Let $A = \begin{bmatrix} 1 & 2 \\ -2 & -5 \end{bmatrix}$. Let $\alpha, \beta \in \mathbb{R}$ be such that $\alpha A^{2} + \beta A = 2I$. Then $\alpha + \beta$ is equal to -

If $z_1 = 2 + 3 \ i$ and $z_2 = 3 + 2 \ i$,where $i = \sqrt{-1}$,then $\begin{bmatrix} z_1 & z_2 \\ -\bar{z}_2 & \bar{z}_1 \end{bmatrix} \begin{bmatrix} \bar{z}_1 & -z_2 \\ \bar{z}_2 & z_1 \end{bmatrix} =$

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 $A$ be a symmetric matrix such that $|A|=2$ and $\begin{bmatrix} 2 & 1 \\ 3 & \frac{3}{2} \end{bmatrix} A = \begin{bmatrix} 1 & 2 \\ \alpha & \beta \end{bmatrix}$. If the sum of the diagonal elements of $A$ is $s$,then $\frac{\beta s}{\alpha^2}$ is equal to $..........$.

If $a, b, c$ are non-zero complex numbers satisfying $a^2 + b^2 + c^2 = 0$ and $\left| \begin{array}{ccc} b^2 + c^2 & ab & ac \\ ab & c^2 + a^2 & bc \\ ac & bc & a^2 + b^2 \end{array} \right| = k a^2 b^2 c^2$,then $k$ 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