Let $A = \begin{bmatrix} 1 & a & a \\ 0 & 1 & b \\ 0 & 0 & 1 \end{bmatrix}$,where $a, b \in \mathbb{R}$. If for some $n \in \mathbb{N}$,$A^n = \begin{bmatrix} 1 & 48 & 2160 \\ 0 & 1 & 96 \\ 0 & 0 & 1 \end{bmatrix}$,then $n + a + b$ is equal to:

The determinant $\left| \begin{array}{ccc} a^2 & a^2 - (b - c)^2 & bc \\ b^2 & b^2 - (c - a)^2 & ca \\ c^2 & c^2 - (a - b)^2 & ab \end{array} \right|$ is divisible by :

Let $A$ be a $3 \times 3$ matrix of non-negative real elements such that $A\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} = 3\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}$. Then the maximum value of $\operatorname{det}(A)$ is:

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:

If $A=\left[\begin{array}{ccc}1 & 1 & 3 \\ 5 & 2 & 6 \\ -2 & -1 & -3\end{array}\right]$,then $A+A^3+A^4+A^5+3 I=$

If $A$ and $B$ are square matrices of order $3$ such that $(A + B)(A - B) = A^2 - B^2$,then $(ABA^{-1})^2$ is equal to