5

1 L 1 1 1 1 1 j 3 1 ] 1Y 1 6 2 L 1 1 1 0 8 3 E 1 V 1 0 1 L 2 1 H 1 1 2 X 0 2 1 3 72} L 1 1 4 1 1 1 51 3 1 1 1 5 1...

Question

1 L 1 1 1 1 1 j 3 1 ] 1Y 1 6 2 L 1 1 1 0 8 3 E 1 V 1 0 1 L 2 1 H 1 1 2 X 0 2 1 3 72} L 1 1 4 1 1 1 51 3 1 1 1 5 1

1 L 1 1 1 1 1 j 3 1 ] 1Y 1 6 2 L 1 1 1 0 8 3 E 1 V 1 0 1 L 2 1 H 1 1 2 X 0 2 1 3 72} L 1 1 4 1 1 1 51 3 1 1 1 5 1



Answers

$\left[\begin{array}{rrr|r}{1} & {-1} & {5} & {-6} \\ {3} & {3} & {-1} & {10} \\ {1} & {3} & {2} & {5}\end{array}\right]-3 R_{1}+R_{2}$

So in this question we have to make two transformations to our one plus Arto Onda minus r one plus R three. So here, only the first tour event as it is untainted. Oneto one toe this will become a zero. Do those are four minus one That this three This is two plus before And this five plus for their this main He had a vin one minus 10 three miners toe that is one minus two minus on that this minus three and minus eight minus. So that is my understanding. So that is my

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

They're. So for this exercise we have this vector B. And the subspace dovey generated by the one, V two and V three that are these vectors that are defined here. So basically we need to calculate the Earth a little projection of you on this space to view. And just remember remember this projection is calculated as the inner proud of the vector V. Each of the generators of this subspace dog. In this case the generators RV one, The two and 3. So we need to calculate the we need to calculate the inner part of me with each of the generator divided the score of the norm of the generators times degenerates. So these for the three vectors B two square plus the interpreter of B would be three. B three. Did the square of the norm of B. Three. Okay, so just to remind you a little bit of the geometric intuition of this, is that the view is generated by these three vectors. So what we're doing is projecting we on each of the generators and then some that together. So we want We t. v. one and V three acts as a basis. Actually in this case they are linearly independent so they form a basis for this. Yeah, subspace of you. So we're writing the in terms of this basis. So we're projecting projecting on this sub space. So let's calculate the correspondent values that we need. So in this case we would be one. The product of B would be to dinner product of the would be three. So this is equal two, one half, There is a constitute and this inner product is equal to zero and then the norms. So because this is the cost to zero means that we don't need this term anymore is going to be equal to zero. So we just need to calculate the score of the norms for B. two and B one. So for me, one square of the norm, remember that there is equal to the inner product of the vector with itself. And in this case this result in one and the inner approach of B two square is equal 2, 1 as well. So these are actually military vectors. And then we just need to put all together on the four. So behalf that the projection of the vector B on the subspace, our view, it's equals to 1/4 times 11 one plus the vector V two. That is equal to one, 1 -1 -1. After some. In these two vectors obtain the action solution that is one half times the vector, three, three minus one minus one. That corresponds to their thermal projection of beyond this subspace of you.

And one more matrix modification here. This time of the times a Roman. eight times I Okay. We will run through what how this works. Okay, a resultant two by two. Matrix again, we're just following our matrix multiplication rules. Row one, column one. One times one. The Stereo Times four will give us a one. Okay, top right. That row one, column two, one times two, zero times three. That will give us a two there. Mhm. Caught him. Right. Row two column oil one time zero. That's four times 1 for their and then I lost out on the bottom right, We know his road to call him too. Zero times two. It was one times 3 for three. His matrix looks familiar. Again, it is a matrix. So multiplying the identity on either side here, either on the left or on the right will not change the matrix that that identity is multiplied.


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