If $A \ne O$ and $B \ne O$ are $ n × n$ matrix such that $AB = O,$ then
$Det(A) = 0$ or $Det(B) = 0$
$Det(A) = 0$ and $Det(B) = 0$
$Det(A) = $ $Det(B) \ne 0$
${A^{ - 1}} = {B^{ - 1}}$
$\left| {\,\begin{array}{*{20}{c}}{1/a}&{{a^2}}&{bc}\\{1/b}&{{b^2}}&{ca}\\{1/c}&{{c^2}}&{ab}\end{array}\,} \right| = $
If $A = \left| {\,\begin{array}{*{20}{c}}{\sin (\theta + \alpha )}&{\cos (\theta + \alpha )}&1\\{\sin (\theta + \beta )}&{\cos (\theta + \beta )}&1\\{\sin (\theta + \gamma )}&{\cos (\theta + \gamma )}&1\end{array}\,} \right|$ ,then
For positive numbers $x,y$ and $z$ the numerical value of the determinant $\left| {\,\begin{array}{*{20}{c}}1&{{{\log }_x}y}&{{{\log }_x}z}\\{{{\log }_y}x}&1&{{{\log }_y}z}\\{{{\log }_z}x}&{{{\log }_z}y}&1\end{array}\,} \right|$is
Let $k_1$, $k_2$ be the maximum and minimum values of $k$ for which the system of equations given by
$x + ky = 1$ ; $kx + y = 2$; $x + y = k$ are consistent then $k_1^2 + k_2^2$ is equal to
The system of equations $kx + y + z =1$ $x + ky + z = k$ and $x + y + zk = k ^{2}$ has no solution if $k$ is equal to