Using traditional method:

(1): 2x+3y+z=0

(2): x-y+2z=0

(3): ax+y-z=0

(4): (2)+(3)=(a+1)x+z=0

(5): (1)+3(2)=5x+7z=0

(6): 7(4)-(5)=(7a+7-5)x=0=(7a+2)x=0. If we choose x=0 we arrive at the trivial solution x=y=z=0. So to avoid this, 7a=-2 and a=-2/7. We cannot calculate x under these conditions.

We can rewrite (3): -2x+7y-7z=0, so we have:

(1): 2x+3y+z=0

(2): x-y+2z=0

(3): -2x+7y-7z=0

(7): (1)+(3)=10y-6z=0, 5y-3z=0

(8): (3)+2(2)=5y-3z=0

(7) and (8) are the same equation, so there are many solutions given by y=3z/5 and x=-7z/5 where z is a parameter for x and y.

Going back to the question and looking at the two given equations, the Gauss Jordan method would produce a matrix with a row of zeroes (for 0x+0y+0z=0), indicating that there is no unique solution (because there are only two equations for three unknowns). The traditional method shows that although there appeared to be three equations and three unknowns, there was in fact a fourth unknown, namely a. To avoid the trivial solution a must be -2/7.

One of the many solutions is when z=5, x=-7 and y=3 (y=-3x/7).