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??

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*}
