If $A$ and $B$ are two square matrices of the same order and $(AB+BA)^{T}+(AB-BA)^{T}=2BA$, then:

Suppose the vectors $x_{1}, x_{2}$ and $x_{3}$ are the solutions of the system of linear equations $Ax = b$ when the vector $b$ on the right side is equal to $b_{1}, b_{2}$ and $b_{3}$ respectively. If $x_{1} = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}, x_{2} = \begin{bmatrix} 0 \\ 2 \\ 1 \end{bmatrix}, x_{3} = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}, b_{1} = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}, b_{2} = \begin{bmatrix} 0 \\ 2 \\ 0 \end{bmatrix}$ and $b_{3} = \begin{bmatrix} 0 \\ 0 \\ 2 \end{bmatrix}$,then the determinant of $A$ is equal to

Let $\alpha, \beta, \gamma, \delta$ be distinct imaginary roots of $z^5=1$. Find the value of the determinant: $\left| \begin{array}{ccc} e^{\alpha} & e^{2\alpha} & e^{3\alpha+1} \\ e^{\beta} & e^{2\beta} & e^{3\beta+1} \\ e^{\gamma} & e^{2\gamma} & e^{3\gamma+1} \end{array} \right|$.

Let $A$ and $B$ be real matrices of the form $\begin{bmatrix} \alpha & 0 \\ 0 & \beta \end{bmatrix}$ and $\begin{bmatrix} 0 & \gamma \\ \delta & 0 \end{bmatrix}$,respectively.
Statement $1$: $AB - BA$ is always an invertible matrix.
Statement $2$: $AB - BA$ is never an identity matrix.

If $A = \begin{bmatrix} 1 & 2 & 1 \\ 0 & 1 & -1 \\ 3 & -1 & 1 \end{bmatrix}$,then:

Let $A, B, C, D$ be square real matrices such that $C^T = DAB$,$D^T = ABC$,and $S = ABCD$. Then $S^2$ is equal to: