If and | b c a c a b | = 0 , then a - a + b - b + - c is equal to: [2024] Matrices and Determinants jee mains questions | JEE Main Maths Matrices and Determinants Previous Year Question

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
If $α \neq a,$ $β \neq b,$ $γ \neq c$ and $| \begin{matrix} α & b & c \\ a & β & c \\ a & b & γ \end{matrix} |$ $= 0,$ then $\frac{a}{α - a} + \frac{b}{β - b} + \frac{γ}{γ - c}$ is equal to: [2024]

1 1

2 2

3 0

4 3

Ans.
(3)

$| \begin{matrix} α & b & c \\ a & β & c \\ a & b & γ \end{matrix} |$ $= 0$

$R_{1} \to R_{1} - R_{2}, R_{2} \to R_{2} - R_{3}$

$\Rightarrow$ $| \begin{matrix} α - a & b - β & 0 \\ 0 & β - b & c - γ \\ a & b & γ \end{matrix} |$ $= 0$

Taking out $α - a, β - b$ and $γ - c$ common from column-1, 2 and 3 respectively,

$\Rightarrow (α - a) (β - b) (γ - c)$ $| \begin{matrix} 1 & - 1 & 0 \\ 0 & 1 & - 1 \\ \frac{a}{α - a} & \frac{b}{β - b} & \frac{γ}{γ - c} \end{matrix} |$ $= 0$

$\Rightarrow (α - a) (β - b) (γ - c) [1 (\frac{γ}{γ - c} + \frac{b}{β - b}) + 1 (0 + \frac{a}{α - a})] = 0$

$\Rightarrow (α - a) (β - b) (γ - c) [\frac{γ}{γ - c} + \frac{b}{β - b} + \frac{a}{α - a}] = 0$

$\Rightarrow [\frac{γ}{γ - c} + \frac{b}{β - b} + \frac{a}{α - a}] = 0$ $(∵ α \neq a, β \neq b, γ \neq c)$

$∴$ $\frac{γ}{γ - c} + \frac{b}{β - b} + \frac{a}{α - a} = 0$

Want Us to Email you About Special Offers & Updates?