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In Exercises $1-4,$ find a least-squares solution of $A \mathbf{x}=\mathbf{b}$ by (a) constructing the normal equations for $\hat{\mathbf{x}}$ and (b) solving for $\hat{\mathbf{x}}$ .$$A=\left[\begin{array}{rr}{1} & {-2} \\ {-1} & {2} \\ {0} & {3} \\ {2} & {5}\end{array}\right], \mathbf{b}=\left[\begin{array}{r}{3} \\ {1} \\ {-4} \\ {2}\end{array}\right]$$

(a) $\left[\begin{array}{cc}{6} & {6} \\ {6} & {42}\end{array}\right]\left[\begin{array}{l}{x_{1}} \\ {x_{2}}\end{array}\right]=\left[\begin{array}{c}{6} \\ {-6}\end{array}\right]$(b) $\hat{x}=\left[\begin{array}{c}{\frac{4}{3}} \\ {-\frac{1}{3}}\end{array}\right]$

Calculus 3

Chapter 6

Orthogonality and Least Square

Section 5

Least-Squares Problems

Vectors

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Hello Today we'll be talking about least squares and using normal equation. So in general, the least squares framework aims to solve the problem of a X equals beef and solving for an optimal X uh, which we did note with a hat to indicate that it's a no approximation. So the normal equations aim to solve this by adding an a transpose to each side's. We have a transpose a X equals a transposed B and then a transpose a. This is what's known as some ah positive semi definite Ah, symmetric positive, semi definite actually. And what this means is that it's always in vertebral because it it's Eigen values. Will all these be greater than zero? So what this allows us to dio is actually take the inverse of this and weaken dio moving this here weaken dio a transpose a inverse a trends Sorry mess that up a transpose a inverse a transpose a X equals a transpose a in verse a transposed Be these cancel and we're left with X hat is equal to a transpose a inverse, a transposed Be So now Ah, let's so for this estimator of X when this is what's called is the OLS estimator. Now we're ready. Toe sulphur This and so are in this case, we have a equals negative 12 to negative three. I could have won three and b is equal to for one to And so we could first will sell for a transpose a a transpose a is equal to I think it if 1 to 2 you negative three will Negative Sorry. Negative 13 transposed times. Negative one to to to give three of 13 and this is going to be equal. Teoh Well, dio in this case, we will do. Let's write this as you have borne Teoh negative to negative 33 And then this is negative for two to negative three negative 13 And so this is a two by three and this is a three by two. The inter dimensions match and the dimensions of the new matrix will be to buy toothy outer dimensions and this will be equal to it will be row by column. So, Toby, negative one times negative one plus negative to just two squared plus negative one squared, which is one plus four plus one is six Ah, and then that's first row by second column, which is going to be negative to negative war. Negative one times two, two times negative. Three negative one times three. That's going to give us negative too minus six minus three. That's negative. 11. And then similarly, the ah diagonals. Will I? I'm sorry. The off diagonal elements will match each other because a transpose a all right, there's appear a transpose a transpose equals a transposed a. So the it's the same thing, which means that often I got the elements will be the same. And then, for the last one, it will be, Ah, the second row temps. The second column, which is two times two plus negative three times negative three plus three times three, which is for two squared plus negative. Three squared Plus three squared, which is four plus nine plus nine, which is 22. So this is a transpose a and now we have to invert this so well, right? A transpose a inverted will be offer to buy two elements. This is one over the determinant of a transposed a times. Um see, I think it's yes, it's a suit to do this will take the determinant of a trans purpose A and then swap. Ah, dear. Negative. Be negative. C and a where these are the four elements of these are the fortunate elements of a tree. It's was a And this will give us Ah, the determinant of this of a transpose a This is going to be equal. Teoh Um a D minus B C, which is six times 22 minus negative, 11 times negative 11 which is going to be 1 30 to minus. Um negative 11 squared. So minus 1 21 which is going to give us 11. Yes, 11. And then this will be 1/11 times. Ah, 22 11 11 and six. So this is, uh, a transpose a inverted. We could see that it indeed is in vertebral. And that again is because the Matrix is symmetric positive, semi definite. So now let's back up for just a second. We have a transpose a in verse just equal to one over 11. 20 to 11 11 6. And then now we just need a transposed Be well so for this here. So we have again that Ah, this is going to be negative one to negative one and just taking the transposed to actually and then to negative 33 and be our column. Vector is going to be 412 Ah, and here the dimensions are again to buy three and through by one. So now, in this case, the resulting dimensions will be the outer dimensions to buy one. So this will be equal to the first row times the column times the secondary times that column. So, first year of times, the only column we have here is negative. One times for plus two times 12 plus negative one times two and that's going to give us negative four. And then, um, the second element of this to buy one column vector is two times for which is eight plus negative. Three times negative types, one which is negative three plus three times to which is six. That's going to give us 11 smell for you. Go down here now back up one more time If you recall our estimator is given by a transpose a inverse, a transposed be. And as we found, we now have ah explicit formula for this. This is given by 1/11 22 11 11 6. This is a two by two matrix and then a trans pose B is given by negative four and 11 Akane Vector of negative four and 11. So now we have a two by two by two by one inner dimensions match or good outer dimensions again will be it to buy one, which is actually what we'd expect because the least scores estimators should match the dimensionality. Um, of thea of the column space of A. This is given by 1/11. Ah, we do the first row by the column and then the second row by the column. So this is given by 22 times negative four plus 11 times, 11 in the first column and then 11 times negative for plus six times 11. The second element and, ah, we can actually just divide everything out by 11 beforehand. So this will be two times negative for, ah cancelling out that 22 11 1/11 and then plus one times 11 just cancelling out one of the 11 and then similarly below we cancel out the 1st 11. We get one times negative for Waas six times 11/11 which is just six times one. And this gives us our final answer for X hat. The least squares estimator, which is negative. Eight plus 11. And the first element, which is equal to three. And then negative four plus six, which is equal to two. And that is our least squares solution. Using the normal equations A calm victor of three. The first element to in the second Thank you.

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