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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 \((i=\sqrt{-1})\)
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Verified Answer
The correct answer is:
\(z\) is purely real
\(\begin{aligned}
& \text { Hints : } z=\left|\begin{array}{ccc}
1 & 1+2 i & -5 i \\
1-2 i & -3 & 5+3 i \\
5 i & 5-3 i & 7
\end{array}\right|=1(-21-64)-((1-2 i)(7(1+2 i)+5 i(5-3 i)))+5 i((1+2 i)(5+3 i)-15 i) \\
& \text { = Real } \\
&
\end{aligned}\)
& \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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