3x+2y+5z=1800
4x+y+3z+1450
2x+4y+z=1900
The method is easier if you don't try to get 1's on the
diagonal, but just get three 0's in the lower left corner.
That way you can keep everything whole numbers:
4R1-3R2 -> R2 12 8 20 | 7200
-12 -3 -9 | -4350
--------------------
0 5 11 | 2850
-2R1+3R3 -> R3 -6 -4 -10 | -3600
6 12 3 | 5700
--------------------
0 8 -7 | 2100
8R2-5R3 -> R3 0 40 88 | 22800
0 -40 35 | -10500
--------------------
0 0 123 | 12300
[Note that we can now easily get 1's on the diagonal from
here, just by dividing each row by the first non-zero
element in the row, like this:
But why bother? Why not just leave all the elements as whole
numbers? Why books and teachers insist that the first
non-zero element in every row must be 1 is just plain dumb!!!
Yes, I'm saying that teachers and books are all dumb to
require that you make the first non-zero element in each row
a 1 in Gaussian elimination. You certainly don't have to as
you go!!!]
I would just not bother making the first non-zero elements in
each row be 1, and just take it as it is:
Now the system is simply:
3x + 2y + 5z = 1800
5y + 11z = 2850
123z = 12300
Solving the bottom equation for z
123z = 12300
z = 100
Substituting z = 100 in the 2nd equation
5y + 11(100) = 2850
5y + 1100 = 2850
5y = 1750
y = 350
Substituting y = 350 and z = 100 in the first equation
3x + 2(350) + 5(100) = 1800
3x + 700 + 500 = 1800
3x + 1200 = 1800
3x = 600
x = 200
So the solution is (x,y,z) = (200,350,100)
Since your teacher is probably one of the dumb-bunnies who
requires that the first non-zero element in each row be 1,
then just use the matrix above where we divided every row
by its first non-zero element and got:
and go from there, dealing with the fractions. This is just
to satisfy the dumb mathematicians who think it's necessary
to have 1's on the diagonal. It isn't at all!
Edwin
