If $A = \begin{bmatrix} 1 & -1 \\ 2 & -1 \end{bmatrix}$,$B = \begin{bmatrix} a & 1 \\ b & -1 \end{bmatrix}$ and $(A + B)^2 = A^2 + B^2$,then the values of $a$ and $b$ are:

If ${\Delta _1} = \left| {\begin{array}{*{20}{c}} x & {\sin \theta } & {\cos \theta } \\ {\sin \theta } & { - x} & 1 \\ {\cos \theta } & 1 & x \end{array}} \right|$ and ${\Delta _2} = \left| {\begin{array}{*{20}{c}} x & {\sin 2\theta } & {\cos 2\theta } \\ {\sin 2\theta } & { - x} & 1 \\ {\cos 2\theta } & 1 & x \end{array}} \right|$,$x \ne 0$; then for all $\theta \in \left( {0, \frac{\pi }{2}} \right)$:

If $A$ and $B$ are two non-zero $n \times n$ matrices such that $A^2 + B = A^2 B$,then:

Let $A = \begin{bmatrix} 1 & a & a \\ 0 & 1 & b \\ 0 & 0 & 1 \end{bmatrix}$,where $a, b \in \mathbb{R}$. If for some $n \in \mathbb{N}$,$A^n = \begin{bmatrix} 1 & 48 & 2160 \\ 0 & 1 & 96 \\ 0 & 0 & 1 \end{bmatrix}$,then $n + a + b$ is equal to:

If $A = \begin{vmatrix} -1 & 2 & 4 \\ 3 & 1 & 0 \\ -2 & 4 & 2 \end{vmatrix}$ and $B = \begin{vmatrix} -2 & 4 & 2 \\ 6 & 2 & 0 \\ -2 & 4 & 8 \end{vmatrix}$,then $B$ is given by

If $a_{r} = \cos \frac{2 r \pi}{9} + i \sin \frac{2 r \pi}{9}$,$r = 1, 2, 3, \ldots$,$i = \sqrt{-1}$,then the determinant $\left|\begin{array}{lll}a_{1} & a_{2} & a_{3} \\ a_{4} & a_{5} & a_{6} \\ a_{7} & a_{8} & a_{9}\end{array}\right|$ is equal to: