Question
Evaluate the determinant $\Delta=\left|\begin{array}{rrr} {1} & {2} & {4} \\ {-1} & {3} & {0} \\ {4} & {1} & {0} \end{array}\right|$

Answer

Note that in the third column, two entries are zero. So expanding along the third column (C3), we get
$\Delta=4\left|\begin{array}{cc} {-1} & {3} \\ {4} & {1} \end{array}\right|-0\left|\begin{array}{cc} {1} & {2} \\ {4} & {1} \end{array}\right|+0\left|\begin{array}{cc} {1} & {2} \\ {-1} & {3} \end{array}\right|$
     = 4 (–1 – 12) – 0 + 0 = – 52

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