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Q. If $\begin{vmatrix}3i&-9i&1\\ 2&9i&-1\\ 10&9&i\end{vmatrix} = x + iy $, then

KEAMKEAM 2017Determinants

Solution:

We have,
$\begin{vmatrix}3 i & -9 i & 1 \\ 2 & 9 i & -1 \\ 10 & 9 & i\end{vmatrix}=x +i y$
$\Rightarrow \begin{vmatrix}3 i+2 & 0 & 0 \\ 2 & 9 i & -1 \\ 10 & 9 & i\end{vmatrix}=x +i y$
$\left[\because R_{1} \rightarrow R_{1}+R_{2}\right]$
$\Rightarrow (3 i+2)\left[9 i^{2}+9\right]=x +i y$
$\Rightarrow (3 i+2)(-9+9)=x +i y\, \left[\because i^{2}=-1\right]$
$\Rightarrow 0=x +i y $
$\Rightarrow x=0, y=0$