Systems of linear equations

Consider the linear system
(*) {x+az=b
....{2x-y-z=0
....{-x+ay+2z=1
,where a and b are real numbers.

Find the range of values of a such that (*) has a unique solution. a=/=1 and a=/=-1
Solve (*) when it has a unique solution.
Is the ans x =(-2b+a+ab)/2(a^2-1), y=(-5b+1)/2(a^2-1), z=(-1+ab-b)/2(a^2-1) ?

x + az = b -------- (1)
2x – y – z = 0 -------- (2)
– x + ay + 2z = 1 -------- (3)
(1) + (3)
a(y + z) + 2z = b + 1 -------- (4)

from (2) 2x = (y + z)
from (4)
2ax + 2z = b + 1 --------- × a
2a^2x + 2az = ab + a-------- (5)
x + az = b -------- (1) --------- × 2
2x + 2az = 2b -------- (6)
(5) – (6)
2x(a^2 – 1) = b(a – 2) + a
x = [b(a – 2) + a]2(a^2 – 1)
= [– 2b + a + ab]/2(a^2 – 1)

x + az = b -------- (1)
[– 2b + a + ab]/2(a^2 – 1) + az = b
az = b – [– 2b + a + ab]/2(a^2 – 1)
az = [b× 2(a^2 – 1) – [– 2b + a + ab]]/2(a^2 – 1)
az = [2a^2b – 2b + 2b – a – ab]]/2(a^2 – 1)
z = [2a^2b – a – ab]]/2(a^2 – 1)a
z = [2ab – 1 – b]]/2(a^2 – 1)
try to find 'y' yourself.
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