# Solving $A\vec x=\vec b$ for matrices

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?

• if the column space is $3$-dimensional, then, yes, it contains any element of $\mathbb R^3$, including $(1,1,1)$ Commented Nov 15, 2020 at 16:23
• Ok. So, what does the solution set become? Commented Nov 15, 2020 at 16:47

$$b\in C(A)$$ 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 $$\mathbb{R}^3$$.