<s>[INST] How to show whether a $3\times 4$ matrix has no solution, a unique solution or infinitely many solutions? The system is :
$$
\begin{matrix}
1 & -4 & 6 & a & | & 0 \\
-2 & 5 & -4 & -1 & | & b \\
1 & -10 & 22 & 8 & | & c
\end{matrix}
$$
After Gaussian elimination, I found that
$$
\begin{array}{cccc|cc}
1 & -4 & 6 & a &  & 0 \\
0 & 1 & -\tfrac{8}{3} & - \left( 2a- \tfrac{1}{3} \right) & & - \tfrac{1}{3}b \\
0 & 0 & 0 & 10-5a & & c-2b
\end{array}
$$
Is it correct and I can continue to determine whether there is no solution, a unique solution or infinitely many solutions?
Here are the operations:
$$
\begin{matrix}
1 & -4 & 6 & a & | & 0 \\
-2 & 5 & -4 & -1 & | & b \\
1 & -10 & 22 & 8 & | & c
\end{matrix}
$$
$R_2+2R_1\rightarrow R_2$
$$
\begin{matrix}
1 & -4 & 6 & a & | & 0 \\
0 & -3 & 8 & 2a-1 & | & b \\
1 & -10 & 22 & 8 & | & c
\end{matrix}
$$
$R_3-R_1\rightarrow R_3$
$$
\begin{matrix}
1 & -4 & 6 & a & | & 0 \\
0 & -3 & 8 & 2a-1 & | & b \\
0 & -6 & 16 & 8-a & | & c
\end{matrix}
$$
$R_3-2R_2\rightarrow R_3$
$$
\begin{matrix}
1 & -4 & 6 & a & | & 0 \\
0 & -3 & 8 & 2a-1 & | & b \\
0 & 0 & 0 & 10-5a & | & c-2b
\end{matrix}
$$
$-\frac 13(R_2)\rightarrow R_2$
$$
\begin{matrix}
1 & -4 & 6 & a & | & 0 \\
0 & 1 & -\frac 83 & -\tfrac{2a-1}{3} & | & -\frac 13b \\
0 & 0 & 0 & 10-5a & | & c-2b
\end{matrix}
$$
 [/INST] There's no solution if there's a row of the form $0 0 0 0 | Q$ where $Q$ is not zero. Which row could possibly look like that? What would have to happen (to $a$, $b$, and $c$) for the row to look like that? 
If there is a solution, then the variable represented by the 3rd column is arbitrary, so there are infinitely many solutions. 
</s>