4-1.Complex numbers
easy

If $\,\left| \begin{array}{l}\,6i\,\,\,\,\, - 3i\,\,\,\,\,\,\,\,\,1\\\,\,4\,\,\,\,\,\,\,\,\,3i\,\,\,\,\,\, - 1\\\,20\,\,\,\,\,\,\,\,3\,\,\,\,\,\,\,\,\,\,i\end{array} \right|\,$=$x + iy$, then $(x, y)$ is

A

$(3, 1)$

B

$(1, 3)$

C

$(0, 3)$

D

$(0, 0)$

Solution

(d) $\left| {\,\begin{array}{*{20}{c}}{6i\,\,\,}&{ – 3i\,\,\,}&{\,\,1}\\{4\,\,}&{3i}&{ – 1}\\{20\,}&3&{\,\,i}\end{array}\,} \right|$=$x + iy$
==> $\left| {\,\begin{array}{*{20}{c}}{6i + 4\,\,\,\,}&{0\,\,\,\,}&{\,\,0}\\{4\,\,\,}&{3i\,\,\,\,}&{ – 1}\\{20\,\,\,}&{3\,\,\,\,}&{\,\,i}\end{array}\,} \right| = x + iy$ $[{R_1} \to {R_1} + {R_2}]$
==> $(6i + 4)\,(3{i^2} + 3)$= $x + iy$
==> $(6i + 4)\,( – 3 + 3) = x + iy$
==> $x + iy = 0\,\, = 0 + i.0$

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

Standard 11
Mathematics

Similar Questions

Start a Free Trial Now

Confusing about what to choose? Our team will schedule a demo shortly.