# Thread: Complex Numbers?(Find value of determinant)

1. ## Complex Numbers?(Find value of determinant)

Find the value of the determinant:
$\left|\begin{array}{ccc}e^{2iA}&e^{-iC}&e^{-iB}\\e^{-iC}&e^{2iB}&e^{-iA}\\e^{-iB}&e^{-iA}&e^{2iC}\end{array}\right|$
where A, B, C are the angles of a triangle.

EDIT: I corrected the typo

2. If d is the determinant, then

$d=e^{2i(A+B+C)}+e^{-i(A+B+C)}+e^{-i(A+B+C)}-1-1-1=$

$=e^{4\pi i}+2e^{-\pi i}-3=1-2-3=-4$

3. Hello,

$\left|\begin{array}{ccc}e^{2iA}&e^{-iC}&e^{-iB}\\e^{{\color{red}-}i{\color{red}C}}&e^{2iB}&e^{-iA}\\e^{-iB}&e^{-iA}&e^{2iC}\end{array}\right|
$
I think there's a typo in red. (which red_dog considered...)

Sarrus' rule gives the value :

$e^{2i(A+B+C)}+e^{i(-C-B-A)}+e^{-i(A+B+C)} \quad - \quad e^{i(2B-B-B)}-e^{i(2C-C-C)}-e^{i(2A-A-A)}$

$A+B+C=\pi$ and recall these values : $e^{i\pi}=e^{-i\pi}=-1$, $e^{2i\pi}=1$

--> $=1-1-1-1-1-1=-4$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Another approach.

Multiply the first row by $e^{i(B+C)}$, the second row by $e^{i(A+C)}$, the third row by $e^{i(A+B)}$.

This gives (implicitly using $e^{i(A+B+C)}=e^{i\pi}=-1$) :

$D=e^{-i(B+C)-i(A+C)-i(A+B)} \begin{vmatrix} -e^{iA} & e^{iB} & e^{iC} \\ e^{iA} & -e^{iB} & e^{iC} \\ e^{iA} & e^{iB} & -e^{iC} \end{vmatrix}=e^{-i(A+B+C)}\begin{vmatrix} -1&1&1\\1&-1&1\\1&1&-1\end{vmatrix}$