Question 173732
<pre><font size = 4, color = "indigo"><b>
Put 1's in front of the single letters

{{{system(-4x-6y-1z=-25, 1x-4y+3z=9, -7x+1y+1z=-7)}}} 

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


{{{system(matrix(3,5,-4x,-6y,-1z,"=",-25,
           1x,-4y,3z, "=",9,
          -7x,1y,1z,"=",-7))}}} 

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

{{{system(matrix(3,5,-4,-6,-1,"|",-25,
 1,-4,3, "|",9,-7,1,1,"|",-7))}}} 

Erase the brace and put parentheses around it:

{{{(matrix(3,5,-4,-6,-1,"|",-25,
           1,-4,3, "|",9,
          -7,1,1,"|",-7))}}}

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,-4,-6,-1,"|",-25,
           1,-4,3, "|",9,
          -7,1,1,"|",-7))}}}

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,-4,3, "|",9,-4,-6,-1,"|",-25,
           
          -7,1,1,"|",-7))}}}

Now we will add 4 times the top row to the 2nd row,
to get a zero where the -4 is. It's easier if you
write 4 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(3,1,4,1,"")}}}{{{(matrix(3,5,1,-4,3,"|",9,
           -4,-6,-1, "|",-25,
          -7,1,1,"|",-7))=(matrix(3,5,1,-4,3,"|",9,
           "","","", "|","",
          -7,1,1,"|",-7))}}}

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

{{{(matrix(3,5,1,-4,3, "|",9,0,-22,11,"|",11,
           
          -7,1,1,"|",-7))}}}

Since all the numbers in the middle row are divisible by
11, we can multiply it through by {{{1/11}}}:

{{{matrix(3,1,"",1/11,"")*(matrix(3,5,1,-4,3, "|",9,0,-22,11,"|",11,
           
          -7,1,1,"|",-7))=(matrix(3,5,1,-4,3, "|",9,0,-2,1,"|",1,
           
          -7,1,1,"|",-7))}}}

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,-4,3,"|",9,
           0,-2,1, "|",1,
          -7,1,1,"|",-7))=(matrix(3,5,1,-4,3,"|",9,
           0,-2,1, "|",1,
          "","","","|",""))}}}

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

{{{(matrix(3,5,1,-4,3,"|",9,
           0,-2,1, "|",1,
          0,-27,22,"|",56))}}}

---

Now we will add -27 times the middle row to 2 times
the 3rd row, to get a zero where the -27 is. It's 
easier if you write -27 to the left of the middle row 
and 2 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,"",-27,2)}}}{{{(matrix(3,5,1,-4,3,"|",9,
           0,-2,1, "|",1,
          0,-27,22,"|",56))=(matrix(3,5,1,-4,3,"|",9,
           0,-2,1, "|",1,
          "","","","|",""))}}}

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

{{{(matrix(3,5,1,-4,3,"|",9,
           0,-2,1, "|",1,
          0,0,17,"|",85))}}}

The bottom row can be multiplied through by {{{1/17}}}

{{{matrix(3,1,"","",1/17)}}}{{{(matrix(3,5,1,-4,3,"|",9,
           0,-2,1, "|",1,
          0,0,17,"|",85))}}}

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


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

{{{(matrix(3,5,1x,-4y, 3z,"=",9,
           0x,-2y,1z, "=",1,
          0x,0y,1z,"=",5))}}}

So we have this system:

{{{system(x-4y+3z=9,-2y+1z=1,z=5)}}}

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

Substitute {{{z=5}}} into the middle equation:

{{{matrix(4,1,
-2y+1(5)=1,
-2y+5=1,
-2y=-4,
y=2)}}}

Finally substitute both {{{z=5}}} and {{{y=2}}} in
the top equation:


{{{matrix(5,1,
x-4y+3z=9,
x-4(2)+3(5)=9,
x-8+15=9,
x+7=9,
x=2)}}}

So {{{x=2}}}, {{{y=2}}}, {{{z=5}}}.

Edwin</pre>