5

8 I2 J 2 1 3 8 I 9 1 3 3 1 2 1844 1 2 1 1 1 2 2 L 1 8 1 3 JN JW 08 3 2 H 1 6 828 { V K 3 M 1 521 1...

Question

8 I2 J 2 1 3 8 I 9 1 3 3 1 2 1844 1 2 1 1 1 2 2 L 1 8 1 3 JN JW 08 3 2 H 1 6 828 { V K 3 M 1 521 1

8 I2 J 2 1 3 8 I 9 1 3 3 1 2 1844 1 2 1 1 1 2 2 L 1 8 1 3 JN JW 08 3 2 H 1 6 828 { V K 3 M 1 521 1



Answers

$\begin{array}{|c|c|c|c|}\hline m_{i} & {8} & {1} & {4} \\ \hline\left(x_{i}, y_{i}\right) & {(-3,-1)} & {(0,0)} & {(-1,2)} \\ \hline\end{array}$

Okay, we're asking to find the norm of the. Now the norm of V is the magnitude of the So it's really, if you consider that these are components, um they would form triangles. And we're really looking for our high pot news because that is going to be the distance Between 00 and the point that we are going to. Okay, so for a we will be finding the magnitude of the by taking the square root of one squared plus negative one squared. And obviously it doesn't matter if we bring our negative in because it is going to get squared. Um but that will end up being squared too. So when I write my problems out, I will put the negatives in there. Um But often on my calculator, I don't, if you forget apprentices, you're taking a negative of a one squared, which can be an issue. So just um, I would say in your calculator because you know, you're squaring it, just put the positive in there and so you don't make any um silly mistakes. So this guy is the square root of negative one squared plus seven squared, which is square at a 50 50 is made of 25 2. So we can take the square root of 25 put a five in front five square too. Now notice our next um vector has three components, and if you have a hypothesis of two of them and you square it and you add it to the third squared and take that square root, um that also works. So what happens is you can again just take the some of your squares, it works fine. So if you're finding the magnitude, we can just do a negative one squared plus a two squared plus a forest squared, and then take the square root of that. So let's say we have 14 16, So that is going to be 21, so square root of 21 and our last one also have the three components. So we will do a three squared a two squared in a one squared, so that's nine plus four plus one. So that will be the square root of 14.

So on this problem, we're given this matrix and were asked to use matrix capability of a graphing utility to find the determinant. So I went to Desmond's dot com and went to math tools matrix calculator and got this matrix calculator. So I have a four x four matrix. So I go new matrix And I got four rows and four columns And the first entry is a zero and then a minus three in an eight and that too. And then eight one a minus one and six And then I -4 in a six. Mhm and zero and a nine. And then uh minus seven, two, zeros zero, zero and 14. They had dinner now to find the determiner of this, I use the D E T button right here go determine it a matrix A And there it is 7000 441 7441.

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.

So in this question, we have to perform an operation. Are one entertained with our do. So it's very simple. We just have to rewrite the rose. So are you will become our one basically one minus three. Why do for on here? I have minus six nine and for so that's my answer.


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