If $A$ and $B$ are two matrices such that $A+B$ and $AB$ are both defined, then

If $A = \frac{1}{\pi} \begin{bmatrix} \sin^{-1}(x\pi) & \tan^{-1}(\frac{x}{\pi}) \\ \sin^{-1}(\frac{x}{\pi}) & \cot^{-1}(\pi x) \end{bmatrix}$ and $B = \begin{bmatrix} -\frac{1}{\pi} \cos^{-1}(x\pi) & \frac{1}{\pi} \tan^{-1}(\frac{x}{\pi}) \\ \frac{1}{\pi} \sin^{-1}(\frac{x}{\pi}) & -\frac{1}{\pi} \tan^{-1}(\pi x) \end{bmatrix}$,then $A-B$ is equal to:

If $A = \begin{bmatrix} 1/3 & 2 \\ 0 & 2x - 3 \end{bmatrix}$,$B = \begin{bmatrix} 3 & 6 \\ 0 & -1 \end{bmatrix}$ and $AB = I$,then $x =$

Let $A$ and $B$ be two symmetric matrices of the same order. Then, the matrix $AB - BA$ is

If $A_1, A_3, \dots, A_{2n-1}$ are $n$ skew-symmetric matrices of the same order,then $B = \sum_{r=1}^n (2r-1)(A_{2r-1})^{2r-1}$ will be:

The value of $\cos \alpha$ is . . . . . . if $A+A^{\prime}=I$,where $A = \begin{bmatrix} \sin \alpha & -\cos \alpha \\ \cos \alpha & \sin \alpha \end{bmatrix}$.