative 1/10 Nuss to fist times two. That's four fists is 8/10 plus 3/10 minus one 89 10 11. It's 11 10th minus one would be 10 night, 10 tents minus one is zero good Z negative. 2/10 minus two fists. That's 4/10 plus 6/10 plus one zero. All right. Did I somehow make another mistake? Negative. 2/10 minus 4/10 plus 6/10. Oh, my goodness. At the very beginning, I wrote the wrong term. That's supposed to be one. Oh my! Wow. But wait a minute. I did read it correctly here. So since I wrote it correctly here, that would mean in theory, that's correct. Theory. That's correct. And theory. That's correct. In theory, that would be correct. That's correct. That's correct. That's correct. Okay, I see what I did right here. 2 50 times. Negative one. No, that's correct. 2/5 times negative. One. His negative 2/5 which is negative for tense. The re tense Times two is 6/10 No. All right. And I have found my mistake again. I believe that it's right here. But let's look, um, what I was doing is I was adding this row in this row together, two plus eight is 10 negative. One plus negative one is negative. Two. So now this is going to be negative. Too tense. And now, which is, by the way, negative 1/5. That's now going to effect these. What I'd like to recommend is try not to get too frustrated. This is pretty frustrating to me right now, but try to maintain a positive attitude because everybody makes mistakes. My advice to you. Okay, So you if I multiply that last road times negative too, and add it to the row above it. Negative. Two times one plus two a zero negative, too. Times negative. 1/5 is two fists, two fists plus negative one. His negative. Three fists, 2/5. Try to do this slowly. Okay, I believe that's good. Now I multiplied the last row times three. That's gonna give me negative 3/5. And I have to add that to one to this. All right. That, of course, changes everything. But I am choosing to be positive, believing that this will be the last time. Okay to this minus 3/5. Minus 1/5. That gives us 1/5 the Revis and negative two fists. Now we erase all these things. First one's negative. All the others are positive. Negative. 2/5. All the other is a positive. 3/5 and 1/5 and we try again. But this time I think I learned my lesson. Maybe not all of my lesson, but least some of it. I am going, Teoh. Oops. It didn't work. What just happened? Okay, that was annoying.

So we have a set of four vectors in four dimensions. And so, ah, when we set up a matrix, M will equal and and so if the determined is zero, then they are dependent. And if the determinant is not zero than the set is independent. So let's go ahead and write the Matrix to negative 101 10 Negative one to 03 12 Negative 11 to what? Okay, Thinking about how I'm going to do this, I guess I'm going to dio two times the determinant of 031 negative 112 to to what? Minus one times to determine it of negative one 31 012 1 to 1 minus negative one, which would be plus one times the determine it of negative. One zero three zero negative one one 1 to 2. All right, so let's just figure out the determinant of 031 negative 11 to 2 to one. I'm going to write the 1st 2 columns again. That will give me zero 12. Negative too. To zero. No negative. Three, 12 11 10 98 It's 11. Now let's do the determine it of the next one. Negative. 131 012 1 to 1. Now I'm wondering, Is it? Is this obvious? The 1st 3 of zeros in different places? The last one doesn't have any zeros at all. So I'm thinking maybe it's not obvious. It's not obvious to me, so just keep going. What I meant was is the relationship obvious, and I don't see how it would be obvious. Okay, negative. One six zero mice one plus four zero 6548 to the determinant of the last one. Negative. 103 zero Negative. 11 1 to 2 to vera zero negative. Three. Negative, too. Zero. Okay, so that gives us two times 11 minus one times eight plus one times seven, which is 22 minus eight plus seven, which is minus one, which is 21. The determinant is not zero. So these vectors are linear. Lee in De Upended

So in this question we want to find some vector that we can add to the set. You want to be to to create a basis of our three. Now the easiest way to do this is just just to try some standard basis factors. That means like E one, E two or E three. I'll just put E. J. To represent. It's one of those. Try these vectors and see which of these vectors gives a linearly independent set. So let's go ahead and put our vectors into a matrix. And then we're going to choose one of our standard basis vectors to try as a vector to add to the set. So let's try E three. So we have a 00 and one and then we're gonna row reduced to see whether these three vectors are linearly independent. Well, well, first use this one in the top row, 20 out the two underneath you multiply the topper by two and add it to the second, giving us following. And then we'll multiply it by -3 and added to the third to give us the following. Well now we see that our second row just has one entry in it. And so we can just scale our second road down by multiplying it by 1 5th And we can change the second row to just be a 010. Well at this point we can then use the second row To cancel out the -3. So our first road stays the same and our second row stays the same. And then we multiply the second row by three and add it to the third, which gives us a matrix and reduced row echelon form. And since we were able to get a one along the diagonal, This implies that are three vectors are linearly independent. So we have the vectors the one v two, An E three make a linearly independent set. And since we're looking at our three, we know there must be three independent vectors. In order to make a basis, we have three independent vectors, so this set must be a basis for our three.

Okay, so we will use determinants to see if these three sets of electors are the newly independence. So we just simply take the tenement 6462 minus seven. There are seven minus three minus five. So the first thing that we do is perform right operations, So we'll take grow one and add that to grow. Three. Let's put that in the right through. So then this is equal. This will be equal to four minus seven months. Three, 60 minus five, four plus two gives you 67 minus seven. Give you 02 minus three gives you minus one soco factor expanding along second column. We need the signs. This is positive and negative. So we negative negative seven. So that politics seven plans by its six minus 56 amorous one, which is equal to seven times with six times on modest one gives you 96 minus by six times went native file, which is needed to 30 which is seven time by 36 which is seven times by 24. Now we just know that this is not equal to zero. And so therefore the vectors are linearly independence


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