<s>[INST] For which values does the Matrix system have a unique solution, infinitely many solutions and no solution? Given the system: 
$$\begin{align}
 & x+3y-3z=4 \\ 
 & y+2z=a \\
 & 2x+5y+(a^2-9)z=9
\end{align}$$
For which values of a (if any) does the system have a unique solution, infinitely many solutions, and no solution?

So I am getting that it has:
infinitely many solutions at: (-1)

No solution at (1)

Unique solution at (-1,1)

AM I RIGHT??
 [/INST] Simply by Gauss-Jordan elimination you get:
$$\left(\begin{array}{ccc|c}
1 & 3 & -3 & 4\\
0 & 1 & 2 & a\\
2 & 5 & a^2-9 & 9
\end{array}\right)\sim
\left(\begin{array}{ccc|c}
1 & 3 & -3 & 4\\
0 & 1 & 2 & a\\
0 &-1 & a^2-3 & 1
\end{array}\right)\sim
\left(\begin{array}{ccc|c}
1 & 3 & -3 & 4\\
0 & 1 & 2 & a\\
0 & 0 & a^2-1 & a+1
\end{array}\right)\sim
\left(\begin{array}{ccc|c}
1 & 3 & -3 & 4\\
0 & 1 & 2 & a\\
0 & 0 & (a-1)(a+1) & a+1
\end{array}\right)$$
Now we would like to divide by $(a-1)(a+1)$. But we can do this only if this expression is non-zero. So we have to deal with the cases $a=\pm1$ separately.
If $a=1$, the last row is 
$\left(\begin{array}{ccc|c}
0 & 0 & 0 & 2
\end{array}\right)$,
which corresponds to the equation $0x+0y+0z=2$. The equation $0=2$ clearly has no solution.
If $a=-1$ you get the system
$$\left(\begin{array}{ccc|c}
1 & 3 &-3 & 4\\
0 & 1 & 2 & -1\\
0 & 0 & 0 & 0
\end{array}\right)$$
which has infinitely many solutions. (You can also compute them if you want.)
In all other cases (i.e., for $a\ne\pm1$) you can continuo and you get
$$
\left(\begin{array}{ccc|c}
1 & 3 & -3 & 4\\
0 & 1 & 2 & a\\
0 & 0 & (a-1)(a+1) & a+1
\end{array}\right)\sim
\left(\begin{array}{ccc|c}
1 & 3 & -3 & 4\\
0 & 1 & 2 & a\\
0 & 0 & 1 & \frac1{a-1}
\end{array}\right)\sim
\left(\begin{array}{ccc|c}
1 & 0 &-9 & 4-3a\\
0 & 1 & 2 & a\\
0 & 0 & 1 & \frac1{a-1}
\end{array}\right)\sim
\left(\begin{array}{ccc|c}
1 & 0 & 0 & \frac{-3a^2+7a+5}{a-1}\\
0 & 1 & 0 & \frac{a^2-a-2}{a-1}\\
0 & 0 & 1 & \frac1{a-1}
\end{array}\right)$$
So in these cases the solution is $x=\frac{-3a^2+7a+5}{a-1}$, $y=\frac{a^2-a-2}{a-1}$, $z=\frac{1}{a-1}$.
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