For each $n \in \mathbf{N}$, let $A_n=\{(n+1) k / k \in \mathbf{N}\}$ and $X=\bigcup_{n \in \mathbf{N}} A_n \cdot A$ mapping $f: X \rightarrow N$ defined by $f(x)=x$, $\forall x \in \mathbf{X}$, is

For $n>2$ and $n \in \mathbf{N}$, the product of the roots of $(x-n)\left(\left(x^2-2 n x\right)^2+\left(2 n^2-5\right)\left(x^2-2 n x\right)\right. \left.+\left(n^4-5 n^2+4\right)\right)=0$ is divisible by

Let $I$ be a unit matrix of order 6 . Let $A=\left(a_{i j}\right)$ be a square matrix of order 6 such that $a_{i j}=\left\{\begin{array}{l}1, \text { if } i+j=7 \\ 0, \text { if } i+j \neq 7\end{array}\right.$ then $\left(A(\operatorname{adj} A) A^{-1}\right) A^2=$

Let $a, b, c \notin\{0,1\}$. If the system of equations

$$ \begin{aligned} & \Pi_1 \equiv x+a y+a z=0 \\ & \Pi_2 \equiv b x+y+b z=0 \\ & \Pi_3 \equiv c x+c y+z=0 \end{aligned} $$

has a non-trivial solution, then the system of equations $\Pi_1=a, \Pi_2=b, \Pi_3=c$ has