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Q.

If $Z = |\begin{array}{l} 1 1 - 2 i 3 + 5 i \\ 1 + 2 i - 5 10 i \\ 3 - 5 i - 10 i 11 \end{array}|$ then

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a

Z is purely imaginary

b

Z is purely real

c

Z = 0

d

arg $Z = \frac{- π}{4}$

answer is B.

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Detailed Solution

$\begin{array}{l} |\begin{array}{l} 1 1 - 2 i 3 + 5 i \\ 1 + 2 i - 5 10 i \\ 3 - 5 i - 10 i 11 \end{array}| \\ z = 1 (- 55 + 100 i^{2}) - (1 - 2 i) (11 + 22 i - 30 i + 50 i^{2}) + (3 + 5 i) (- 10 i - 20 i^{2} + 15 - 25 i) \\ z = (- 55 - 100) - (1 - 2 i) (- 39 - 8 i) + (3 + 5 i) (35 - 35 i) \\ z = - 155 + 39 + 8 i - 78 i + 16 + 105 - 105 i + 175 i + 175 \\ z = 180 + o i \end{array}$

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