Question 206096
{{{(matrix(3,4,1,1,1,9,2,-3,4,7,1,-4,3,-2))}}}
<pre><font size = 4 color = "indigo"><b>
Multiply row 1 thru by -2 and add to row 2

{{{(matrix(3,4,1,1,1,9,0,-5,2,-11,1,-4,3,-2))}}}

Multiply row 1 by -1 and add to row 3

{{{(matrix(3,4,1,1,1,9,0,-5,2,-11,0,-5,2,-11))}}}

Multiply row 2 by -1 and add to row 3

{{{(matrix(3,4,1,1,1,9,0,-5,2,-11,0,0,0,0))}}}

Now we write the system of equations that this
matrix represents:

{{{system(1x + 1y + 1z =   9,
0x - 5y + 2z = -11,
0x + 0y + 0z =   0)}}}

The bottom equation is all 0's so any values of
x, y, and z would be solutions to it, so we eliminate
it

Choose an arbitrary value for z, we will use {{{z=a}}}

Or, after erasing the 0s and 1 coefficients:

{{{system(x + y + z =   9,
 -5y + 2z = -11)}}}

Substitute {{{a}}} for {{{z}}} in the 2nd equation:

{{{-5y + 2a = -11}}}

Solve for y

{{{-5y=-11-2a}}}

{{{y=11/5+2a/5}}}

Substitute {{{a}}} for {{{z}}} and {{{11/5+2a/5}}} 
for {{{y}}} in the first equation:

{{{x + y + z =   9}}}

{{{x + (11/5 +2a/5) + a = 9}}}

{{{x + 11/5 + 2a/5 + a = 9}}}

Clear of fractions:

{{{5x + 11 + 2a + 5a = 45}}}

{{{5x + 11 + 7a = 45}}}

{{{5x = 34 - 7a}}}

{{{ x = 34/5 - 7a/5}}}

So the solution is:

(x, y, z) = ({{{34/5-7a/5}}},{{{11/5+2a/5}}},{{{a}}}) 

Edwin</pre>