Let $$z=11$$−$$2i3+5i1+2i$$−$$510i3$$−$$5i$$−$$10i11$$, then

Correct option is 2z is purely real

Questions

Let $z = |\begin{array}{ccc} 1 & 1 - 2 i & 3 + 5 i \\ 1 + 2 i & - 5 & 10 i \\ 3 - 5 i & - 10 i & 11 \end{array}|,$ then

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z is purely imaginary

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detailed solution

Correct option is B

Conjugate of zequals the determinant obtained by taking conjugate of each of its element. Therefore,z¯=11+2i3−5i1−2i−5−10i3+5i10i11=11−2i3+5i1+2i−510i3−5i−10i11=zThus, z is purely real.

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Let z=11−2i3+5i1+2i−510i3−5i−10i11, then

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Let $z = |\begin{array}{ccc} 1 & 1 - 2 i & 3 + 5 i \\ 1 + 2 i & - 5 & 10 i \\ 3 - 5 i & - 10 i & 11 \end{array}|,$ then