WBJEE · Maths · Determinants
If \(z=\left(\begin{array}{ccc}1 & 1+2 i & -5 i \\ 1-2 i & -3 & 5+3 i \\ 5 i & 5-3 i & 7\end{array}\right)\) then \((i=\sqrt{-1})\)
- A \(z\) is purely real
- B \(z\) is purely imaginary
- C \(z+\bar{z}=0\)
- D \((z-\bar{z})\) i is purely imaginary
Answer & Solution
Correct Answer
(A) \(z\) is purely real
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \text { Hints : } z=\left|\begin{array}{ccc} 1 & 1+2 i & -5 i \\ 1-2 i & -3 & 5+3 i \\ 5 i & 5-3 i & 7 \end{array}\right|=1(-21-64)-((1-2 i)(7(1+2 i)+5 i(5-3 i)))+5 i((1+2 i)(5+3 i)-15 i) \\ & \text { = Real } \\ & \end{aligned}\)
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