Cramer's Method uses a determinant of coefficients which must not be zero if there's to be a solution. So let's check:
⎡ 1 a-1 0 1 ⎤
⎢ a-2 -1 0 -1⎥
⎢ 1 a-1 a 1⎥
⎣ a a-1 a+4 1⎦
To evaluate this we need to evaluate 3 3×3 determinants:
⎡ -1 0 -1⎥
⎢a-1 a 1⎥= -1(-4)-1[(a-1)(a+4)-a2+a]=4-(4a-4)=8-4a, let A=8-4a,
⎣ a-1 a+4 1⎦
⎡ a-2 0 -1⎤
⎢ 1 a 1⎥=(a-2)(-4)-(-a2+a+4)=-4a+8+a2-a-4=a2-5a+4, let B=a2-5a+4,
⎣ a a+4 1⎦
⎡ a-2 -1 0 ⎤
⎢ 1 a-1 a ⎥=(a-2)(4a-4)+a+4-a2=3a2-11a+12, let C=3a2-11a+12.
⎣ a a-1 a+4⎦
Now we can evaluate Cramer's determinant: A-(a-1)B-C=8-4a-(a-1)(a2-5a+4)-3a2+11a-12=
8-4a-a3+5a2-4a+a2-5a+4-3a2+11a-12=-a3+3a2-2a=-a(a2-3a+2)=-a(a-2)(a-1).
The determinant is zero when a=0, 1 or 2, and under these conditions Cramer's Method would find inconsistency in the system of equations (or many solutions).
It follows that if a is none of these values, Cramer's Method would find a unique solution.