Question 246519
Solve the system using augmented matrix. Show your work.

{{{system(5x + 4y - z= 1,
2x - 2y + z= 1,
-x - y + z= 2)}}}


I just don't know what to do!
<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,
     5, 4,-1,"|",1,
     red(2),-2, 1,"|",1,
    red(-1),red(-1), 1,"|",2  ))}}}

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 is, multiply R3
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,"",1,2)}}}{{{(matrix(3, 5,
     5, 4,-1,"|",1,
     red(2),-2, 1,"|",1,
    red(-1),red(-1), 1,"|",2  ))}}}

We are going to change only R2.  Although R3 gets multiplied
by 2 we are going to just do that mentally and add it to R2, but
not really change R3.

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

-----

To get a 0 where the lower left red -1 is, multiply R1
by 1 and add it to 5 times R3.  That's written as

1R1+5R3->R3

Write the multipliers to the left of the two rows you're 
working with; that is, put a 1 by R1 and a 5 by R3

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

We are going to change only R3. 

{{{(matrix(3, 5,
     5, 4,-1,"|",1,
     red(0),-4, 3,"|",5,
    red(0),red(-1), 4,"|",11  ))}}}

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

To get a 0 where the remaining red -1 is, multiply R2
by 1 and add it to -4 times R3.  That's written as

1R2-4R3->R3

Write the multipliers to the left of the two
rows you're working with; that is, put a +1 by R2 and a -4 by R3

{{{matrix(3,1,"","+1",-4)}}}{{{(matrix(3, 5,
     5, 4,-1,"|",1,
     red(0),-4, 3,"|",5,
    red(0),red(-1), 4,"|",11  ))}}}

We are going to change only R3. 

{{{(matrix(3, 5,
     5, 4,-1,"|",1,
     red(0),-4, 3,"|",5,
    red(0),red(0),-13,"|",-39  ))}}}

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(5x+4y-z=1,-4y+3z=5,-13z=-39)}}}

Solve the third equation for z:

{{{-13z=-39}}}
{{{z=(-39)/(-13)}}}
{{{z=3}}}

Substitute 3 for z in the middle equation:

{{{-4y+3z=5}}}
{{{-4y+3(3)=5}}}
{{{-4y+9=5}}}
{{{-4y=-4}}}
{{{y=1}}}

Substitute 3 for z and 1 for y in the top equation:

{{{5x+4y-z=1}}}
{{{5x+4(1)-(3)=1}}}
{{{5x+4-3=1}}}
{{{5x+1=1}}}
{{{5x=0}}}
{{{x=0/5}}}
{{{x=0}}}

So the solution is {{{"(x,y,z)"="(0,1,3)"}}}

Edwin</pre>