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??
 A: Note that your system is equivalent to the matrix equation
$$
\begin{bmatrix}
1 & 3 & -3\\ 0 & 1 & 2 \\ 2 & 5 & a^2-9
\end{bmatrix}
\begin{bmatrix}
x\\ y\\ z
\end{bmatrix}
=
\begin{bmatrix}
4\\ a\\9
\end{bmatrix}
$$
Since
$$
\det\begin{bmatrix}
1 & 3 & -3\\ 0 & 1 & 2 \\ 2 & 5 & a^2-9
\end{bmatrix}=a^2-1
$$
this system is guaranteed a unique solution for $a\neq\pm 1$ (do you know why?).
Now the augmented systems for $a=1$ is
$$
\begin{bmatrix}
1&3&-3&4\\ 0&1&2&1\\ 2&5&-8&9
\end{bmatrix}
$$
Row-reducing this matrix gives
$$
\DeclareMathOperator{rref}{rref}\rref\begin{bmatrix}
1&3&-3&4\\ 0&1&2&1\\ 2&5&-8&9
\end{bmatrix}
=
\begin{bmatrix}
1&0&-9&0\\ 0&1&2&0\\ 0&0&0&1
\end{bmatrix}
$$
This system is not consistent (why?) so the original system has no solution for $a=1$.
Can you repeat the process for $a=-1$?
Addendum. You mention in your question that you're having trouble taking determinants. To find the determinant computed above we can expand about the first column:
\begin{align*}
\det\begin{bmatrix}
\color{blue}1 & \color{red}3 & \color{red}{-3}\\ \color{blue}0 & \color{green}1 & \color{green}2 \\ \color{blue}2 & \color{purple}5 & \color{purple}{a^2-9}
\end{bmatrix}
&= (\color{blue}{1})\cdot\det
\begin{bmatrix}\color{green}1&\color{green}2\\\color{purple}5 & \color{purple}{a^2-9} \end{bmatrix}-(\color{blue}0)\det\begin{bmatrix}\color{red}3 & \color{red}{-3}\\ \color{purple}5 & \color{purple}{a^2-9}\end{bmatrix}+(\color{blue}{2})\det\begin{bmatrix}\color{red}3 & \color{red}{-3}\\ \color{green}1 & \color{green}2 \end{bmatrix} \\
&= \left(a^2-9-10\right)-(0)+2\,\left(6+3\right) \\
&= a^2-19+18 \\
&= a^2-1
\end{align*}
A: 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 $x=\frac{-3a^2+7a+5}{a-1}$, $y=\frac{a^2-a-2}{a-1}$, $z=\frac{1}{a-1}$.
A: the determinant of a 3X3 matrix can be calculated by adding the downward cross products and subtracting the upward cross products, to see what I mean ..
$$\det\begin{bmatrix}
1 & 3 & -3\\ 0 & 1 & 2 \\ 2 & 5 & a^2-9
\end{bmatrix} \\$$ $$
 = (1)(1)(a^2-9)+(3)(2)(2)+(-3)(5)(0) \\- (2)(1)(-3)- (5)(2)(1)-(a^2-9)((3)(0)\\$$ $$
=(a^2-9)+12+0-(-6)-(10)-0 $$ $$=a^2-1 $$
