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3 and 4 .Determinants and Matrices
easy
If $\left| {\,\begin{array}{*{20}{c}}a&b&c\\m&n&p\\x&y&z\end{array}\,} \right| = k$, then $\left| {\,\begin{array}{*{20}{c}}{6a}&{2b}&{2c}\\{3m}&n&p\\{3x}&y&z\end{array}\,} \right| = $
A
$k/6$
B
$2k$
C
$3k$
D
$6k$
Solution
(d)$\left| {\,\begin{array}{*{20}{c}}{6a}&{2b}&{2c}\\{3m}&n&p\\{3x}&y&z\end{array}\,} \right|= 2 × 3$ $\left| {\,\begin{array}{*{20}{c}}a&b&c\\m&n&p\\x&y&z\end{array}\,} \right| = 6k$.
Standard 12
Mathematics