The roots of the equation $\left| {\,\begin{array}{*{20}{c}}0&x&{16}\\x&5&7\\0&9&x\end{array}\,} \right| = 0$  are

Let $M$ and $N$ be two $3 \times 3$ matrices such that $M N=N M$. Further, if $M \neq N^2$ and $M^2=N^4$, then

$(A)$ determinant of $\left( M ^2+ MN ^2\right)$ is $0$

$(B)$ there is a $3 \times 3$ non-zero matrix $U$ such that $\left( M ^2+ MN ^2\right) U$ is the zero matrix

$(C)$ determinant of $\left( M ^2+ MN ^2\right) \geq 1$

$(D)$ for a $3 \times 3$ matrix $U$, if $\left( M ^2+ MN ^2\right) U$ equals the zero matrix then $U$ is the zero matrix

If the system of equations  $x +y + z = 6$ ; $x + 2y + 3z= 10$ ; $x + 2y + \lambda z = 0$ has a unique solution, then $\lambda $ is not equal to

$\left| {\,\begin{array}{*{20}{c}}{a - 1}&a&{bc}\\{b - 1}&b&{ca}\\{c - 1}&c&{ab}\end{array}\,} \right| = $

For $\alpha, \beta \in \mathrm{R}$ and a natural number $\mathrm{n}$, let

$A_r=\left|\begin{array}{ccc}r & 1 & \frac{n^2}{2}+\alpha \\ 2 r & 2 & n^2-\beta \\ 3 r-2 & 3 & \frac{n(3 n-1)}{2}\end{array}\right|$. Then $2 A_{10}-A_8$

If $\left| {\,\begin{array}{*{20}{c}}{x - 1}&3&0\\2&{x - 3}&4\\3&5&6\end{array}\,} \right| = 0$, then $x =$