Let $A$ be a $3 \times 3$ matrix made from the variable coefficient of the following system. Let $B$ be a $3 \times 1$ matrix made from the coefficients of the right hand side. Solve the system by computing $A^{-1}B$.
\begin{eqnarray*} 8x+8y-z & = &117\\ x+5y-9z & = &24\\ 7x+y+z & = & 54 \end{eqnarray*}
On
Here's a freebie. The left sides of the equations above can be written in matrix notation by the following:
$$ \left [ \begin{array}{ccc} 8 & 8 & -1 \\ 1 & 5 & -9 \\ 7 & 1 & 1 \\ \end{array} \right ] \left [ \begin{array}{c} x\\ y\\ z\\ \end{array} \right ] \;\; =\;\; A\textbf{x}. $$
$B$ is the column vector you get from right-hand sides of the equations above. Lastly, to compute $A^{-1}$ you will need to perform Gaussian elimination to invert $A$. Do these concepts make sense?
First learn to multiply matrices of any compatible shape together:
If $A$ is $m\times k$ and $B$ is $k \times n$ then these can be multiplied in the order $AB = C$ and $C$ is $m\times n$. So in your problem $A$ is going to be $3 \times 3$ and $b$ is $3\times 1$ and $AB$ is $3\times 1$ and is the list of answers for $x,y,z$.
TO multiply matrices $AB$ , you find the sums of products, going across the rows of $A$ and the columns of $B$ -- that is why the number of columns of $A$ needs to match the number of rows of $B$.
For instance $$\left(\matrix{1&2\\3&4 }\right) \left( \matrix{5\\6} \right) = \left(\matrix{1\cdot 5 + 2 \cdot 6 = 17\\3\cdot 5 + 4 \cdot 6 = 39}\right) =\left(\matrix{17\\39}\right) $$
Then you can look up how to invert a matrix. The best way for you is probably going to be Gaussian elimination. The matrix to start with is the coefficients in your equations: $$ \left(\matrix{ 8&8&-1\\1&5&-1\\7&1&1} \right) $$
ALthough shortcuts would be taken by experienced people, the easy way to see Gaussian elimination is that you are doing the same operatoins of adding a multiple of one row to another row, on the original matrix -- with the aim of reaching the identity matrix -- and on the identity matrix (which will end up being the inverse of the original. (the operations can include dividing a whole row by some number).
For example: $$ \left(\matrix{ 8&8&-1\\1&5&-1\\7&1&1} \right)\left(\matrix{ 1&0&0\\0&1&0\\0&0&1} \right) $$ $$ \left(\matrix{ 15&9&0\\1&5&-1\\7&1&1} \right)\left(\matrix{ 1&0&1\\0&1&0\\0&0&1} \right) $$ $$ \left(\matrix{ 15&9&0\\8&6&0\\7&1&1} \right)\left(\matrix{ 1&0&1\\0&1&1\\0&0&1} \right) $$ $$ \left(\matrix{ 15&9&0\\\frac 43&1&0\\7&1&1} \right)\left(\matrix{ 1&0&1\\0&\frac 16&\frac16\\0&0&1} \right) $$ $$ \left(\matrix{ 15&9&0\\\frac 43&1&0\\\frac {17}{3}&0&1} \right)\left(\matrix{ 1&0&1\\0&\frac 16&\frac16\\0&-\frac 16&\frac 56} \right) $$ $$ \left(\matrix{ 3&0&0\\\frac 43&1&0\\\frac {17}{3}&0&1} \right)\left(\matrix{ 1&-\frac 32&-\frac12\\0&\frac 16&\frac16\\0&-\frac 16&\frac 56} \right) $$ $$ \left(\matrix{ 1&0&0\\\frac 43&1&0\\\frac {17}{3}&0&1} \right)\left(\matrix{ \frac13&-\frac 12&-\frac16\\0&\frac 16&\frac16\\0&-\frac 16&\frac 56} \right) $$ $$ \left(\matrix{ 1&0&0\\0&1&0\\\frac {17}{3}&0&1} \right)\left(\matrix{ \frac13&-\frac 12&-\frac16\\-\frac49&\frac 56&-\frac1{18}\\0&-\frac 16&\frac 56} \right) $$ $$ \left(\matrix{ 1&0&0\\0&1&0\\0&0&1} \right)\left(\matrix{ \frac13&-\frac 12&-\frac16\\-\frac49&\frac 56&-\frac1{18}\\-\frac{17}{9}&\frac 83&\frac {16}{9}} \right) $$ $$ \left(\matrix{ \frac13&-\frac 12&-\frac16\\-\frac49&\frac 56&-\frac1{18}\\-\frac{17}{9}&\frac 83&\frac {16}{9}} \right)\left(\matrix{117\\24\\54} \right) = $$ and that last product gives you your answer. I have not done the work for you because I havve left two errors in the sequence of steps -- but you can see how the inversion is done.
ANoother way to invert a 3 by 3 matrix involves determinants, it you have learned about them.