If - 6i -3i 1 4 3i -1 20 3 i -x-iy then

The correct option is D x=0, y=0 Expanding the Determinant #160; we get 6i(3i^2+3) ( 3i)(4i+20)+1(12 60i)=18i^3+18i+12i^2+60i+12 60i= 18i+18i 12+12+ ...

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Byju's Answer

Standard XII

Mathematics

Definition of a Determinant

If | 6i -...

Question

If $∣ ∣ ∣ ∣ \begin{matrix} 6 i & - 3 i & 1 4 & 3 i & - 1 20 & 3 & i \end{matrix} ∣ ∣ ∣ ∣ = x + i y$ then

x = 3, y = 1

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x = 1, y = 3

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x = 0, y = 3

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x = 0, y = 0

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Solution

The correct option is D $x = 0, y = 0$
Expanding the Determinant we get
$6 i (3 i^{2} + 3) - (- 3 i) (4 i + 20) + 1 (12 - 60 i) = 18 i^{3} + 18 i + 12 i^{2} + 60 i + 12 - 60 i = - 18 i + 18 i - 12 + 12 + 0 = 0$

Hence,
$x + i y = 0 + i 0$

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If ∣∣ ∣∣6i−3i143i−1203i∣∣ ∣∣=x+iy then

The correct option is D x=0,y=0Expanding the Determinant we get6i(3i2+3)−(−3i)(4i+20)+1(12−60i)=18i3+18i+12i2+60i+12−60i=−18i+18i−12+12+0=0Hence,x+iy=0+i0

If $∣ ∣ ∣ ∣ \begin{matrix} 6 i & - 3 i & 1 4 & 3 i & - 1 20 & 3 & i \end{matrix} ∣ ∣ ∣ ∣ = x + i y$ then

The correct option is D $x = 0, y = 0$
Expanding the Determinant we get
$6 i (3 i^{2} + 3) - (- 3 i) (4 i + 20) + 1 (12 - 60 i) = 18 i^{3} + 18 i + 12 i^{2} + 60 i + 12 - 60 i = - 18 i + 18 i - 12 + 12 + 0 = 0$

Hence,
$x + i y = 0 + i 0$