Let A be a square matrix of order 3 times 3,t | vedclass

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(C) Given that $A$ is a square matrix of order $n 	imes n$,the property of the determinant states that $|kA| = k^n |A|$.Here,the order of the matrix

Vedclass · Accepted Answer

$k^3|A|$

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(C) Given that $A$ is a square matrix of order $n 	imes n$,the property of the determinant states that $|kA| = k^n |A|$.Here,the order of the matrix $A$ is $3 	imes 3$,so $n = 3$.Therefore,$|kA| = k^3 |A|$.Alternatively,if $A = \begin{bmatrix} a_1 & b_1 & c_1 \ a_2 & b_2 & c_2 \ a_3 & b_3 & c_3 \end{bmatrix}$,then $kA = \begin{bmatrix} ka_1 & kb_1 & kc_1 \ ka_2 & kb_2 & kc_2 \ ka_3 & kb_3 & kc_3 \end{bmatrix}$.Taking the determinant,we can factor out $k$ from each of the $3$ rows:$|kA| = k \cdot k \cdot k \cdot \begin{vmatrix} a_1 & b_1 & c_1 \ a_2 & b_2 & c_2 \ a_3 & b_3 & c_3 \end{vmatrix} = k^3 |A|$.Hence,the correct answer is $C$.

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

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