solve homogeneous system of equations
x - 2y + 3z = 0
3x - 7y - 4z = 0
4x - 4y + z = 0
Every homogeneous system of 3 equations in
3 unknowns has solution (x,y,z) = (0,0,0).
But sometimes homogeneous systems have other
solutions. So we go through Gauss elimination:
x - 2y + 3z = 0
3x - 7y - 4z = 0
4x - 4y + z = 0
Make this augmented matrix of coefficients:
[1 -2 3 | 0]
[3 -7 -4 | 0]
[4 -4 1 | 0]
The idea is to get 0's where the three red
numbers are.
[1 -2 3 | 0]
[3 -7 -4 | 0]
[4 -4 1 | 0]
[1 -2 3 | 0]
[3 -7 -4 | 0]
[4 -2 1 | 0]
Get a 0 where the red 3 is by multiplying row 1 by
-3 and adding it to 1 times row 2, but restore row 1:
-3×[1 -2 3 | 0]
1×[3 -7 -4 | 0]
[4 -2 1 | 0]
[1 -2 3 | 0]
[0 -1 -13 | 0]
[4 -2 1 | 0]
Get a 0 where the red 4 is by multiplying row 1 by
-4 and adding it to 1 times row 3, but restore row 1:
-4×[1 -2 3 | 0]
[0 -1 -13 | 0]
1×[4 -2 1 | 0]
[1 -2 3 | 0]
[0 -1 -13 | 0]
[0 6 -11 | 0]
Get a 0 where the red 6 is by multiplying row 2 by
6 and adding it to 1 times row 3, but restore row 2:
[1 -2 3 | 0]
6×[0 -1 -13 | 0]
1×[0 6 -11 | 0]
[1 -2 3 | 0]
[0 -1 -13 | 0]
[0 0 -89 | 0]
Get a 1 where the red -1 is by dividing row 2 thru by -1
Get a 1 where the red -89 is by dividing row 3 thru by -89
[1 -2 3 | 0]
[0 1 13 | 0]
[0 0 1 | 0]
Convert back into a system of equations:
1x - 2y + 3z = 0
0x + 1y + 13z = 0
0x + 0y + 1z = 0
Erase the first 0 term of the 2nd equation,
Erase the first two 0 terms of the third equation:
Erase the 1 coefficients
x + 2y + 3z = 0
y + 13z = 0
z = 0
The last equation tells us the only value of z is 0
Substitute in the 2nd equation:
y + 13(0) = 0
y = 0
Substitute z = 0 and y = 0 in the first equation:
x + 2(0) + 3(0) = 0
x + 0 + 0 = 0
x = 0
So there is but one solution,
the obvious one.
(x, y, z) = (0, 0, 0)
Edwin
