This is probably the most confusing thing I've ever stumbled across in my academic career.
if you write the augmented matrices for these problems, you will end up with matrix equations of the form: $\displaystyle A \bold{x} = b$
to solve for $\displaystyle \bold{x}$ using the inverse matrix method, we simply multiply both sides by the matrix $\displaystyle A^{-1}$ to obtain $\displaystyle \bold{x} = A^{-1}b$
For example, you first system of equations can be represented by:
$\displaystyle \left( \begin{array}{ccc} 1 & 3 & 0 \\ 0 & 1 & -3 \\ -2 & -1 & 3 \end{array} \right) \left( \begin{array}{c} x \\ y \\ z \end{array} \right) = \left( \begin{array}{c} 9 \\ 10 \\ -4 \end{array} \right)$
............$\displaystyle \downarrow$..................$\displaystyle \downarrow$...............$\displaystyle \downarrow$
...........$\displaystyle A$..................$\displaystyle \bold{x}$...............$\displaystyle b$
I hope you see how we got this. now, find the inverse of $\displaystyle A$ and multiply the right side by it (on the left)
Why is this so confusing? You should have examples in your class notes and textbook to follow .....?
The matrix equation that represents the first system of linear equations is $\displaystyle A X = B$ where
$\displaystyle A = \left[ \begin{tabular}{c c c } 1 & 3 & 0 \\ 0 & 1 & -3 \\ -2 & -1 & 3 \\ \end{tabular} \right]$
$\displaystyle X = \left[ \begin{tabular}{c} x \\ y \\ z \\ \end{tabular} \right]$
$\displaystyle B = \left[ \begin{tabular}{c} 9 \\ 10 \\ - 4 \\ \end{tabular} \right]$.
The solution is given by $\displaystyle X = A^{-1} B$ where I leave it to you find $\displaystyle A^{-1}$ (the inverse of matrix A) and do the matrix multiplication.
Read this: Matrix Inversion: Finding the Inverse of a Matrix.
Or this: The inverse of an nxn matrix.
Or this: Inverse of a Matrix.
And of course you can always just use technology such as a TI-89 calculator.
You should already know how to find the inverse of a 3x3 matrix if you're trying to use matrices to solve systems of linear equations.
You need to go back and revise the basic prerequisite material.
why did you wait till the last minute. of course we can't do the problem for you, but here is some direction
do you know how to bring a matrix to reduced row-echelon form?
set this up:
...........$\displaystyle A$...............$\displaystyle I_n$
...........$\displaystyle \downarrow$................$\displaystyle \downarrow$
$\displaystyle \left( \begin{array}{ccc|ccc} 1 & 3 & 0 & 1 & 0 & 0 \\ 0 & 1 & -3 & 0 & 1 & 0 \\ -2 & -1 & 3 & 0 & 0 & 1 \end{array} \right)$
now, perform row operations (on the entire matrix) to get the matrix $\displaystyle A$ to look like $\displaystyle I_n$. once this is done, the matrix in the $\displaystyle I_n$ position will end up being the matrix $\displaystyle A^{-1}$