3 and 4 .Determinants and Matrices
easy

माना $\left| {\,\begin{array}{*{20}{c}}{6i}&{ - 3i}&1\\4&{3i}&{ - 1}\\{20}&3&i\end{array}\,} \right| = x + iy$, तो

A

$x = 3,y = 1$

B

$x = 0,y = 0$

C

$x = 0,y = 3$

D

$x = 1,y = 3$

(IIT-1998)

Solution

(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 $ $x = 0,\,y = 0$.

Standard 12
Mathematics

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