If $D_1$ and $D_2$ are two $3 \times 3$ diagonal matrices,then

If $A = \begin{bmatrix} 3 & -2 \\ 4 & 2 \end{bmatrix}$, then $A^2 - 5A + 14I = 0$. Which of the following is equivalent to $A^2$?

Let $ABC = I$. Then $tr(ABC + BCA + CAB)$ is (where the order of matrices $A, B, C$ is $3 \times 3$ and $tr(A)$ is the sum of the diagonal elements in $A$).

$A = \frac{1}{\pi} \begin{bmatrix} \sin^{-1}(\pi x) & \tan^{-1}(\frac{x}{\pi}) \\ \sin^{-1}(\frac{x}{\pi}) & \cot^{-1}(\pi x) \end{bmatrix}$ and $B = \frac{1}{\pi} \begin{bmatrix} -\cos^{-1}(\pi x) & \tan^{-1}(\frac{x}{\pi}) \\ \sin^{-1}(\frac{x}{\pi}) & -\tan^{-1}(\pi x) \end{bmatrix}$. Then,$A - B = $ . . . . . . .

In $\triangle ABC$,if $\left|\begin{array}{lll}a & b & c \\ b & c & a \\ c & a & b\end{array}\right|=0$,then $\cos A \cos B+\cos B \cos C+\cos C \cos A=$

If $p, q, r, s$ are in $A.P.$ and $f(x) = \left| \begin{array}{ccc} p + \sin x & q + \sin x & p - r + \sin x \\ q + \sin x & r + \sin x & -1 + \sin x \\ r + \sin x & s + \sin x & s - q + \sin x \end{array} \right|$ such that $\int_{0}^{\pi} f(x) dx = -4$,then the common difference of the $A.P.$ can be: