# Linear Algebra Solution Set Question

1. Sep 25, 2008

### pimpalicous

1. The problem statement, all variables and given/known data
Find all solutions of a for which the resulting linear system has (a) no solution, (b) a unique solution, (c) infinetly many solutions.

x+z=4
2x+y+3z=5
-3x-3y+(a^2-5a)=a-8

2. Relevant equations

...Row reduction

3. The attempt at a solution

I solved the augmented matrix and got z=a-5/[(a-2)(a-3)], then got x and y in terms of z. I solved the homogeneous and only got the trivial solution regardless of a's value. I'm thinking that the system has no solution when a=2 or a=3, and has a unique solution when a equals any other real number. Is this correct?

2. Sep 25, 2008

### HallsofIvy

Staff Emeritus
Yes, that is correct. The only way you could have "infinitely many solutions" would be if you arrived at something like "z= (a-2)/[(a-2)(a-3)]". In that case, any value of z would satisfy (2-2)(2-3)z= (2-2) so there would be infinitely many solutions for a= 2.. (I haven't checked your calculations. I am assuming you did not cancel something like (a-b)/(a-b) in your solution.)