Describe all least squares solutions of the equation ax b a...

Describe all least-squares solutions of the equation $Ax = b$. $$A = \begin{bmatrix} 1 & 0 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \\ 1 & 1 & 0 \end{bmatrix}, b = \begin{bmatrix} 9 \\ 1 \\ 3 \\ 3 \end{bmatrix}$$ The least-squares solutions are all vectors of the form $\hat{x} = \begin{bmatrix} \end{bmatrix} + x_3 \begin{bmatrix} \end{bmatrix}$ with $x_3$ free. (Simplify your answers.)

Transcript

00:02 In this video, i will be talking about how to find the least squares solution to this inconsistent system ax equals b, where a and b are given here.

00:20 So first of all, recall that the least squares solution is the solution to the guaranteed consistent system a transpose ax equals a transpose b.

00:38 It is guaranteed to be consistent by construction using the fact, well, using facts from previous linear algebra topics about the null space, orthogonal complements, and the column space.

00:59 So first of all, we need to find a transpose a.

01:08 You'll notice because of the rotation that we did to transpose a, the diagonal entries of this matrix that we want to calculate will be the sums of the squares of the columns of a.

01:24 So for example, the top left diagonal entry is 1 squared plus negative 2 squared plus 3 squared.

01:38 Similarly, the other diagonal entry is the sum of the squares of 3, 6, and 9.

01:52 And then lastly, the other two entries will be given by the dot product of the two columns.

02:01 So 1 times 3 plus negative 2 times negative 6 plus 3 times 9.

02:09 And that gives us 42.

02:11 So again, there is a little bit of a trick that we can use to make our, make the way we think about this calculation easier.

02:30 The results or the entries here will just be dot products of b with each of the columns of a.

02:44 And because the numbers in b are simple, that results in easy multiplication.

02:56 So now that we have a transpose a and a transpose b, we can solve the normal system, which is this one here.

03:13 So we can solve this system using elimination, substitution, or gaussian elimination, or whatever solving method you want.

03:27 But i'm going to point out that it should be quite easy to see that the second row is just the first row times 3.

03:36 In other words, a linear multiple, a scalar multiple of the first row...

Question

Please give Ace some feedback