Question 182659
Your first step is to write out the coefficient matrix:
{{{ matrix(3,4, 1,1,1,2, 2,-3,2,4, 4, 1, -3, 1) }}}
<br>
You want to end up with a matrix that looks like this (triangular form):
{{{ matrix(3,4, "#", "#", "#", "#", 0, "#", "#", "#",  0, 0, "#", "#") }}}
<br>
To do that, you perform a combination of the elementary row operations:
1) Switch any 2 rows
2) Multiply each row element by a non-zero constant
3) Replace a row by adding its values to a multiple of another row
<br>
Since our first row is all 1's, it's easy to pick multipliers that will 'zero' out the required fields in rows 2 & 3 when we add the rows together (using row operation 3).  So let's replace row 2 with (-2 * Row 1) + Row 2 -- here's the shorthand way to say that:
{{{ -2R[1] + R[2] -> R[2] }}}
So basically, Row 1 and Row 3 remain the same, only row 2 changes:
{{{ matrix(3,4, 1, 1, 1, 2, 0, -5, 0, 0, 4, 1, -3, 1) }}}
<br>
So that got us a zero under the 1 in the first column which is why we picked -2 as the multiplier.  Next, we'll replace row 3 with (-4 * Row 1) + Row 3:
{{{ -4R[1] + R[3] -> R[3] }}}
{{{ matrix(3,4, 1, 1, 1, 2, 0, -5, 0, 0, 0, -3, -7, -7) }}}
<br>
So now, our 1st column looks like we want it.  We only have one more zero to create (in the 3rd row, 2nd column), so we'll choose {{{ -3/5 }}} as our multiplier and use it against row 2 this time:
{{{ (-3/5)R[2] + R[3] -> R[3] }}}
{{{ matrix(3,4, 1, 1, 1, 2, 0, -5, 0, 0, 0, 0, -7, -7) }}}
<br>
We now have our matrix in triangular form -- let's write the equations using the coefficients from it:
{{{ x + y + z = 2 }}}
{{{ -5y = 0 }}}
{{{ -7z = -7 }}}
<br>
So solving from the bottom up (reverse substitution), we know <b>z = 1</b> and <b>y = 0</b>.  Substituting these values into the top equation gives us <b>x = 1</b>.  As a final check, you can substitute these values in each of the 3 original equations:
{{{ 1 + 0 + 1 = 2 }}}
{{{ 2(1) - 3(0) + 2(1) = 4 }}}
{{{ 4(1) + 0 - 3(1) = 1 }}}
<br>
Hope this helps.  ~ Joe