First slide

Algebra of complex numbers

Question

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

Moderate

A

$z$ is purely imaginary
B

$z$ is purely real
C

$z = 0$
D

none of these

Solution

Conjugate of $z$ equals the determinant obtained by taking conjugate of each of its element. Therefore,
$\bar{z} = |\begin{array}{ccc} 1 & 1 + 2 i & 3 - 5 i \\ 1 - 2 i & - 5 & - 10 i \\ 3 + 5 i & 10 i & 11 \end{array}| = |\begin{array}{ccc} 1 & 1 - 2 i & 3 + 5 i \\ 1 + 2 i & - 5 & 10 i \\ 3 - 5 i & - 10 i & 11 \end{array}| = z$
Thus, $z$ is purely real.

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