<s>[INST] For which values of $a$ does this system of equations $\mathbf{{not}}$ have a unique solution? Here's my system of linear equations:
$\begin{cases}
x + 2y + 2z = 1\\x + ay + 3z = 3\\x + 11y +az = 0\\
\end{cases}$
Thus I have the augmented matrix $\left[\begin{array}{ccc|c}1&2&2&1\\1&a&3&3\\1&11&a&0\end{array}\right]$
By row reduction, I obtain:
$\left[\begin{array}
{ccc|c}1&2&2&1\\0&a-2&1&2\\0&9&a-2&-1\end{array}\right]$
Unfortunately, I am stuck at this stage. I have tried swapping rows around but I didn't have much luck.
Update: I have managed to solve this with the use of the determinant.
Matrix of minors:
$\left[\begin{array}
{ccc}a^2-33&a-3&11-a\\2a-22&a-2&9\\6-2a&1&a-2\end{array}\right]$
Matrix of cofactors:
$\left[\begin{array}
{ccc}a^2-33&3-a&11-a\\22-a&9&a-2\\6-2a&-1&a-2\end{array}\right]$
Adjugate matrix:
$\left[\begin{array}
{ccc}a^2-33&22-2a&6-2a\\3-a&a-2&-1\\11-a&-9&a-2\end{array}\right]$
Det(A) = $1(a^2 - 33) + 2(3 - a) + 2(11 - a) = a^2 - 4a - 5$
$(a - 5)(a + 1) = 0$
Thank you all for your help!
 [/INST] Well if $a = 2$ you have $z=2; y=-\frac 19$ and $x = whatever$.
But if $a-2\ne 0$ keep at it.
Multiply row 2 by $\frac 9{a-2}$ and subtract getting rows.
$\begin{array}\\
1&2&2&|\,\,1\\
0&a-2&1&|\,\,2\\
0&0&(a-2)-\frac 9{a-2}&|-\frac 9{a-2}\\
\end{array}$
Now if $(a-2)-\frac 9{a-2} = 0$ we have a case with not solutions.
That is if $(a-2)^2 = 9$ of $a = 5,-1$.
If we assume $a\ne 2,5,-1$ then we can reduce
$\begin{array}\\
1&2&2&|\,\,1\\
0&1&\frac 1{a-2}&|\,\,\frac 2{a-2}\\
0&0&1&|-\frac 9{a-2}\cdot\frac 1{(a-2)-\frac 9{a-2}}=-\frac{9}{(a-2)^2-9}\\
\end{array}$
..... and that's in the whatchamacallit form "upper triangle"?) so we know this results in a unique solution.
So $a = 5$ or $1$ are the cases without solutions.
(because we'd get $((a-2)-\frac 9{a-2})z=-\frac 9{a-2}$ which if $a = 5$ is
$(3-3)z = -3$ or $0 = 3$ and if $a=-1$ we get $(-3 + 3)z = 3$ or $0 = 3$.
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