For $i=1, 2, 3$ and $j=1, 2, 3$. If $a_i^2+b_i^2+c_i^2=1$,$a_i a_j+b_i b_j+c_i c_j=0$,$\forall i \neq j$ and $A=\begin{bmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{bmatrix}$,then $\det(AA^T)=$

Let $A$ be a $2 \times 2$ matrix with real entries. Let $I$ be the $2 \times 2$ identity matrix. $\operatorname{Tr}(A)$ denotes the sum of diagonal entries of $A$. Assume that $A^2=I$.
Statement $I$: If $A \neq I$ and $A \neq -I$,then $\operatorname{det}(A) = -1$.
Statement $II$: If $A \neq I$ and $A \neq -I$,then $\operatorname{Tr}(A) \neq 0$.

Let $A$ be the set of all $3 \times 3$ determinants with entries $0$ or $1$ only and $B$ be the subset of $A$ consisting of all determinants with value $1$. If $C$ is the subset of $A$ consisting of all determinants with value $-1$,then:

Let $A = [a_{ij}]_{2 \times 2}$ where $a_{ij} \neq 0$ for all $i, j$ and $A^2 = I$. Let $a$ be the sum of all diagonal elements of $A$ and $b = |A|$. Then $3a^2 + 4b^2$ is equal to:

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).

The solutions of the equation $\left|\begin{array}{ccc}1+\sin ^{2} x & \sin ^{2} x & \sin ^{2} x \\ \cos ^{2} x & 1+\cos ^{2} x & \cos ^{2} x \\ 4 \sin 2 x & 4 \sin 2 x & 1+4 \sin 2 x\end{array}\right|=0$ for $(0 < x < \pi)$ are: