If $\frac{x^2+7}{(x^2+1)(x-2)}=\frac{A}{x-2}+\frac{Bx+C}{x^2+1}$, then the determinant of the matrix $\begin{bmatrix} A & B \\ C & \frac{2}{5} \end{bmatrix}$ is

If $\begin{bmatrix} x & 4 & -1 \end{bmatrix} \begin{bmatrix} 2 & 1 & 0 \\ 1 & 0 & 2 \\ 0 & 2 & 4 \end{bmatrix} \begin{bmatrix} x \\ 4 \\ -1 \end{bmatrix} = 0$,then $x=$

If $A=\left[\begin{array}{ccc}1 & 1 & 3 \\ 5 & 2 & 6 \\ -2 & -1 & -3\end{array}\right]$,then $A+A^3+A^4+A^5+3 I=$

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

Let $A = \begin{bmatrix} 1 & -1 \\ 2 & \alpha \end{bmatrix}$ and $B = \begin{bmatrix} \beta & 1 \\ 1 & 0 \end{bmatrix}$,where $\alpha, \beta \in \mathbb{R}$. Let $\alpha_{1}$ be the value of $\alpha$ which satisfies $(A + B)^{2} = A^{2} + \begin{bmatrix} 2 & 2 \\ 2 & 2 \end{bmatrix}$ and $\alpha_{2}$ be the value of $\alpha$ which satisfies $(A + B)^{2} = B^{2}$. Then $|\alpha_{1} - \alpha_{2}|$ is equal to:

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 $\theta \in \left( {0, \frac{\pi }{2}} \right)$: