If A is a square matrix of order 3,then |oper | 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 $n = 3$. We know the property of the adjoint of a matrix: $|\operatorname{adj}(\operatorname{adj} M)| =

Vedclass · Accepted Answer

$|A|^8$

Vedclass · Answer

(C) Given that $A$ is a square matrix of order $n = 3$. We know the property of the adjoint of a matrix: $|\operatorname{adj}(\operatorname{adj} M)| = |M|^{(n-1)^2}$. Here,we need to find $|\operatorname{adj}(\operatorname{adj} A^2)|$. Let $M = A^2$. Then,$|\operatorname{adj}(\operatorname{adj} A^2)| = |A^2|^{(3-1)^2}$. Since $(3-1)^2 = 2^2 = 4$,we have $|\operatorname{adj}(\operatorname{adj} A^2)| = |A^2|^4$. Using the property $|A^k| = |A|^k$,we get $|A^2|^4 = (|A|^2)^4 = |A|^8$. Therefore,the correct option is $|A|^8$.

If $A$ is a square matrix of order $3$,then $|\operatorname{adj}(\operatorname{adj} A^2)|$ is equal to

Similar Questions

Which hormone is responsible for apical dominance?

Plaster of Paris,when mixed with the correct amount of water,sets into a solid mass due to the formation of:

The structure of $PCC$ is

In an $LCR$ circuit,the capacitance is changed from $C$ to $2C$. For the resonant frequency to remain unchanged,the inductance should be changed from $L$ to:

If $\frac{1}{2} \leq x \leq 1$,then $\cos ^{-1} x+\cos ^{-1}\left(\frac{x}{2}+\frac{1}{2} \sqrt{3-3 x^2}\right)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module