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:

The least positive integer $n$ such that $\left(\begin{array}{cc}\cos \frac{\pi}{4} & \sin \frac{\pi}{4} \\ -\sin \frac{\pi}{4} & \cos \frac{\pi}{4}\end{array}\right)^{n}$ is an identity matrix of order $2$ is

Matrix $A$ is such that ${A^2} = 2A - I$,where $I$ is the identity matrix. Then for $n \ge 2$,${A^n} = $

If $K = \left|\begin{array}{ll}3 & 4 \\ 5 & 4\end{array}\right| + \left|\begin{array}{cc}1 & -1 \\ 5 & 4\end{array}\right| + \left|\begin{array}{cc}\frac{1}{3} & \frac{1}{4} \\ 5 & 4\end{array}\right| + \left|\begin{array}{cc}\frac{1}{9} & -\frac{1}{16} \\ 5 & 4\end{array}\right| + \ldots \text{ to } \infty$, then $K = $

If $f(\theta ) = \left| \begin{array}{ccc} 1 & \cos \theta & 1 \\ - \sin \theta & 1 & - \cos \theta \\ - 1 & \sin \theta & 1 \end{array} \right|$ and $A$ and $B$ are respectively the maximum and the minimum values of $f(\theta )$,then $(A, B)$ is equal to

If $a, b, c$ and $d$ are complex numbers,then the determinant $\Delta = \begin{vmatrix} 2 & a+b+c+d & ab+cd \\ a+b+c+d & 2(a+b)(c+d) & ab(c+d)+cd(a+b) \\ ab+cd & ab(c+d)+cd(a+b) & 2abcd \end{vmatrix}$ is