Determining which values a coefficient can take

Determining which values a coefficient can take

pI have the following system and matrix:/p p$-x_1 + 3x_2-2x_3 = 7\\4x_1 +
2x_2-6x_3 = 14\\4x_1 + 5x_2+\textit{a}x_3 = 23\\$/p p$\begin{pmatrix} -1
amp; 3 amp; -2\\4 amp; 2 amp; -6\\4 amp; 5 amp; \textit{a}
\end{pmatrix}$/p pThe question is to determine for all values of
$\textit{a}$ how many solutions the system has: none, exactly one,
infinitely many./p pI have two ways of solving this, the first one by
finding the determinant:/p p$((-1)\cdot 2 \cdot \textit{a}) + (3\cdot (-6)
\cdot 4) + ((-2)\cdot 4\cdot5) - (4\cdot 2\cdot(-2))-(5\cdot (-6)\cdot
(-1))-(\textit{a}\cdot 4\cdot 3)=-14\textit{a}-126=\textit{a}-9$/p pFrom
this, I would say there are infinitely many solutions, namely $(-\infty,
9)\cup(9, \infty)$./p pBut if this is the case, how can I determine for
all values of $\textit{a}$?/p pThe second one is by doing elimination. I
will end up with this matrix:/p p$\begin{pmatrix} -1 amp; 3 amp;
-2amp;|amp;7\\0 amp; 1 amp; -1amp;|amp;3\\0 amp; 0
amp;\textit{a}+9amp;|amp; 0 \end{pmatrix}$/p pBut then I get stuck,
because I don't know how to continue from this. Should I do the
following?:/p p$\textit{a}+9 = 0$, so $\textit{a} = -9$ and thus getting
the following matrix:/p p$\begin{pmatrix} -1 amp; 3 amp; -2amp;|amp;7\\0
amp; 1 amp; -1amp;|amp;3\\0 amp; 0 amp;-9amp;|amp; 0 \end{pmatrix}$/p