If $\omega (\neq 1)$ is a cube root of unity,then the value of the determinant $\left| \begin{array}{ccc} 1 & 1 + i + \omega^2 & \omega^2 \\ 1 - i & -1 & \omega^2 - 1 \\ -i & -i + \omega - 1 & -1 \end{array} \right|$ is equal to

Let $\omega$ be a complex cube root of unity with $\omega \neq 1$ and $P = [p_{ij}]$ be an $n \times n$ matrix with $p_{ij} = \omega^{i+j}$. Then $P^2 \neq 0$ when $n =$

Let $A$ and $B$ be two square matrices of order $3$ such that $|A|=3$ and $|B|=2$. Then $\left|A^{T} A(\operatorname{adj}(2A))^{-1}(\operatorname{adj}(4B))(\operatorname{adj}(AB))^{-1} AA^{T}\right|$ is equal to:

The number of solutions of the matrix equation $X^2 = \begin{bmatrix} 1 & 1 \\ 2 & 3 \end{bmatrix}$ is

If the inverse of $\begin{bmatrix} -x & 14x & 7x \\ 0 & 1 & 0 \\ x & -4x & -2x \end{bmatrix}$ is $\begin{bmatrix} 2 & 0 & 7 \\ 0 & 1 & 0 \\ 1 & -2 & 1 \end{bmatrix}$,then $\left|\begin{array}{ccc} x & x+1 & x+2 \\ x+1 & x+2 & x+3 \\ x+2 & x+3 & x+4 \end{array}\right| = $