Answer
$\mathrm{A}\mathrm{x}=\mathrm{b}$ does not have a solution for all $\mathrm{b}\in \mathbb{R}^{3}$
$\mathrm{A}\mathrm{x}=\mathrm{b}$ has solutions when $\mathrm{b} =(\mathrm{b}_{1},\mathrm{b}_{2},\mathrm{b}_{3})$ such that $\mathrm{b}_{1}+2\mathrm{b}_{2}+\mathrm{b}_{3}=0$
Work Step by Step
Reduce the augmented matrix $[\mathrm{A}\ \mathrm{x}]$
$\left[\begin{array}{llll}
1 & -3 & -4 & b_{1}\\
-3 & 2 & 6 & b_{2}\\
5 & -1 & -8 & \mathrm{b}_{3}
\end{array}\right]\left(\begin{array}{l}
.\\
+3r_{1}.\\
-5\mathrm{r}_{1}
\end{array}\right)$
$\sim\left[\begin{array}{llll}
1 & -3 & -4 & b_{1}\\
0 & -7 & -6 & 3\mathrm{b}_{1}+b_{2}\\
0 & 14 & 12 & -5\mathrm{b}_{1}+\mathrm{b}_{3}
\end{array}\right]\left(\begin{array}{l}
.\\
.\\
+2\mathrm{r}_{2}
\end{array}\right)\sim$
$\sim\left[\begin{array}{llll}
1 & -3 & -4 & b_{1}\\
0 & -7 & -6 & 3\mathrm{b}_{1}+b_{2}\\
0 & 0 & 0 & -5\mathrm{b}_{1}+2\mathrm{b}_{2}+\mathrm{b}_{3}
\end{array}\right]$
The last row represents the equation
$0=\mathrm{b}_{1}+2\mathrm{b}_{2}+\mathrm{b}_{3}$
When $-5\mathrm{b}_{1}+2\mathrm{b}_{2}+\mathrm{b}_{3}\neq 0,\ \mathrm{A}\mathrm{x}=\mathrm{b}$ does not have a solution.
When $-5\mathrm{b}_{1}+2\mathrm{b}_{2}+\mathrm{b}_{3}=0,\ \mathrm{A}\mathrm{x}=\mathrm{b}$ has a solution.
