This is more (linear) algebra than calculus, but okey...
Because of the variables used, it looks to me that you have to treat a as a parameter.
First, find out for which values of a, the coŽfficiŽnt matrix isn't regular (hint: think determinant).
I have tried and tried to solve this set of 3 linear equations. The problem is that instead of there being 3 equations and 3 unknowns, there are 3 equations with 4 unknowns. I have tried row reduction and keep getting into a mess with squares of the 4th unknown etc. any help would be greatly appreciated.
Q. Solve completely the following set of linear equations.
x + 2y + az = 0
2x + 3y - 2z = a
ax + y + z =3
I found a=-1 or a = 1/3, perhaps one of us made a little mistake.
Now, you have three cases:
1) a = -1
2) a = 1/3
3) a isn't -1 or 1/3
For 1 and 2: substitute the value for a and solve, it'll either be underdetermined or false.
For 3: the system is regular, thus you can uniquely solve x,y,z as a function of a (Cramer, Gauss, ...)
It IS going to be long and messy. My suggestion is to write out EVERYTHING: every tiny little step by step change that you make. It will take a while, and you'll probably make some mistakes and have to start over, but it's the only way. If you write out all your steps you will be able to find mistakes more easily.
Multiply the first equation by 2 and subtarct the second to get:
y + (2a+2)z=-a. ...(A)
Now multiply the second equation by a and the third by 2 and subtarct:
(2ax + 3ay - 2az)-(2ax + 2y + 2z) = a^2-6
(3a-2)y - (2a+2)z = a^2-6 ...(B)
Now add equationa (A) and (B) to get:
(3a-1)y = a^2-a-6,
y = (a^2-a-6)/(3a-1) ...(C).
Substituting this into equation (A) gives:
Then substitute (C) and (D) into the first equation to find x.
(Check the algebra it can always go wrong)