MCQ
Let $\left| {\,\begin{array}{*{20}{c}}{6i}&{ - 3i}&1\\4&{3i}&{ - 1}\\{20}&3&i\end{array}\,} \right| = x + iy$, then
  • A
    $x = 3,y = 1$
  • $x = 0,y = 0$
  • C
    $x = 0,y = 3$
  • D
    $x = 1,y = 3$

Answer

Correct option: B.
$x = 0,y = 0$
b
(b) $\,\left| {\,\begin{array}{*{20}{c}}
  {6i}&{ - 3i}&1 \\ 
  4&{3i}&{ - 1} \\ 
  {20}&3&i 
\end{array}\,} \right| = x + iy$

$ \Rightarrow $ $6i( - 3 + 3) + 3i(4i + 20) + 1(12 - 60i) = x + iy$

$ \Rightarrow $ $(0+60i-12+12-60i)$ $ \Rightarrow $ $x = 0,\,y = 0$.

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