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

The correct option is D x = 0, y = 0 6i amp; 3i amp; 1 4 amp; 3i amp; 1 20 amp; 3 amp; i =x+iy ⇒ 6i( 3 i ^ 2 +3 ...

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Determinant

If 6i -3i...

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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I f ∣ ∣ ∣ ∣ \begin{matrix} 6 i & - 3 i & 1 4 & 3 i & - 1 20 & 3 & i \end{matrix} ∣ ∣ ∣ ∣ = x + y i

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

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

i = \sqrt{- 1}

x = t^{2}

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