Let $A = \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix}$,$\alpha \in R$ such that $A^{32} = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$. Then a value of $\alpha$ is

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 $A = \begin{bmatrix} 1 & 2 \\ 1 & \alpha \end{bmatrix}$ and $B = \begin{bmatrix} 3 & 3 \\ \beta & 2 \end{bmatrix}$. If $A^2 - 4A + I = O$ and $B^2 - 5B - 6I = O$,then among the two statements:
(S1): $[(B - A)(B + A)]^T = \begin{bmatrix} 13 & 15 \\ 7 & 10 \end{bmatrix}$
and
(S2): $\det(\text{adj}(A + B)) = -5$.

If $A$ and $B$ are both $3 \times 3$ matrices,then which of the following statements are true?
$(i)$ $AB=0 \Rightarrow A=0$ or $B=0$
(ii) $AB=I_3 \Rightarrow A^{-1}=B$
(iii) $(A-B)^2=A^2-2AB+B^2$

If the polynomial $f(x) = \left|\begin{array}{ccc} (1+x)^{a} & (2+x)^{b} & 1 \\ 1 & (1+x)^{a} & (2+x)^{b} \\ (2+x)^{b} & 1 & (1+x)^{a} \end{array}\right|$,then the constant term of $f(x)$ is ($a$ and $b$ are positive integers).

If $A = \int_{1}^{\sin \theta} \frac{t}{1+t^2} dt$ and $B = \int_{1}^{\operatorname{cosec} \theta} \frac{1}{t(1+t^2)} dt$,then the value of $\left| \begin{array}{ccc} A & A^2 & B \\ e^{A+B} & B^2 & -1 \\ 1 & A^2+B^2 & -1 \end{array} \right| = $