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
$A$ is a singular matrix of order five. $B$ is another matrix having the rank $\rho(B)$ equal to the $\operatorname{rank} \rho(A)$ and $B$ has a non-zero minor of order 3. Then which one of the following is true?
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