Question 976106
Instead of doing your problem for you, I'll do one just like it and
you can use it as a guide:

{{{system(x + y + z= 4,
2x - y + 3z= 4,
4x + 2y - z= -15)}}}

<pre><font size = 4 color = "indigo"><b>
Write that as a matrix by dropping the letters
and putting vertical line instead of equal signs:

{{{(matrix(3, 5,
     1, 1, 1,"|",4,
     red(2),-1, 3,"|",4,
    red(4),red(2), -1,"|",-15  ))}}}

The idea is to get three zeros in the three positions
in the lower left corner of the matrix, where the elements
I've colored red are:

To get a 0 where the red 2 on the left of the middle row is,
multiply R1 by -2 and add it to 1 times R2, and put it in place 
of the present R2.  That's written as

-2R1+1R2->R2

To make it easy, write the multipliers to the left of the two
rows you're working with; that is, put a -2 by R1 and a 1 by R2

{{{matrix(3,1,-2,1,"")}}}{{{(matrix(3, 5,
     1, 1, 1,"|",4,
     red(2),-1, 3,"|",4,
    red(4),red(2), -1,"|",-15  ))}}}


We are going to change only R2.  Although R1 gets multiplied
by -2 we are going to just do that mentally and add it to R2, but
not really change R1.

{{{(matrix(3, 5,
     1, 1,1,"|",4,
     red(0),-3, 1,"|",-4,
    red(4),red(2), -1,"|",-15  ))}}}

-----

To get a 0 where the lower left red 4 is, multiply R1
by -4 and add it to 1 times R3.  That's written as

-4R1+1R3->R3

Write the multipliers to the left of the two rows you're 
working with; that is, put a -4 by R1 and a 1 by R3

{{{matrix(3,1,-4,"",1)}}}{{{(matrix(3, 5,
     1, 1,1,"|",4,
     red(0),-3, 1,"|",-4,
    red(4),red(2), -1,"|",-15  ))}}}


We are going to change only R3. 

{{{(matrix(3, 5,
     1, 1,1,"|",4,
     red(0),-3, 1,"|",-4,
    red(0),red(-2), -5,"|",-31  ))}}}


---------------

To get a 0 where the red -2 is, multiply R2
by -2 and add it to 3 times R3.  That's written as

-2R2+3R3->R3

Write the multipliers to the left of the two
rows you're working with; that is, put a -2 by R2 and a 3 by R3

{{{matrix(3,1,"","-2",3)}}}{{{(matrix(3, 5,
     1, 1,1,"|",4,
     red(0),-3, 1,"|",-4,
    red(0),red(-2), -5,"|",-31  ))}}}

We are going to change only R3. 

{{{(matrix(3, 5,
     1, 1,1,"|",4,
     red(0),-3, 1,"|",-4,
    red(0),red(0), -17,"|",-85  ))}}}

Now that we have 0's in the three positions in the
lower left corner of the matrix, we change the matrix
back to equations:

{{{system(x+y+z=4,-3y+z=-4,-17z=-85)}}}

Solve the third equation for z:

{{{-17z=-85}}}
{{{z=(-85)/(-17)}}}
{{{z=5}}}

Substitute 5 for z in the middle equation:

{{{-3y+z=-4}}}
{{{-3y+(5)=-4}}}
{{{-3y+5=-4}}}
{{{-3y=-9}}}
{{{y=3}}}

Substitute 5 for z and 3 for y in the top equation:

{{{x+y+z=4}}}
{{{x+(3)+(5)=4}}}
{{{x+3+5=4}}}
{{{x+8=4}}}
{{{x=-4}}}

So the solution is {{{"(x,y,z)"="(-4,3,5)"}}}

Edwin</pre>