If $A = \begin{bmatrix} -2 & 6 \\ -5 & 7 \end{bmatrix}$,then find $adj(A)$.

If ${I_3}$ is the identity matrix of order $3$,then ${I_3}^{-1}$ is

If $A=\begin{bmatrix} \frac{1}{\sqrt{5}} & \frac{2}{\sqrt{5}} \\ \frac{-2}{\sqrt{5}} & \frac{1}{\sqrt{5}} \end{bmatrix}$,$B=\begin{bmatrix} 1 & 0 \\ i & 1 \end{bmatrix}$,$i=\sqrt{-1}$,and $Q=A^{T}BA$,then the inverse of the matrix $AQ^{2021}A^{T}$ is equal to:

Let $A$ be a $3 \times 3$ matrix such that $|\operatorname{adj}(\operatorname{adj}(\operatorname{adj} A ))|=81$. If $S =\{ n \in \mathbb{Z} :(|\operatorname{adj}(\operatorname{adj} A)|)^{\frac{(n-1)^2}{2}}=|A|^{(3n^2-5n-4)}\}$,then $\sum_{n \in S}|A^{(n^2+n)}|$ is equal to

If $A = [a_{ij}]_{3 \times 3}$ is a matrix such that $a_{ij} = |2i - 5j|$, where $|.|$ denotes the modulus function, then the element in the $2^{nd}$ row and $3^{rd}$ column of $A^{-1}$ is ...

If matrix $A = \begin{bmatrix} 3 & 2 & 4 \\ 1 & 2 & -1 \\ 0 & 1 & 1 \end{bmatrix}$ and $A^{-1} = \frac{1}{K} \text{adj}(A)$,then $K$ is: