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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.

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1 Answer

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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.

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