linear systems in three variables
2x + 3y - 6z = 4
3x - 2y - 9z = -7
2x + 5y - 6z = 8
Make this augmented matrix of coefficients:
[2 3 -6 | 4]
[3 -2 -9 | -7]
[2 5 -6 | 8]
The idea is to get 0's where the three red
numbers are.
[2 3 -6 | 4]
[3 -2 -9 | -7]
[2 5 -6 | 8]
[2 3 -6 | 4]
[3 -2 -9 | -7]
[2 5 -6 | 8]
Get a 0 where the red 3 is by multiplying row 1 by
-3 and adding it to 2 times row 2, but restore row 1:
-3×[2 3 -6 | 4]
2×[3 -2 -9 | -7]
[2 5 -6 | 8]
[2 3 -6 | 4]
[0 -13 0 | -26]
[2 5 -6 | 8]
Get a 0 where the red 2 is by multiplying row 1 by
-1 and adding it to 1 times row 3, but restore row 1:
-1×[2 3 -6 | 4]
[0 -13 0 | -26]
1×[2 5 -6 | 8]
[2 3 -6 | 4]
[0 -13 0 | -26]
[0 2 0 | 4]
Get a 0 where the red 2 is by multiplying row 2 by
2 and adding it to 13 times row 3, but restore row 2:
[2 3 -6 | 4]
2×[0 -13 0 | -26]
13×[0 2 0 | 4]
[2 3 -6 | 4]
[0 -13 0 | -26]
[0 0 0 | 0]
Get a 1 where the red 2 is by dividing row 1 thru by 2
Get a 1 where the red -13 is by dividing row 2 thru by -13
[1 
-3 | 2]
[0 1 0 | 2]
[0 0 0 | 0]
Convert back into a system of equations:
1x + y - 3z = 2
0x + 1y + 0z = 2
0x + 0y + 0z = 0
Erase the first 0 term of the 2nd equation,
Erase the first two 0 terms of the third equation:
x + y - 3z = 2
y + 0z = 2
0z = 0
The last equation is true for any value of z
So we let z be arbitrary. Then z = a
Substitute in the 2nd equation:
y + 0(a) = 2
y = 2
Substitute z = a and y = 2 in the first equation:
x + (2) - 3(a) = 2
x + 3 - 3a = 2
x = 3a-1
So the set of solutions is
(x, y, z) = (3a-1, 2, a)
Edwin
