First slide

Introduction to Determinants

Question

Suppose $a, b, c$ are distinct real numbers and $Δ = |\begin{array}{ccc} a & a^{2} & b + c \\ b & b^{2} & c + a \\ c & c^{2} & a + b \end{array}| = 0$ .Then $a + b + c$ equals

Moderate

A

-1
B

2
C

0
D

-5

Solution

Using $C_{3} \to C_{3} + C_{1}$ and taking $(a + b + c)$
common from $C_{3}$ we get
$Δ = (a + b + c) Δ_{1}$ (1)
Where
$Δ_{1} = |\begin{array}{ccc} a & a^{2} & 1 \\ b & b^{2} & 1 \\ c & c^{2} & 1 \end{array}|$
Using $R_{1} \to R_{1} - R_{2}$ and $R_{2} \to R_{2} - R_{3}$ , we get
$\begin{aligned} Δ_{1} = |\begin{array}{ccc} a - b & a^{2} - b^{2} & 0 \\ b - c & b^{2} - c^{2} & 0 \\ c & c^{2} & 1 \end{array}| \\ = (a - b) (b - c) |\begin{array}{ccc} 1 & a + b & 0 \\ 1 & b + c & 0 \\ c & c^{2} & 1 \end{array}| \\ = (a - b) (b - c) |\begin{array}{cc} 1 & a + b \\ 1 & b + c \end{array}| \end{aligned}$
[Expand along $C_{3}$ ]
$= (a - b) (b - c) (c - a)$ (2)
From (1) and (2)
$Δ = (a + b + c) (a - b) (b - c) (c - a)$
As $Δ = 0$ and $a, b, c$ are distinct
we get $a + b + c = 0$

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