Answer
-4, Yes.
Work Step by Step
Evaluate the determinant with row-3 cofactors: (it is easier to use row3 instead of row1 as it contains two zeros, you can also use row2.)
$\begin{array}( \\|A|= \\ \\ \end{array}
\begin{vmatrix} 1 & 3 &3&0\\0 &2 &0&1\\-1 &0 &0&2\\1 &6 &4&1 \end{vmatrix} \begin{array}( \\=(-1) \\ \\\end{array}\begin{vmatrix} 3 &3&0\\2 &0&1 \\6 &4&1\end{vmatrix} \begin{array}( \\-2 \\ \\\end{array}\begin{vmatrix} 1 &3&3\\0 &2&2 \\1 &6&4\end{vmatrix} $
$|A|=(-1)[3(-4)-3(2-6)]-2[(8-12)-3(-2)+3(-2)]=0-4=-4$
As $|A|\ne0$, the matrix should have an inverse.
