First slide

Introduction to Determinants

Question

A root of the equation $Δ = |\begin{array}{ccc} 0 & x - a & x - b \\ x + a & 0 & x - c \\ x + b & x + c & 0 \end{array}| = 0$ is

Moderate

A

$\frac{1}{2} (a + b + c)$
B

0
C

-1
D

1

Solution

When we substitute $x = 0, ∆$ becomes
$|\begin{array}{ccc} 0 & - a & - b \\ a & 0 & - c \\ b & c & 0 \end{array}|$
which is equal to $0$ as $∆$ is skew symmetric determinant of odd order.
Alternative Solution
Evaluating $∆$ along $R_{1}$ , we get
$- (x - a) |\begin{array}{cc} x + a & x - c \\ x + b & 0 \end{array}| + (x - b) |\begin{array}{cc} x + a & 0 \\ x + b & x + c \end{array}| = 0$
$\begin{array}{l} \Rightarrow (x - a) (x + b) (x - c) + (x - b) (x + a) (x + c) = 0 \\ \Rightarrow x^{3} + (- a + b - c) x^{2} + (- a b + a c - b c) x + a b c + x^{3} \\ + (a - b + c) x^{2} + (- a b + a c - b c) - a b c = 0 \\ \Rightarrow 2 x^{3} - 2 (a b - a c + b c) x = 0 \\ \Rightarrow 2 x [x^{2} - (a b - a c + b c)] = 0 \end{array}$
$\Rightarrow x = 0$ or $x = \pm \sqrt{a b - a c + b c}$
$∴$ one of the roots of the equation is $0 .$

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