If A is a square matrix of order 3, then |ope | vedclass

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

Question

(C) We know that for a square matrix $M$ of order $n$, $|\operatorname{adj} M| = |M|^{n-1}$.Given $A$ is a square matrix of order $n=3$.First, conside

Vedclass · Accepted Answer

$|A|^8$

Vedclass · Answer

(C) We know that for a square matrix $M$ of order $n$, $|\operatorname{adj} M| = |M|^{n-1}$.Given $A$ is a square matrix of order $n=3$.First, consider the matrix $M = A^2$. The order of $M$ is $3$.Then, $|\operatorname{adj} M| = |M|^{3-1} = |M|^2 = (|A|^2)^2 = |A|^4$.Now, we need to find $|\operatorname{adj}(\operatorname{adj} A^2)|$.Let $K = \operatorname{adj} A^2$. Then $|K| = |A|^4$.Using the property $|\operatorname{adj} K| = |K|^{n-1}$, where $n=3$:$|\operatorname{adj} K| = |K|^{3-1} = |K|^2$.Substituting $|K| = |A|^4$:$|\operatorname{adj}(\operatorname{adj} A^2)| = (|A|^4)^2 = |A|^8$.

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

Similar Questions

If $A = \begin{bmatrix} -2 & 2 \\ -3 & 2 \end{bmatrix}$ and $B = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$, then $(B^{-1} A^{-1})^{-1}$ is equal to

If $A$ is a square matrix such that $A(\operatorname{adj} A) = \begin{bmatrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{bmatrix}$, then $\operatorname{det}(\operatorname{adj} A)$ is equal to

If $\omega$ is a complex cube root of unity and $A=\begin{bmatrix} \omega & 0 \\ 0 & \omega \end{bmatrix}$,then $A^{-1}=$

If $A = \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 1 \\ 1 & 2 & 1 \end{bmatrix}$, then $|\operatorname{Adj}(A^2)| = $

If $A = \begin{bmatrix} \sec \theta & -\tan \theta \\ -\tan \theta & \sec \theta \end{bmatrix}$, $\theta \in (0, \frac{\pi}{2})$ such that $A + \text{adj } A = 4I$, then $\theta =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module