If $A = \frac{1}{2} \begin{bmatrix} -1 & -\sqrt{3} \\ \sqrt{3} & -1 \end{bmatrix}$, then $A^{-1} - A^2$ is not:

If $A=\left[\begin{array}{cc}2 & 3 \\ 1 & -4\end{array}\right]$ and $B=\left[\begin{array}{cc}1 & -2 \\ -1 & 3\end{array}\right],$ then verify that $(AB)^{-1}=B^{-1} A^{-1}$.

Let $A = \begin{bmatrix} 2 & 3 \\ -4 & -6 \end{bmatrix}$. Verify that $A(\text{adj } A) = (\text{adj } A) A = |A| I$.

Assertion $(A)$: If $B$ is a $3 \times 3$ matrix and $|B|=6$,then $|\operatorname{Adj}(B)|=36$.
Reason $(R)$: If $B$ is a square matrix of order $n$,then $|\operatorname{Adj}(B)|=|B|^{n}$.

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 ...

Matrices $A$ and $B$ will be inverse of each other only if