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Question: If \(\left| \begin{matrix} 6i & - 3i & 1 \\ 4 & 3i & - 1 \\ 20 & 3 & i \end{matrix} \right|\) = x + ...

If ∣6i−3i143i−1203i∣\left| \begin{matrix} 6i & - 3i & 1 \\ 4 & 3i & - 1 \\ 20 & 3 & i \end{matrix} \right| = x + iy, then

A

x = 3, y = 1

B

x = 1, y = 3

C

x = 0, y = 0

D

x = 0, y = 3

Answer

x = 0, y = 0

Explanation

Solution

R1 ® R1 + R2

∣6i+40043i−1203i∣\left| \begin{matrix} 6i + 4 & 0 & 0 \\ 4 & 3i & - 1 \\ 20 & 3 & i \end{matrix} \right| = x + iy Ž (6i + 4) (–3 + 3) = x + iy

0 = x + iy Ž 0 + 0i = x + iy Ž x = 0, y = 0