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Solve for each $a \in \Bbb R$ the following linear system:

$3x+5y-4z=7$

$-3x-2y+4z=-1$

$6x+1y-8z=a$

So I put everything in a matrix en eventually ended with

\begin{pmatrix} 3& 0&-4 \\ 0 & 1 & 0 \\ 0& 0 & 0 \end{pmatrix}

So

$3x-4z=-3$

$ y=2$

$0x+0y+0z=a+4$

I thought, you have to distinguish two possibilities, $a \neq -4$ and $a=-4$. For $a\neq-4$ there are no solutions, and for $a=-4$ I ended up with $(-1+4\lambda,2,3\lambda$).

I'm not sure if this is correct.

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    The last step, the special value should be where $a=-4$, because that is where $a+4=0$. So in the case where $a\neq -4$ you have no solutions, and then calculate for $a=-4$. – 2017-01-25

1 Answers 1

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You can easily verify that everything is ok by yourself. First of all, your system is of rank $2$, and thus solution set is one-dimensional, so having one parameter is correct. What you should now do is evaluate $(-1+4\lambda, 2,3\lambda)$ in the original system and verify that you get $(7,-1,a)$ (i.e., $(7,-1,-4)$ since you already discussed that there are no solutions for $a\neq -4$). I checked with Wolfram and it turns out your solution is correct.