Complex Matrix and its Conjugate terms

Question

If $z = \left| {\begin{array}{*{20}{c}} {3 + 2i}&1&i\\ 2&{3 - 2i}&{1 + i}\\ {1 - i}&{ - i}&3 \end{array}} \right|\& \left| {z + \overline z } \right| = k\left| z \right|$, find the value of k

My approach is as follow

$ \Rightarrow z = - \left| {\begin{array}{*{20}{c}} 1&{3 + 2i}&i\\ {3 - 2i}&2&{1 + i}\\ { - i}&{1 - i}&3 \end{array}} \right|$ where $\left( {{C_1} \leftrightarrow {C_2}} \right)$

$ \Rightarrow \overline z = - \left| {\begin{array}{*{20}{c}} 1&{3 - 2i}&{ - i}\\ {3 + 2i}&2&{1 - i}\\ i&{1 + i}&3 \end{array}} \right|$

How do we proceed from here.

score 3 · Accepted Answer · answered Sep 13 '21 at 18:49

You already exchanged the first two columns, which is a good start. The corresponding matrix $$ A = \begin{pmatrix} 1&{3 + 2i}&i\\ {3 - 2i}&2&{1 + i}\\ { - i}&{1 - i}&3\end{pmatrix} $$ has the property that $A^T$ is the elementwise conjugate of $A$, so that $$ \det(A) = \det(A^T) = \overline{\det(A)} \, . $$

It follows that $z = -\det(A)$ is a real number, and therefore is $|z + \bar z| =2|z|$.

score 1 · Answer 2 · answered Sep 13 '21 at 18:37

1

Since $z=27$, $\left|z+\overline z\right|=2\times27$, and therefore $k=2$.

answered Sep 13 '21 at 18:37

José Carlos Santos

427,504

can you elaborate what property you use, I also got this answer but I would like to solve it by using determinant propertry – Samar Imam Zaidi Sep 13 '21 at 18:39
I simply computed the determinant. – José Carlos Santos Sep 13 '21 at 18:43

Complex Matrix and its Conjugate terms

2 Answers2