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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