Question

Solve $\left|\begin{array}{ccc}-3& 6& 2\\ -2& 4& 2\\ 5& -10& 2\end{array}\right|$

Answer 1

$\left|\begin{array}{ccc}-3& 6& 2\\ -2& 4& 2\\ 5& -10& 2\end{array}\right|$

$C_3\to \frac{C_3}{2}$

$=2\left|\begin{array}{ccc}-3& 6& 1\\ -2& 4& 1\\ 5& -10& 1\end{array}\right|$

$C_2\to \frac{C_2}{2}$

$=4\left|\begin{array}{ccc}-3& 3& 1\\ -2& 2& 1\\ 5& -5& 1\end{array}\right|$

$C_1\to C_1+C_2$

$=4\left|\begin{array}{ccc}-3+3& 3& 1\\ -2+2& 2& 1\\ 5-5& -5& 1\end{array}\right|$

$=4\left|\begin{array}{ccc}0& 3& 1\\ 0& 2& 1\\ 0& -5& 1\end{array}\right|$

Since $C_1=0$ the determinant$=0$