Let $A$ and $B$ be two square matrices of order $3$ such that $|A|=3$ and $|B|=2$. Then $\left|A^{T} A(\operatorname{adj}(2A))^{-1}(\operatorname{adj}(4B))(\operatorname{adj}(AB))^{-1} AA^{T}\right|$ is equal to:

Let $A$ be a $3 \times 3$ real matrix such that $A^2(A-2I) - 4(A-I) = O$,where $I$ and $O$ are the identity and null matrices,respectively. If $A^5 = \alpha A^2 + \beta A + \gamma I$,where $\alpha, \beta$ and $\gamma$ are real constants,then $\alpha + \beta + \gamma$ is equal to:

$A=\begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 1 & 0 \end{bmatrix} \Rightarrow A^2-2A=$

If $A = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$,then which one of the following statements is not correct?

Match the statements given in Column $I$ with the intervals/union of intervals given in Column $II$.

Column $I$	Column $II$
$(A)$ The set $\{\operatorname{Re}(\frac{2 i z}{1-z^2}): |z|=1, z \neq \pm 1\}$ is	$(p)$ $(-\infty,-1) \cup(1, \infty)$
$(B)$ The domain of $f(x)=\sin ^{-1}(\frac{8(3)^{x-2}}{1-3^{2(x-1)}})$ is	$(q)$ $(-\infty, 0) \cup(0, \infty)$
$(C)$ If $f(\theta)=\left|\begin{array}{ccc}1 & \tan \theta & 1 \\ -\tan \theta & 1 & \tan \theta \\ -1 & -\tan \theta & 1\end{array}\right|$,then the set $\{f(\theta): 0 \leq \theta < \frac{\pi}{2}\}$ is	$(r)$ $[2, \infty)$
$(D)$ If $f(x)=x^{3 / 2}(3 x-10), x \geq 0$,then $f(x)$ is increasing in	$(s)$ $(-\infty,-1] \cup[1, \infty)$
	$(t)$ $(-\infty, 0] \cup[2, \infty)$

Let $A$ be a symmetric matrix and $B$ be a skew-symmetric matrix,such that $A - B = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$. Then $|A|$ is: