x+y=7
y+z=8
x+2z=2
Write it neatly lined up so that
the like terms are under each other
and the coefficients, + and = signs
line up vertically:
x + y = 7
y + z = 8
x + 2z = 2
Indicate missing terms with the letters
with 0 coefficients and put 1's for the
coefficients of the letters that don't show
their coefficients:
1x + 1y + 0z = 7
0x + 1y + 1z = 8
1x + 0y + 2z = 2
Erase the letters and the + signs.
Replace the equal signs by a vertical
line. Enclose the array in a bracket:
[1 1 0 | 7]
[0 1 1 | 8]
[1 0 2 | 2]
The opect is to get that with 1's on
the diagonal (left of the vertical bars,
and 0's elsewhere.
Multiply the top row temporarily by -1,
getting:
-1 -1 0 -7
and add it to the 3rd row to
get a 0 where the 1 is on the
bottom left. We denote that
operation by -R1+R3->R3
[1 1 0 | 7]
[0 1 1 | 8]
[0 -1 2 | -5]
Add row 2 to row 3 to get a 0 where
the -1 is. We denote that operation
by R2+R3->R3
[1 1 0 | 7]
[0 1 1 | 8]
[0 0 3 | 3]
Divide the bottom row through by 3
We denote that operation by
1/3R3->R3
[1 1 0 | 7]
[0 1 1 | 8]
[0 0 1 | 1]
Multiply the bottom row temporarily
by -1, getting:
0 0 -1 -1
and add it to the middle row to get a
0 where the 1, which is the 3rd
element on the second row, is.
That operation is denoted by
-R3+R2->R2
[1 1 0 | 7]
[0 1 0 | 7]
[0 0 1 | 1]
Multiply the middle row by -1
0 -1 0 -7
and add it to the top row to get a
0 where the 1, which is the 2nd
element on the top row, is.
This is denoted by -R2+R1->R1
[1 0 0 | 0]
[0 1 0 | 7]
[0 0 1 | 1]
That is in row reduced echelon form
If we were to convert that back to a
system of equations we would have:
1x + 0z + 0z = 0
0x + 1y + 0z = 7
0x + 0y + 1z = 1
which is the same as
x = 0
y = 7
z = 1
And the solution is
(x,y,z) = (0,7,1)
which we could have told from
the last column of the final
matrix above.
This was an easy one. Others
are not so easy, but the
principle is the same.
Edwin