Question 171695
<pre><font size = 4, color = "indigo"><b>
Put 1's in front of the single letters

{{{system(7x+8y-1z=9, 1x-2y-5z=-31, -7x+1y+1z=0)}}} 

Separate into terms, erasing the plus signs, keeping
the minus signs as negative signs:


{{{system(matrix(3,5,7x,8y,-1z,"=",9,
           1x,-2y,-5z, "=",-31,
          -7x,1y,1z,"=",0))}}} 

Erase all the letters and replace the equal signs
with "|"'s:

{{{system(matrix(3,5,7,8,-1,"|",9,
 1,-2,-5, "|",-31,-7,1,1,"|",0))}}} 

Erase the brace and put parentheses around it:

{{{(matrix(3,5,7,8,-1,"|",9,
           1,-2,-5, "|",-31,
          -7,1,1,"|",0))}}}

Now we want to end up with a matrix like this,
with three zeros on the the bottom left, and
numbers everywhere else:

{{{(matrix(3,5,"#","#","#","|","#",
           0,"#","#", "|","#",
          0,0,"#","|","#"))}}}

Start with this:

{{{(matrix(3,5,7,8,-1,"|",9,
           1,-2,-5, "|",-31,
          -7,1,1,"|",0))}}}

Swap the rows so that the smallest number in absolute
value in the first column is on the far left of the 
top row.  Since 1 is the smallest number in absolute
value in row 1, I will swap rows 1 and 2:

{{{(matrix(3,5,1,-2,-5, "|",-31,7,8,-1,"|",9,
           
          -7,1,1,"|",0))}}}

Now we will add -7 times the top row to the 2nd row,
to get a zero where the 7 is. It's easier if you
write -7 to the left of the top row and 1 to the left
of the second row,and write that equal to a new matrix
with the same 1st and 3rd rows, with a blank middle row:

{{{matrix(3,1,-7,1,"")}}}{{{(matrix(3,5,1,-2,-5,"|",-31,
           7,8,-1, "|",9,
          -7,1,1,"|",0))=(matrix(3,5,1,-2,-5,"|",-31,
           "","","", "|","",
          -7,1,1,"|",0))}}}

Then you can easily fill in the blank row term by term as:

{{{(matrix(3,5,1,-2,-5, "|",-31,0,22,34,"|",226,
           
          -7,1,1,"|",0))}}}

Since all the numbers in the middle row are even, we can
multiply it through by {{{1/2}}}:

{{{matrix(3,1,"",1/2,"")*(matrix(3,5,1,-2,-5, "|",-31,0,22,34,"|",226,
           
          -7,1,1,"|",0))=(matrix(3,5,1,-2,-5, "|",-31,0,11,17,"|",113,
           
          -7,1,1,"|",0))}}}

Now we will add 7 times the top row to the 3rd row,
to get a zero where the -7 is. It's easier if you
write 7 to the left of the top row and 1 to the left
of the bottom row,and write that equal to a new matrix
with the same 1st and 2nd rows, with a blank bottom row:

{{{matrix(3,1,-7,"",1)}}}{{{(matrix(3,5,1,-2,-5,"|",-31,
           0,22,34, "|",226,
          -7,1,1,"|",0))=(matrix(3,5,1,-2,-5,"|",-31,
           0,11,17, "|",113,
          "","","","|",""))}}}

Then you can easily fill in the blank row term by term as:

{{{(matrix(3,5,1,-2,-5,"|",-31,
           0,11,17, "|",113,
          0,-13,-34,"|",-217))}}}

---

Now we will add 13 times the middle row to 11 times
the 3rd row, to get a zero where the -13 is. It's 
easier if you write 13 to the left of the middle row 
and 11 to the left of the bottom row,and write that 
equal to a new matrix with the same 1st and 2nd rows,
with a blank bottom row:

{{{matrix(3,1,"",13,11)}}}{{{(matrix(3,5,1,-2,-5,"|",-31,
           0,11,17, "|",113,
          0,-13,-34,"|",-217))=(matrix(3,5,1,-2,-5,"|",-31,
           0,11,17, "|",113,
          "","","","|",""))}}}

Then you can easily fill in the blank row term by term as:

{{{(matrix(3,5,1,-2,-5,"|",-31,
           0,11,17, "|",113,
          0,0,-153,"|",-918))}}}

The bottom row can be multiplied through by {{{-1/153}}}

{{{matrix(3,1,"","",-1/153)}}}{{{(matrix(3,5,1,-2,-5,"|",-31,
           0,11,17, "|",113,
          0,0,-153,"|",-918))}}}

{{{(matrix(3,5,1,-2,-5,"|",-31,
           0,11,17, "|",113,
          0,0,1,"|",6))}}}


Now we put the letters back as we took them out, and
put equal signs where the "|"'s are:

{{{(matrix(3,5,1x,-2y,-5z,"=",-31,
           0x,11y,17z, "=",113,
          0x,0y,1z,"=",6))}}}

So we have this system:

{{{system(x-2y-5z=-31,11y+17z=113,z=6)}}}

Now we do what is called "back-substitution":

Substitute {{{z=6}}} into the middle equation:

{{{matrix(4,1,
11y+17(6)=113,
11y+102=113,
11y=11,
y=1)}}}

Finally substitute both {{{z=6}}} and {{{y=1}}} in
the top equation:


{{{matrix(5,1,
x-2y-5z=-31,
x-2(1)-5(6)=-31,
x-2-30=-31,
x-32=-31,
x=1)}}}

So {{{x=1}}}, {{{y=1}}}, {{{z=6}}}.

Edwin</pre>