# Matrix - solving linear system

1. Sep 14, 2009

### Amy-Lee

linear system:
x + y +z = 2
2x + 3y+ 2z = 3
2x + 3y+ (a2 - 2)z = a+1

when reducing it to row echelon form, the last step looks like the following (if my calculations are right)

1 1 1 2
0 0 1 1
0, 1, a2-5, a-4

the question is to determine all values of a for which he system has
(a)no solution, (b) infinitely many solutions, (c) only one solution

for (a) to happen a2-5 = 0 and a-4 not=0

but I can't seem to factorize a2-5=0 or are my calculations just wrong?

thanks Amy-Lee

2. Sep 14, 2009

### tiny-tim

Welcome to PF!

Hi Amy-Lee! Welcome to PF!
I haven't checked how you got there, but a2 - 5 = 0 is just (a + √5)(a - √5) = 0, or a = ±√5

3. Sep 15, 2009

### Amy-Lee

Thank you Tiny Tim!!!

4. Sep 15, 2009

### HallsofIvy

Staff Emeritus
Did it occur to you that $a^2- 5= 0$ is the same as $a^2= 5$ and so $a= \pm\sqrt{5}$?