Choose the correct answer.
Let $A$ be a square matrix of order $3 \times 3$, then $|k A|$ is equal to
A. $k|A|$
B. $k^{2}|A|$
C. $k^{3}|A|$
D. $3 k|A|$
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.