Gaussian Elimination for larger systems
Consider the set of equations
x - y - z = 1
3x + y + z = 15
2x + 2y + 4z = 18
The first equation implies x = 1+ y + z. We now replace x in the other
two equations to get three equations
x = 1+ y + z
3(1+ y + z) + y + z = 15
2(1+ y + z) + 2y + 4z = 18
We now get (justification via distributive law)
x = 1+ y + z
3 + 3y + 3z + y + z = 15
2 + 2y + 2z + 2y + 4z = 18
Now we add like terms (justification follows from distributive law in reverse
and the commutatively of addition) to obtain
x = 1+ y + z
3 + 4y + 4z = 15
2 + 4y + 6z = 18
or
x = 1+ y + z
4y + 4z = 12
4y + 6z = 16
The second equation 4y + 4z = 12 gives 4y = 12-4z
x = 1+ y + z
4y = 12 - 4z
4y + 6z = 16
We replace the 4y in the third equation by what it should equal 12-4z when
x, y and z are known. That gives
x = 1+ y + z
4y = 12 - 4z
12 - 4z + 6z = 16
or
x = 1+ y + z
4y = 12 - 4z
12 + 2z = 16
So 2z= 4 and z = 2. Hence 4y = 12 - 4z gives y = 3- z = 1 and finally x = 1+ y + z = 1 +
1 + 2 = 4
So the solution should be given by x = 4, y = 1 and z = 2.
Check: Substitution of x = 4, y = 1 and z = 2 implies
x - y - z = 1 (Yes as 4 - 1 - 2 =1)
3x + y + z = 15 (Yes as 12 + 1 + 2 = 15)
2x + 2y + 4z = 18 (Yes 8+2+8=18)
Note: I made an error the first time I tried to solve the system and could
not see it. When error is made in a long calculation, if you cannot see
it, redo the calculation elsewhere.
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