the determinant we want is:

for a 3x3 matrix, the easiest method to employ is the "rule of sarrus" (form 3 triples product by "wrap-around diagonals" from upper-left to lower right, and sum these, form three more triple products from "wrap around diagonals" from upper-right to lower-left, subtract these from the first sum).

in the matrix at hand that gives us:

if the determinant is non-zero, it means the matrix IS invertible, if we call our matrix A, then the unique solution is A^{-1}(4,1,12).

the determinant will be non-zero whenever:

that is, when

note that what happens when a = -1/2, is the first column is: twice the second column plus the third. this means that A is of rank 2 (since the 2nd and 3rd columns are clearly LI), and that the image (column space) of A is:

span({(1/2,2,-1),(2,-3,1)}). in this particular case, when a = -1/2, the system has NO solution, as (4,1,12) is not IN the column space.

since the problem does not ask you to specifically give a solution when a ≠ -1/2, this is all that needs be done.