Solve Least Squares Problem for Matrix A and B | Homework Equations

[chuy52506]

Homework Statement

Let
A=
|2 -1 -1|
|-1 2 -1|
|-1 -1 2|
and
B=
|1|
|2|
|3|

Homework Equations

Find the x in which minimizes ||Ax-b||²

The Attempt at a Solution

I tried to solve it by using this formula (A^**A)-1A^**b=x but i get the inverse of A^*A equal 0

 

[Undoubtedly0]

Greetings! Right, since A does not have linearly independent columns, $$A^TA$$ is not invertible. Call $$B = A^TA$$ and $$\vec{y} = A^T\vec{b}$$ and try using row reduction to solve the matrix equation $$B\vec{x} = \vec{y}$$ for $$\vec{x}$$.

 

[chuy52506]

When i try to solve it i get the last row in rref of B=A*A to be a row of 0's equal to 3=/
Is there any other way to solve this??

 

[Undoubtedly0]

I got that

$$B = A^TA = \[ \left[ \begin{array}{ccc}<br /> 6 & -3 & -3 \\<br /> -3 & 6 & -3 \\<br /> -3 & -3 & 6 \end{array} \right]\]$$

and

$$\vec{y} = A^T\vec{b} = \[ \left[ \begin{array}{c}-3 \\<br /> 0 \\3 \end{array} \right]\]$$.

Is this what you got? Then we can row reduce

$$[ B\ \vec{y} ] = \[ \left[ \begin{array}{cccc}<br /> 6 & -3 & -3 & -3 \\<br /> -3 & 6 & -3 & 0 \\<br /> -3 & -3 & 6 & 3 \end{array} \right]\]$$

In fact, reducing this shows that there is a free variable, meaning that there is a whole line worth of solutions that give the best approximation.

 

[chuy52506]

thanks!