Least Square Solution(zeros in one row)

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Homework Statement


Give a least squares solution to Ax=C and give the residual error

A=
-1, 1, 2;
1, -1, 0;
1, -1, 2;

C=
-1;
-1;
2;


Homework Equations



Residual Error= |Ax-C|

The Attempt at a Solution


I have done an RREF on the Transpose of A times A augmented with the Transpose of A time C and got this:
1, -1, 0, 2/3;
0, 0, 1, 1/4;
0, 0, 0, 0;

I'm not sure what I should do about the bottom row... I know that SHOULD be always an LSQ but it doesn't seem to come out. I know that bottom row indicates an infinite amount of solutions but not sure how to proceed next. Please help
 
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Does anyone have any clue?

If it helps, I tried to set the third variable as a number and first and second as equations in terms of each other (i.e 1st in terms of 2nd).
 
I presume this is a linear algebra class and you've learned about rank and rank deficiency. Is A full rank?
 
not in precise terms...I know the columm space of A is 3x2. I looked up rank in wikipedia...
 
I suggest you look at the 2nd equation. What does that tell you? Then, look at the 1st equation. It may help to write them out.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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