Evaluate the determinant:-

    \[\left|\begin{array}{ccc} 3 & -1 & -2 \\ 0 & 0 & -1 \\ 3 & -5 & 0 \end{array}\right|\]

Evaluate the determinant:-

    \[\left|\begin{array}{ccc} 3 & -1 & -2 \\ 0 & 0 & -1 \\ 3 & -5 & 0 \end{array}\right|\]

Solution:-

    \[\left|\begin{array}{ccc} 3 & -1 & -2 \\ 0 & 0 & -1 \\ 3 & -5 & 0 \end{array}\right|\]

Expanding the above determinant along \mathrm{C}_{1}, we have

    \[\begin{array}{l} =3(0-5)-0+3(1-0) \\ =-15+3 \\ =-12 \end{array}\]

Hence \left|\begin{array}{ccc}3 & -1 & -2 \\ 0 & 0 & -1 \\ 3 & -5 & 0\end{array}\right|=-12.