Question

finding the value of a in the given equations through gaussian elimination
x+y+z=2
x+3y+2z=5
x+y+(a^2-1)z=a

Answer 1

(a^2)z + 2z - a - 2 = 0.

Answer 2

You can't find a specific numerical value for a,
since there will be infinitely many solutions, a different solution parametrized by a different value of a.

Let’s write this system in matrix form:
1 1 1 2
1 3 2 5
1 1 (a^2 – 1) a.

Then reduce.
1 1 1 2
0 2 1 3
0 0 (a^2 – 2) (a – 2).

So we get:
1 1 1 2
0 2 1 3
0 0 1 (a – 2)/(a^2 – 2).

We can let “a” be anything such that the denominator isn’t equal to 0.
For example, let a = 2.
Then that forces z to be 0.

Then back substitute into 2y + z = 3 to see that y = 3/2.
Then back substitute into x + y + z = 2 to get x = 1/2.
We can check that (1/2 , 3/2, 0) solves our system.

But we can let a = 1 as well.
If we do, we get z = 1. Back substituting into 2y + z = 3 gives y = 1.
Back subbing into x + y + z = 3 gives x = 0.
So (0, 1, 1) is another solution corresponding to a = 1.