If $A = \left| {\,\begin{array}{*{20}{c}}{ - 1}&2&4\\3&1&0\\{ - 2}&4&2\end{array}\,} \right|$and $B = \left| {\,\begin{array}{*{20}{c}}{ - 2}&4&2\\6&2&0\\{ - 2}&4&8\end{array}\,} \right|$, then $B$ is given by
$B = 4A$
$B = - 4A$
$B = - A$
$B = 6A$
If ${\left| {\,\begin{array}{*{20}{c}}4&1\\2&1\end{array}\,} \right|^2} = \left| {\,\begin{array}{*{20}{c}}3&2\\1&x\end{array}\,} \right| - \left| {\,\begin{array}{*{20}{c}}x&3\\{ - 2}&1\end{array}\,} \right|$, then $ x =$
$\left| {\,\begin{array}{*{20}{c}}{{a_1}}&{m{a_1}}&{{b_1}}\\{{a_2}}&{m{a_2}}&{{b_2}}\\{{a_3}}&{m{a_3}}&{{b_3}}\end{array}\,} \right| = $
A root of the equation $\left| {\,\begin{array}{*{20}{c}}{3 - x}&{ - 6}&3\\{ - 6}&{3 - x}&3\\3&3&{ - 6 - x}\end{array}\,} \right| = 0$ is
If $\left| {\begin{array}{*{20}{c}}{a\, + \,1}&{a\, + \,2}&{a\, + \,p}\\{a\, + \,2}&{a\, +\,3}&{a\, + \,q}\\{a\, + \,3}&{a\, + \,4}&{a\, + \,r}\end{array}} \right|$ $= 0$ , then $p, q, r$ are in :
An ordered pair $(\alpha , \beta )$ for which the system of linear equations
$\left( {1 + \alpha } \right)x + \beta y + z = 2$ ; $\alpha x + \left( {1 + \beta } \right)y + z = 3$ ; $\alpha x + \beta y + 2z = 2$ has a unique solution, is