The product of the cofactors of 3 and -2 in the matrix $$\begin{bmatrix}1 & 0 & -2 \\3 & -1 & 2 \\4 & 5 & 6\end{bmatrix}$$​ is:

Correct option is C

​$$\textbf{Cofactor of 3 (element at (2,1))} \\\text{The cofactor of element } a_{21} = 3 \text{ is:} \\C_{21} = (-1)^{2+1} \cdot \det \left( \begin{vmatrix} 0 & -2 \\ 5 & 6 \end{vmatrix} \right) \\\det \left( \begin{vmatrix} 0 & -2 \\ 5 & 6 \end{vmatrix} \right) = (0)(6) - (-2)(5) = 10 \\\text{Thus,} \\C_{21} = (-1)^3 \cdot 10 = -10 \\\textbf{Cofactor of -2 (element at (1,3))} \\\text{The cofactor of element } a_{13} = -2 \text{ is:} \\C_{13} = (-1)^{1+3} \cdot \det \left( \begin{vmatrix} 3 & -1 \\ 4 & 5 \end{vmatrix} \right) \\\det \left( \begin{vmatrix} 3 & -1 \\ 4 & 5 \end{vmatrix} \right) = (3)(5) - (-1)(4) = 15 + 4 = 19 \\\text{Thus,} \\C_{13} = (-1)^4 \cdot 19 = 19 \\\textbf{Product of the cofactors} \\\text{The product of the cofactors is:} \\C_{21} \cdot C_{13} = (-10) \cdot 19 = -190$$​​