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Let $\mathrm{A}$ be a square matrix of order $3 \times 3$ , then $|\mathrm{k A}|$ is equal to
$k|A|$
$k^2|A|$
$k^3|A|$
$3k|A|$
Solution
$A$ is a square matrix of order $3 \times 3$
Let $A=\left[\begin{array}{lll}a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3}\end{array}\right]$
Then, $k A=\left[\begin{array}{lll}k a_{1} & k b_{1} & k c_{1} \\ k a_{2} & k b_{2} & k c_{2} \\ k a_{3} & k b_{3} & k c_{3}\end{array}\right]$
$\therefore|k A|=\left|\begin{array}{lll}k a_{1} & k b_{1} & k c_{1} \\ k a_{2} & k b_{2} & k c_{2} \\ k a_{3} & k b_{3} & k c_{3}\end{array}\right|$
$=k^{3}\left|\begin{array}{lll}a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3}\end{array}\right|$ (Taking out common factors $k$ from each row)
$=k^{3}|A|$
$\therefore|k A|=k^{3}|A|$
Hence, the correct answer is $C$.
