First slide

Introduction to Determinants

Question

Suppose $A, B, C$ are angles of a triangle, and let $Δ = |\begin{array}{ccc} e^{2 i A} & e^{- i C} & e^{- i B} \\ e^{- i C} & e^{2 i B} & e^{- i A} \\ e^{- i B} & e^{- i A} & e^{2 i C} \end{array}|$ then value of $∆$ is

Moderate

A

-1
B

-4
C

0
D

4

Solution

Talking $e^{i A},$ common from $R_{1}, e^{i B}$ from $R_{2}$ and $e^{i C}$ from $R_{3}$ we get
$Δ = e^{i (A + B + C)} Δ_{1}$
where
$Δ_{1} = |\begin{array}{ccc} e^{i A} & e^{i (A + C)} & e^{i (A + B)} \\ e^{- i (B + C)} & e^{i B} & e^{- i (A + B)} \\ e^{- i (B + C)} & e^{- i (A + C)} & e^{i C} \end{array}|$
But $A + B + C = π,$ so that $e^{i (A + R + C)} = e^{i π}$
$= \cos π + i \sin π = - 1$ .Also
$A + C = π - B \Rightarrow e^{- i (A + C)} = e^{- π i} e^{i B} = - e^{i B}$
Thus, $Δ_{1} = |\begin{array}{ccc} e^{i A} & - e^{i B} & - e^{i C} \\ - e^{- i A} & e^{i B} & - e^{i C} \\ - e^{i A} & - e^{i B} & e^{i C} \end{array}|$
$= e^{i (A + B + C)} |\begin{array}{ccc} 1 & - 1 & - 1 \\ - 1 & 1 & - 1 \\ - 1 & - 1 & 1 \end{array}|$
Using $C_{1} \to C_{1} + C_{2}$ we get
$Δ_{1} = (- 1) |\begin{array}{ccc} 0 & - 1 & - 1 \\ 0 & 1 & - 1 \\ - 2 & - 1 & 1 \end{array}| = (- 1) (- 2) (2) = 4$
Therefor $Δ = (- 1) Δ_{1} = - 4$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App