Choose the correct answer.

Answer: C

A is a square matrix of order 3 × 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.