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I've the following matrix. I shall solve the system of equations for all real number of a. $$ \begin{vmatrix} 1 & 2 & -3 & |4\\ 3 & -1 & 5 & |2\\ 4 & 1 & a^2-14 &|a+2 \end{vmatrix}\Leftrightarrow \begin{vmatrix} 1 & 2 & -3 & |4\\ 0 & -7 & 14 & |-10\\ 0 & -7 & a^2-2 &|a-14 \end{vmatrix} \Leftrightarrow \begin{vmatrix} 1 & 2 & -3 & |4\\ 0 & -7 & 14 & |-10\\ 0 & 0 & a^2-16 &|a-4 \end{vmatrix}\\ z(a^2-14)=a+2\Rightarrow z=\frac{a-4}{a^2-16}\\ a\neq\pm4\\ z=\frac{a-4}{a^2-16}\Rightarrow z=\frac{1}{a+4}\\ $$ Then i put z in the second row to get $y=\frac{2}{a+4}+\frac{10}{7}$.

Then i put y and z in the first row to get $x=\frac{8}{7}- \frac{1}{a+4}$.

The answer is therefor: $$(x,y,z)=(\frac{8}{7}- \frac{1}{a+4},\frac{2}{a+4}+\frac{10}{7},\frac{1}{a+4}) \ where \ a\neq\pm4$$

Have i understood this correct? What can i improve in the solution or in my thinking?

Alessio K
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user9060784
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    Looks good to me. One thing you should also do is consider the cases $a = \pm 4$. Let $a = 4$ in the original matrix and row reduce that. Also, when $a = -4$, there is no solution. – Moo Sep 02 '20 at 14:52
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    You could inspect the cases $a=4$ and $a=-4$. –  Sep 02 '20 at 14:54

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