If $A = \begin{bmatrix} \alpha^2 & 5 \\ 5 & -\alpha \end{bmatrix}$ and $\det(A^{10}) = 1024$,then $\alpha = $

If $a, b, c$ are real numbers such that $a-b=1$ and $b-c=3$, then the number of matrices of the form $A=\begin{bmatrix} 1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2 \end{bmatrix}$ such that $|A|=-12$ is:

Find the values of $k$ if the area of the triangle is $4$ square units and the vertices are $(-2, 0), (0, 4), (0, k)$.

Evaluate the determinant: $\left|\begin{array}{cc}\sin \frac{11 \pi}{36} & \cos \frac{11 \pi}{36} \\\sin \frac{2 \pi}{9} & \cos \frac{2 \pi}{9}\end{array}\right|$.

If $\left[\begin{array}{ccc}1 & -1 & x \\ 1 & x & 1 \\ x & -1 & 1\end{array}\right]$ has no inverse,then the real value of $x$ is

Let $A = \begin{bmatrix} [x+1] & [x+2] & [x+3] \\ [x] & [x+3] & [x+3] \\ [x] & [x+2] & [x+4] \end{bmatrix}$,where $[t]$ denotes the greatest integer less than or equal to $t$. If $\operatorname{det}(A) = 192$,then the set of values of $x$ is the interval: