(B) Given the determinant: $\Delta = \left| {\begin{array}{*{20}{c}}{6i}&{ - 3i}&1\\4&{3i}&{ - 1}\\{20}&3&i\end{array}} \right| = x + iy$Expanding alo

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(B) Given the determinant: $\Delta = \left| {\begin{array}{*{20}{c}}{6i}&{ - 3i}&1\\4&{3i}&{ - 1}\\{20}&3&i\end{array}} \right| = x + iy$Expanding along the first row:$\Delta = 6i((3i)(i) - (3)(-1)) - (-3i)((4)(i) - (20)(-1)) + 1((4)(3) - (20)(3i))$$\Delta = 6i(3i^2 + 3) + 3i(4i + 20) + (12 - 60i)$Since $i^2 = -1$,we have $3i^2 + 3 = 3(-1) + 3 = 0$.$\Delta = 6i(0) + 12i^2 + 60i + 12 - 60i$$\Delta = 0 + 12(-1) + 60i + 12 - 60i$$\Delta = -12 + 12 + 60i - 60i = 0$Thus,$x + iy = 0 + 0i$,which implies $x = 0$ and $y = 0$.

Class 12 Mathematics 3 and 4 .Determinants and Matrices Expansion of determinants, Solution of equation in the form of determinants and area of triangle and Equation of Line

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

IIT 1998 All IIT

A
$x = 3, y = 1$
B
$x = 0, y = 0$
C
$x = 0, y = 3$
D
$x = 1, y = 3$

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3 and 4 .Determinants and Matrices

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Expansion of determinants, Solution of equation in the form of determinants and area of triangle and Equation of Line

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The value of $a$ for which the system of equations has a non-zero solution is
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Let $\left| {\begin{array}{*{20}{c}}{6i}&{ - 3i}&1\\4&{3i}&{ - 1}\\{20}&3&i\end{array}} \right| = x + iy$,then

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