If $f(x) = \left| \begin{array}{ccc} 1 & 6+x & 36+x^2 \\ 0 & x-3 & 3x^2-27 \\ 0 & 2x-4 & 8x^2-32 \end{array} \right|$,then $\lim_{x \rightarrow 1} \frac{f(x)}{f(-x)} = $

If $\left|\begin{array}{lll}1 & 2 & 3 \\ 2 & 3 & 4 \\ 3 & 4 & 5\end{array}\right|=2016 k$,then $k=$ . . . . . .

Given that, $a \alpha^2+2 b \alpha+c \neq 0$ and that the system of equations
$\begin{aligned} & (a \alpha+b) x+a y+b z=0 \\ & (b \alpha+c) x+b y+c z=0 \\ & (a \alpha+b) y+(b \alpha+c) z=0\end{aligned}$
has a non-trivial solution, then $a, b$ and $c$ lie in

The value of the determinant $\left| \begin{array}{ccc} 31 & 37 & 92 \\ 31 & 58 & 71 \\ 31 & 105 & 24 \end{array} \right|$ is

Let $\left| {\begin{array}{*{20}{c}}{6i}&{ - 3i}&1\\4&{3i}&{ - 1}\\{20}&3&i\end{array}} \right| = x + iy$,then

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