I have the following matrix $$ A= \begin{bmatrix} 1 & 3 & 0 & 1 & 0\\ 1 & 2 & 1 &-2 & 1\\ -1 & 1 &-1 & 3 & 2\\ \end{bmatrix} $$ I have calculated the column space and I've got $\text{ Col}(A) = \big\{(1,1,-1)^T,(3,2,1)^T,(0,1,-1)^T\big\}$
Now the question is if the vector $\vec b = (1,1,1)^T$ is a part of the column space of $A$?
And what is the solution set for $A\vec x = \vec b$?
Tried to do Gauss-elimination with the values of the vector on the right side. Is that the correct first step? If not, can someone explain how it is done?
$b\in C(A)$ if and only if you can write it as a linear combination of vectors that are in $C(A)$.
In that case, you can set up a system of equations $\alpha_1 v_1 + \alpha_2 v_2 + \alpha_3 v_3=(1,1,1)$ ($\alpha_i\in\mathbb{R}, v_i \in C(A), i\in[|1,3|]$).
Another approuch would be that $C(A)$ you got are linearly indpendent vectors (it should be if you got it right), and then the answer is yes, because it spans $\mathbb{R}^3$.