Get a 0 where the 3 is by multiplying row 1 by -3,
and row 2 by 5, adding them and replacing row 2 by
the result. That's written -3R1+5R2->R2
-15 6 9 | 0
15 -5 -20 | 0
---------------------
0 1 -11 | 0
Get a 0 where the 4 is by multiplying row 1 by -4,
and row 3 by 5, adding them and replacing row 3 by
the result. That's written -4R1+5R3->R3
-20 8 12 | 0
20 -5 -20 | 0
---------------------
0 1 -11 | 0
Get a 0 where the 1 on the bottom row is by
multiplying row 2 by -1, and row 3 by 1,
adding them and replacing row 3 by
the result. That's written -R2+R3->R3
0 1 -11 | 0
0 -1 11 | 0
---------------------
0 0 0 | 0
Convert this back to a system of equations:
Which when simplified is:
z can be assigned any value, some people set z equal
to some other letter, and other just leave it as z.
I'll just leave it as z.
Solve the second equation for y
-y - 11z = 0
-y = -11z
y = 11z
Substitute 11z for y in the first equation:
5x - 2y - 3z = 0
5x - 2(11z) - 3z = 0
5x - 22z - 3z = 0
5x - 25z = 0
5x = 25z
x = 5z
We write the solution as
(x, y, z) = (5z, 11z, z)
Some books and teachers require students to use a different letter
for z, such as
(x, y, z) = (5a, 11a, a) or (5k, 11k, k), etc.
This system is dependent and has infinitely many solutions. Substitute
different numbers for z (or for a or k) and get different solutions.
Edwin
