So I have this question:
In $\mathbb R^3$ with inner product calculate all the least square solutions, and choose the one with shorter length, of the system:
$ x + y + z = 1 $
$ x + z = 0 $
$ y = 0 $
My attempt: So I applied the formula $ A^T A x = A^T b $ with A as being the matrix with row 1 (1,1,1) row 2 (1,0,1) and row 3 (0,1,0); x being the column $(x_1,x_2,x_3)$ and b being the column (1,0,0).
So I did it and I reached to the solution $(x_1, \frac {1}{3}, \frac {1}{3} + x_1)$
And I expanded this solution in two vectors $(0, \frac {1}{3}, \frac {1}{3}) $ and $(1,0,1)$.
So these are the least square solutions and the one with shorter length is the first one.
My doubt is if I'm doing this correctly or if I made any mistake because I used an online calculator that only give one least square solution. Can someone help me to verify my attempt? Thanks!