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MCQ

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
$(0, 3)$
✓
$(0, 0)$

✓

Answer

Correct option: D.

$(0, 0)$

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

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