let the matrix be $\begin{bmatrix} a & b\\ c&d \end{bmatrix}$
determinant of matrix$=ad-bc$
since each of a,b,c,d can be 0,1.So determinant can be one of -1,0,1.
we want the determinant to be non zero.
so Required probability=probability that determinant is -1 $+$ probability that determinant is 1.
Also each element has 2 choices.
$\therefore$ There are 16 possible matrices.
probability that determinant is -1:
for the determinant to be -1,$ad=0,bc=1$.
for bc=1,b=1 and c=1 is the only possibility.
for ad=0,there are 3 possibilities:$a=0,d=1$ and $a=1,d=0$and $a=0,d=0$(at least one of a or d has to be 0).
Therefore, there are 3 matrices with determinant=-1.
Hence, probability that determinant$=-1$ is $3/16$
Similarly, there are 3 matrices with determinant=1.(here ad=1,bc=0)
Hence, probability that determinant$=1$ is $3/16$.
Total probability is $3/16+3/16=3/8$