Answer
$\mathrm{A}\mathrm{x}=\mathrm{b}$ does not have a solution for all $\mathrm{b}\in \mathbb{R}^{2}$
$\mathrm{A}\mathrm{x}=\mathrm{b}$ has solutions when $\mathrm{b}$ has the form $(\mathrm{t},-3\mathrm{t}), \mathrm{t}\in \mathbb{R}.$
Work Step by Step
Reduce the augmented matrix $[\mathrm{A}\ \mathrm{x}]$
$\left[\begin{array}{lll}
2 & -1 & b_{1}\\
-6 & 3 & b_{2}
\end{array}\right]\left(\begin{array}{l}
.\\
+3r_{1}.
\end{array}\right)\sim\left[\begin{array}{lll}
2 & -1 & b_{1}\\
0 & 0 & 3b_{1}+b_{2}
\end{array}\right]$
The last row represents the equation
$0=3b_{1}+b_{2}$
When $3\mathrm{b}_{1}+\mathrm{b}_{2}\neq 0,\ \mathrm{A}\mathrm{x}=\mathrm{b}$ does not have a solution.
When $\mathrm{b}_{2}=-3\mathrm{b}_{1},\ \mathrm{A}\mathrm{x}=\mathrm{b}$ has a solution.
