<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] The easiest thing is to use determinants: the system has a unique solution if and only if the matrix is invertible, i.e., if and only if the matrix has determinant not equal to $0$. So you can compute the determinant (either in your original or row-reduced matrix) in terms of $a$, set it equal to $0$, and solve for $a$. But if you do not know this yet, you can proceed as follows:
If $a=2$, then the system clearly has a unique solution, so you may assume $a\neq 2$. The reason to do this is so that we can continue the row reduction by multiplying by $a-2$. Multiply the second row by $9$ and the last row by $a-2$ to get:
$$\left(\begin{array}{ccc|c}
1 & 2 & 2 & 1\\
0 & 9(a-2) & 9 & 18\\
0 & 9(a-2) & (a-2)^2 & 2-a
\end{array}\right).$$
Subtracting the second row from the third row, we get
$$\left(\begin{array}{ccc|c}
1 & 2 & 2 & 1\\
0 & 9(a-2) & 9 & 18\\
0 & 0 & (a-2)^2-9 & -16-a
\end{array}\right).$$
So: you will have a unique solution if $(a-2)^2-9\neq 0$. If $(a-2)^2-9=0$, you either get infinitely many solutions (if $16+a=0$ as well), or no solutions (if $-16-a\neq 0$). So now we just need to figure out when that bottom right entry of the coefficient matrix is $0$.
</s>