Question
Without expanding, show that the values of the following determinant are zero:
$\begin{vmatrix}49&1&6\\39&7&4\\26&2&3 \end{vmatrix}$

Answer

$\begin{vmatrix}49&1&6\\39&7&4\\26&2&3 \end{vmatrix}$
Applying: C1 → C1 + (-8)C3
$=\begin{vmatrix}1&1&6\\7&7&4\\2&2&3 \end{vmatrix}=0$
$\because\text{C}_1=\text{C}_2$

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