Solution Note that in the third column, two entries \((C_{3})\), are zero. So expanding along third column, we get \(\Delta=4|\begin{smallmatrix}-1&3\\ 4&1\end{smallmatrix}|-0|\begin{smallmatrix}1&2\\ 4&1\end{smallmatrix}|+0|\begin{smallmatrix}1&2\\ -1&3\end{smallmatrix}|\) \(=4(-1-12)-0+0=-52\)