Matrix - solving linear system

[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

 

[tiny-tim]

Welcome to PF!

Hi Amy-Lee! Welcome to PF!

… 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?

I haven't checked how you got there, but a² - 5 = 0 is just (a + √5)(a - √5) = 0, or a = ±√5

 

[Amy-Lee]

Thank you Tiny Tim!

 

[HallsofIvy]

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