Answer
Singular.
Work Step by Step
We know that for a matrix
\[
\left[\begin{array}{rr}
a & b \\
c &d \\
\end{array} \right]
\]
the determinant $D$ is given by the formula $D=ad-bc.$
We also know that if $D=0$, then the matrix is singular.
The given matrix has
$D=(4\cdot2)-[(-8)\cdot(-1)]\\
D=8-8\\
D=0$
The square matrix has a determinant of zero therefore it is singular.
