Linear Algebra Solution Set Question

  • #1

Homework Statement


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


Homework Equations



...Row reduction

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?
 

Answers and Replies

  • #2
HallsofIvy
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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.)
 

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