Answer
$2$, invertible
Work Step by Step
We know that for a matrix
$
\left[\begin{array}{rrr}
a & b & c \\
d &e & f \\
g &h & i \\
\end{array} \right]
$
the determinant is given as
$D=a(ei-fh)-b(di-fg)+c(dh-eg).$
Thus, we have:
$D=0(6\cdot3-4\cdot0)-(-1)(2\cdot3-4\cdot1)+0(2\cdot3-4\cdot1)=0(18)+1(2)+0(2)=2.$
The determinant is not $0$, so the matrix is invertible.