First slide

Introduction to Determinants

Question

Let
$Δ = |\begin{array}{ccc} 0 & b - a & c - a \\ a - b & 0 & c - b \\ a - c & b - c & 0 \end{array}|$ , then
$∆$ equals

Moderate

A

0
B

$a b c$
C

$a^{2} + b^{2} + c^{2}$
D

$b c + c a + a b$

Solution

Interchanging the rows and columns, we get
$Δ = |\begin{array}{ccc} 0 & a - b & a - c \\ b - a & 0 & b - c \\ c - a & c - b & 0 \end{array}|$
Taking –1 common from each of $R_{1}, R_{2}, R_{3}$ we get
$\begin{aligned} Δ = (- 1)^{3} |\begin{array}{ccc} 0 & b - a & c - a \\ a - b & 0 & c - b \\ a - c & b - c & 0 \end{array}| \\ Δ = - Δ \Rightarrow 2 Δ = 0 or Δ = 0 \end{aligned}$
Alternative Solution
$∆$ is a skew symmetric determinant of odd order, therefore $∆$ = 0

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