यदि $\,\left| \begin{array}{l}\,6i\,\,\,\,\, - 3i\,\,\,\,\,\,\,\,\,1\\\,\,4\,\,\,\,\,\,\,\,\,3i\,\,\,\,\,\, - 1\\\,20\,\,\,\,\,\,\,\,3\,\,\,\,\,\,\,\,\,\,i\end{array} \right|\,= x + iy,$ तो $(x, y) =$
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Answer
$\left| {\,\begin{array}{*{20}{c}}{6i\,\,\,}&{ - 3i\,\,\,}&{\,\,1}\\{4\,\,}&{3i}&{ - 1}\\{20\,}&3&{\,\,i}\end{array}\,} \right|$
$=x + iy$
$\Rightarrow \left| {\,\begin{array}{*{20}{c}}{6i + 4\,\,\,\,}&{0\,\,\,\,}&{\,\,0}\\{4\,\,\,}&{3i\,\,\,\,}&{ - 1}\\{20\,\,\,}&{3\,\,\,\,}&{\,\,i}\end{array}\,} \right| $
$= x + iy$
$[{R_1} \to {R_1} + {R_2}]$
$\Rightarrow (6i + 4)\,(3{i^2} + 3)$= $x + iy$
$\Rightarrow (6i + 4)\,( - 3 + 3) = x + iy$
$\Rightarrow x + iy = 0\,\, = 0 + i.0$
$ \Rightarrow (x,\,y)\, = (0,\,0)$.
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