If $\Delta = \begin{vmatrix} a & b & c \\ x & y & z \\ p & q & r \end{vmatrix}$,then $\begin{vmatrix} ka & kb & kc \\ kx & ky & kz \\ kp & kq & kr \end{vmatrix}$ =

If $x, y, z$ are all positive and are the $p$-th, $q$-th, and $r$-th terms of a geometric progression respectively, then the value of the determinant $\left|\begin{array}{lll} \log x & p & 1 \\ \log y & q & 1 \\ \log z & r & 1 \end{array}\right|$ equals:

Statement-$1$: The determinant of a skew-symmetric matrix of order $3$ is zero.
Statement-$2$: For any square matrix $A$ of order $n$,$\det(A^T) = \det(A)$ and $\det(-A) = (-1)^n \det(A)$.

If $\left|\begin{array}{ccc}1+\sin ^{2} \theta & \cos ^{2} \theta & 4 \sin 2 \theta \\ \sin ^{2} \theta & 1+\cos ^{2} \theta & 4 \sin 2 \theta \\ \sin ^{2} \theta & \cos ^{2} \theta & 4 \sin 2 \theta-1\end{array}\right|=0$ and $0 < \theta < \frac{\pi}{2}$,then $\cos 4 \theta$ is equal to

If $A$ and $B$ are two square matrices with $\det(A) = 5$ and $\det(B^T \cdot A^T) = -15$,then $\det(B)$ is equal to

If $A, B$ and $C$ are $n \times n$ matrices and $\det(A) = 2$,$\det(B) = 3$ and $\det(C) = 5$,then the value of $\det(A^2BC^{-1})$ is equal to