A singular value decomposition of a is as follows a usigma...

A singular value decomposition of A is as follows: $A = U\Sigma V^T = \begin{bmatrix} -0.5 & -0.5 & 0.5 & 0.5 \\ 0.5 & -0.5 & 0.5 & -0.5 \\ 0.5 & 0.5 & 0.5 & 0.5 \\ -0.5 & 0.5 & 0.5 & -0.5 \end{bmatrix} \begin{bmatrix} 20 & 0 & 0 & 0 \\ 0 & 10 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} 0.6 & -0.8 \\ 0.8 & 0.6 \end{bmatrix}$ Find the least-squares solution of the linear system $Ax = b$, where $b = \begin{bmatrix} 1 \\ -2 \\ 1 \\ -3 \end{bmatrix}$ $\hat{x}_1 = $ $\hat{x}_2 = $

Transcript

00:01 We need to start this problem by multiplying the given matrix.

00:03 We have a triple multiplication here that we have to do.

00:07 So let's start by multiplying the first two matrices.

00:10 So we're good.

00:10 We can multiply them because the two inside values are four, those are equal, and we're going to end up with a four by two matrix.

00:17 How do we do this multiplication? we go, row times column.

00:23 So it's going to be 0 .5 times 5, that's 2 .5, and then the others are 0.

00:28 So we don't really add them.

00:31 So the first one, let's use a different color.

00:33 The first one, it's going to be 2 .5.

00:39 2 .5.

00:40 Then we do it for the next one.

00:42 We have all of them are just going to be.

00:45 So the first one is 0 .5 times 5, which is 2 .5.

00:48 And then the rest are going to be this times this, which is 0.

00:53 Following, if we repeat, it's going to be the same operation with all of them.

00:57 We're going to get 2 .5 because that's.

00:59 Where the 5 is so 2 .5 2 .5 we do it now for the second column so we have this times this the first one is 0 .5 times 0 so we don't count it plus 7 .5 times 5 that's 2 .5 and then the remaining are going to be times 0 so 2 .5 that means we count this row now so this is again 2 .5 this is going to be now negative 0 .5 negative 2 .5 and now negative 2 .5 again.

01:36 Then we multiply this matrix by the remaining one, which is 0 .6 times 0 .8, 0 .8 times negative 0 .6 .6.

01:52 And we repeat the operation.

01:54 This is a 4 by 4 matrix, and this is a a 4 by 2 matrix.

02:05 Sorry, i actually miswrote them.

02:08 This is a 2, this is a 4 by 2 matrix, and this is a 2 by 2 matrix.

02:16 Yes, it's good because the two inside values are the same, and we're going to get a 4 by 2 matrix as the answer.

02:22 So if we do the operation, we get, let's use a different color, we have to do row times column...