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# Choose the correct answer. Q : 15 Let A be a square matrix of order 3 × 3, then |kA| is equal to (A) k|A| (B) k^2|A| (C) k^3|A| (D) 3k|A|

Q : 15        Let A be a square matrix of order $3\times 3$ , then $|kA|$ is equal to

(A) $k|A|$          (B) $k^2|A|$        (C) $k^3|A|$        (D)  $3k|A|$

Views

Assume a square matrix A of order of $3\times3$.

$A = \begin{bmatrix} a_1 & b_1&c_1 \\ a_2& b_2& c_2\\ a_3& b_3 & c_3 \end{bmatrix}$

Then we have;

$kA = \begin{bmatrix} ka_1 & kb_1&kc_1 \\ ka_2& kb_2& kc_2\\ ka_3& kb_3 & kc_3 \end{bmatrix}$

(Taking the common factors k from each row.)

$|kA| = \begin{vmatrix} ka_1 & kb_1&kc_1 \\ ka_2& kb_2& kc_2\\ ka_3& kb_3 & kc_3 \end{vmatrix} = k^3 \begin{vmatrix} a_1 & b_1&c_1 \\a_2& b_2& c_2\\ a_3& b_3 & c_3 \end{vmatrix}$

$= k^3 |A|$

Therefore correct option is (C).

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