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
$B = 4A$
$B = - 4A$
$B = - A$
$B = 6A$
For non zero, $a,b,c$ if $\Delta = \left| {\,\begin{array}{*{20}{c}}{1 + a}&1&1\\1&{1 + b}&1\\1&1&{1 + c}\end{array}} \right| = 0$, then the value of $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = $
If $\left| {\,\begin{array}{*{20}{c}}{{a_1}}&{{b_1}}&{{c_1}}\\{{a_2}}&{{b_2}}&{{c_2}}\\{{a_3}}&{{b_3}}&{{c_3}}\end{array}\,} \right| = 5$; then the value of $\left| {\,\begin{array}{*{20}{c}}{{b_2}{c_3} - {b_3}{c_2}}&{{c_2}{a_3} - {c_3}{a_2}}&{{a_2}{b_3} - {a_3}{b_2}}\\{{b_3}{c_1} - {b_1}{c_3}}&{{c_3}{a_1} - {c_1}{a_3}}&{{a_3}{b_1} - {a_1}{b_3}}\\{{b_1}{c_2} - {b_2}{c_1}}&{{c_1}{a_2} - {c_2}{a_1}}&{{a_1}{b_2} - {a_2}{b_1}}\end{array}\,} \right|$is
Let $A=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & \alpha & \beta \\ 0 & \beta & \alpha\end{array}\right]$ and $|2 A|^3=2^{21}$ where $\alpha, \beta \in Z$, Then a value of $\alpha $ is
If $a \ne 6,b,c$ satisfy $\left| {\,\begin{array}{*{20}{c}}a&{2b}&{2c}\\3&b&c\\4&a&b\end{array}\,} \right| = 0,$then $abc = $
If $n$ be the number of values of $x$ for which
matrix $\Delta (x) =\left[ {\begin{array}{*{20}{c}}
{ - x}&x&2\\
2&x&{ - x}\\
x&{ - 2}&{ - x}
\end{array}} \right]$ will be singular, then $det(\Delta\,(n))$ is
$($ where $det(B)$ denotes determinant of Matrix $B) -$