Let $A$ be a $3 \times 3$ invertible matrix. If $|\operatorname{adj}(24A)| = |\operatorname{adj}(3 \operatorname{adj}(2A))|$,then $|A^2|$ is equal to

  • A
    $6^6$
  • B
    $2^{12}$
  • C
    $2^6$
  • D
    $1$

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Similar Questions

Let $A = \begin{bmatrix} x & y & z \\ y & z & x \\ z & x & y \end{bmatrix}$,where $x, y$ and $z$ are real numbers such that $x + y + z > 0$ and $xyz = 2$. If $A^2 = I_3$,then the value of $x^3 + y^3 + z^3$ is ............

Let $\alpha \in(0, \infty)$ and $A=\begin{bmatrix} 1 & 2 & \alpha \\ 1 & 0 & 1 \\ 0 & 1 & 2 \end{bmatrix}$. If $\operatorname{det}(\operatorname{adj}(2A-A^{T}) \cdot \operatorname{adj}(A-2A^{T}))=2^8$,then $(\operatorname{det}(A))^2$ is equal to:

Let $A = \begin{bmatrix} 1 & 2 \\ 1 & \alpha \end{bmatrix}$ and $B = \begin{bmatrix} 3 & 3 \\ \beta & 2 \end{bmatrix}$. If $A^2 - 4A + I = O$ and $B^2 - 5B - 6I = O$,then among the two statements:
(S1): $[(B - A)(B + A)]^T = \begin{bmatrix} 13 & 15 \\ 7 & 10 \end{bmatrix}$
and
(S2): $\det(\text{adj}(A + B)) = -5$.

Let $A=\begin{bmatrix} \frac{1}{\sqrt{2}} & -2 \\ 0 & 1 \end{bmatrix}$ and $P=\begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}, \theta > 0$. If $B=P A P^T$,$C=P^T B^{10} P$ and the sum of the diagonal elements of $C$ is $\frac{m}{n}$,where $\operatorname{gcd}(m, n)=1$,then $m+n$ is:

Let $P$ be an $m \times m$ matrix such that $P^2=P$. Then,$(I+P)^n$ equals

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