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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

$k^3|A|$

Vedclass · Answer

(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

Similar Questions

By using properties of determinants,show that:
$\left|\begin{array}{ccc} 1 & 1 & 1 \\ a & b & c \\ a^{3} & b^{3} & c^{3} \end{array}\right|=(a-b)(b-c)(c-a)(a+b+c)$

If $A$ and $B$ are invertible square matrices of order $n$, then which of the following is not correct?

If $\Delta = \begin{vmatrix} a & b & c \\ x & y & z \\ p & q & r \end{vmatrix}$,then $\begin{vmatrix} ka & kb & kc \\ kx & ky & kz \\ kp & kq & kr \end{vmatrix}$ =

Let $a-2b+c=1$. If $f(x) = \begin{vmatrix} x+a & x+2 & x+1 \\ x+b & x+3 & x+2 \\ x+c & x+4 & x+3 \end{vmatrix}$,then:

If $2^{a_1}, 2^{a_2}, 2^{a_3}, \dots, 2^{a_n}$ are in $G.P.$,then the value of the determinant $\left| \begin{array}{ccc} a_1 & a_2 & a_3 \\ a_{n+1} & a_{n+2} & a_{n+3} \\ a_{2n+1} & a_{2n+2} & a_{2n+3} \end{array} \right|$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module