[Math] When does this matrix have zero, one and infinite solution

Question

So I've forgotten what the conditions for when a matrix has zero, one and infinitely many solutions. Starting with this matrix:
$$
\begin{align}
&\left[\begin{array}{rrr|r}
1 & 1 & -1 & 0 \\
2 & -1 & -5 & 3\\
-1 & 2 & a^2+3a & -3a
\end{array}\right]
\end{align}
$$

I reduced this to:
$$
\left[\begin{array}{rrr|r}
1 & 1 & -1 & 0 \\
0 & -1 & -1 & 1\\
0 & 0 & a^2+3a-4 & -3a+3
\end{array}\right]
$$
But I'm not sure when this would have zero, one, or infinitely solutions.

Thanks in advance.

Answer 1

If $a^2 + 3a - 4 = 0$ and $-3a +3 =0$ then you will have infinitely many solutions.

If $a^2 + 3a - 4 = 0$ and $-3a+3\neq 0$, then you will not have any solutions. The point is that a row $(0\quad 0\quad 0\quad 1)$ will correspond to the equation $0x_1 + 0x_2 + 0x_3 = 1$.

If $a^2 + 3a - 4 \neq 0$ then you will have exactly one solution.