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# If $\begin{vmatrix}6i&-3i&1\\4&3i&-1\\20&3&i\end{vmatrix}=x+iy$ then

$\begin{array}{1 1}(a)\;x=3,y=1&(b)\;x=1,y=3\\(c)\;x=0,y=3&(d)\;x=0,y=0\end{array}$

Given :
$\begin{vmatrix}6i &-3i&1\\4 &3i&-1\\20 &3 &i\end{vmatrix}=x+iy$
$\Rightarrow -3i\begin{vmatrix}6i &1&1\\4 &-1&-1\\20&i&i\end{vmatrix}$=0
$C_2$ and $C_3$ are identical.
$\Rightarrow x+iy=0$
$\Rightarrow x=0,y=0$
Hence (d) is the correct answer.