Evaluate $\begin{vmatrix}102 & 18 & 36\\1 & 3 & 4\\17&3&6\end{vmatrix}$.

0

The correct answer is Option (3) → 0 ##

$\begin{vmatrix}102 & 18 & 36\\1 & 3 & 4\\17&3&6\end{vmatrix}$

$=\begin{vmatrix}6(17) & 6(3) & 6(6)\\1 & 3 & 4\\17&3&6\end{vmatrix}$

Taking out the common factor $6$ from $R_1$:

$= 6 \begin{vmatrix} 17 & 3 & 6 \\ 1 & 3 & 4 \\ 17 & 3 & 6 \end{vmatrix}$

Since $R_1$ and $R_3$ are now identical:

$ = 6 \times 0= 0$