Let the matrix $A = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{bmatrix}$ and the matrix $B_{0} = A^{49} + 2A^{98}$. If $B_{n} = \text{Adj}(B_{n-1})$ for all $n \geq 1$,then $\det(B_{4})$ is equal to:

Let $A$ be a non-singular matrix of order $n$ and $|A|=k$, then $(\operatorname{adj} A)^{-1}$ is

The set of all $2 \times 2$ matrices over the real numbers is not a group under matrix multiplication because

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 $\theta =$

If $A = \begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix}$,then $\text{adj } A$ is equal to

Let $a \in R$ and $A$ be a matrix of order $3 \times 3$ such that $\det(A)=-4$ and $A+I=\begin{bmatrix} 1 & a & 1 \\ 2 & 1 & 0 \\ a & 1 & 2 \end{bmatrix}$,where $I$ is the identity matrix of order $3 \times 3$. If $\det((a+1) \operatorname{adj}((a-1) A)) = 2^m 3^n$,where $m, n \in \{0, 1, 2, \ldots, 20\}$,then $m+n$ is equal to: