Let $A$ and $B$ be two $3 \times 3$ real matrices such that $(A^{2}-B^{2})$ is an invertible matrix. If $A^{5}=B^{5}$ and $A^{3} B^{2}=A^{2} B^{3}$,then the value of the determinant of the matrix $A^{3}+B^{3}$ is equal to:

If the minimum and the maximum values of the function $f : \left[\frac{\pi}{4}, \frac{\pi}{2}\right] \rightarrow \mathbb{R}$,defined by $f(\theta) = \left|\begin{array}{ccc} -\sin^2 \theta & -1-\sin^2 \theta & 1 \\ -\cos^2 \theta & -1-\cos^2 \theta & 1 \\ 12 & 10 & -2 \end{array}\right|$ are $m$ and $M$ respectively,then the ordered pair $(m, M)$ is equal to:

If $P$ is a square matrix with $P^2=P$ and if $I$ is the unit matrix of the same order as of $P$,then $(P+I)^4=$

If $A$ is a square matrix such that $A^2=A$,then $(I-A)^3$ is

Let $A = \begin{bmatrix} 5 & \sin^2 \theta & \cos^2 \theta \\ -\sin^2 \theta & -5 & 1 \\ \cos^2 \theta & 1 & 5 \end{bmatrix}$. Then the maximum value of $\det(A)$ is

Let $M = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$ and $N = \begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix}$. Then $N M^{10} N^{-1} =$