Construct a $2 \times 3$ system $Ax = b$ with solution $(1 -\alpha, 3 + \alpha, \alpha)$

Question

I came across this problem. I'm trying to figure out how to go about answering it.

My first thought is that the solution $x̄$ is going to look like

$$\begin{pmatrix}1\\3\\0 \end{pmatrix} + \alpha\begin{pmatrix}-1\\1\\1 \end{pmatrix}$$ and if $Ax = b$, then $$b = A\begin{pmatrix}1\\3\\0 \end{pmatrix} + αA\begin{pmatrix}-1\\1\\1 \end{pmatrix}$$

...right?

I need some help or verification. I don't know if this is satisfactory. Thanks for any help!

Answer 1

Note that$$3(1-\alpha)-(3+\alpha)+6\alpha=0$$and that$$(1-\alpha)+(3+\alpha)=4.$$So, consider the system$$\left\{\begin{array}{l}3x-y+6z=0\\x+y=4.\end{array}\right.$$