The roots of the equation $\left| {\,\begin{array}{*{20}{c}}1&4&{20}\\1&{ - 2}&5\\1&{2x}&{5{x^2}}\end{array}\,} \right| = 0$ are

The values of $a$ and $b$, for which the system of equations    $2 x+3 y+6 z=8$   ;  $x+2 y+a z=5$     ;  $3 x+5 y+9 z=b$  has no solution, are:

Let $ \alpha _1, \alpha _2$ are two values of $\alpha $ for which the system $2 \alpha x + y = 5, x - 6y = \alpha $ and $x + y = 2$ is consistent, then $ |2(\alpha _1 + \alpha _2)| $ is -

If ${A_\lambda } = \left( {\begin{array}{*{20}{c}}
\lambda &{\lambda  - 1}\\
{\lambda  - 1}&\lambda 
\end{array}} \right);\,\lambda  \in N$ then $|A_1| + |A_2| + ..... + |A_{300}|$ is equal to

If the following system of linear equations

$2 x+y+z=5$

$x-y+z=3$

$x+y+a z=b$

has no solution, then :

$A=\left[\begin{array}{lll}1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4\end{array}\right],$ then show that $|3 A|=27|A|$.