Solve the following equations: x + 1 3 5 2 x + 2 5 2 3 x + 4 = 0

Given, x + 1 3 5 2 x + 2 5 2 3 x + 4 = 0 ⇒ x + 9 3 5 x + 9 x + 2 5 x + 9 3 x + 4 = 0 [Operating C 1 → C 1 + C 2 + C 3 ] ⇒ x + 9 1 3 5 1 x + 2 5 1 3 x + 4 = 0 [Taking ( x + 9 ) common from C 1 ] ⇒ x + 9 0 0 1 - x 0 x - 1 1 - x 1 3 x + 4 = 0 [ Operating R 1 → R 1 - R 3 , R 2 → R 2 - R 3 ] Expanding along C 1 , we get 0 = ( x + 9 ) 0 - x + x 2 + 1 - x ⇒ 0 = ( x + 9 ) x 2 - 2 x + 1 ⇒ 0 = ( x + 9 ) ( x - 1 ) 2 ∴ Either x + 9 = 0 o r x - 1 = 0 x = - 9 , x = 1

Solve the following equations: x 3 7 2 x 2 7 6 x = 0

Given, x 3 7 2 x 2 7 6 x = 0 ⇒ x + 9 x + 9 x + 9 2 x 2 7 6 x = 0 [Operating R 1 → R 1 + R 2 + R 3 ] ⇒ x + 9 1 1 1 2 x 2 7 6 x = 0 [ Taking ( x + 9 ) common from R 1 ] ⇒ x + 9 0 0 1 0 x - 2 2 7 - x 6 - x x = 0 [ Operating C 1 → C 1 - C 3 , C 2 → C 2 - C 3 ] Expanding along R 1 , we get 0 = ( x + 9 ) ( 0 - ( x - 2 ) ( 7 - x ) ) ⇒ 0 = ( x + 9 ) ( 7 - x ) ( 2 - x ) Either x + 9 = 0 o r 7 - x = 0 o r 2 - x = 0 ∴ x = - 9 o r x = 7 o r x = 2

Solve the following equations: x - 6 - 1 2 - 3 x x - 3 - 3 2 x x + 2 = 0

x − 6 − 1 2 − 3 x x − 3 − 3 2 x x + 2 = 0 ⇒ x − 1 − 6 − 1 x − 1 − 3 x x − 3 x − 1 2 x x + 2 = 0 [Applying C 1 → C 1 + C 3 ] ⇒ ( x − 1 ) 1 − 6 − 1 1 − 3 x x − 3 1 2 x x + 2 = 0 [Taking x - 1 common from C 1 ] ⇒ ( x − 1 ) 1 − 6 − 1 0 − 3 ( x − 2 ) x − 2 0 2 ( x + 3 ) x + 3 = 0 [Applying R 2 → R 2 - R 1 , R 3 → R 3 - R 1 ] ⇒ ( x - 1 ) ( x - 2 ) ( x + 3 ) 1 − 6 − 1 0 − 3 1 0 2 1 = 0 [Taking x - 2 , x + 3 common from R 2 , R 3 respectively] Expanding along C 1 , we get ⇒ ( x - 1 ) ( x - 2 ) ( x + 3 ) 1 ( - 3 - 2 ) = 0 ⇒ x - 1 = 0 , x - 2 = 0 , x + 3 = 0 ⇒ x = 1 , x = 2 , x = - 3

Prove that a b - c c + b a + c b c - a a - b b + a c = a + b + c a 2 + b 2 + c 2

Given, a b - c c + b a + c b c - a a - b b + a c = 1 a a 2 b - c c + b a 2 + a c b c - a a 2 - a b b + a c [Multiplying C 1 by a ] = 1 a a 2 + b 2 + c 2 b - c c + b a 2 + b 2 + c 2 b c - a a 2 + b 2 + c 2 b + a c [ Operating C 1 → C 1 + b C 2 + c C 3 ] = a 2 + b 2 + c 2 a 1 b - c c + b 1 b c - a 1 b + a c [ Taking a 2 + b 2 + c 2 common from C 1 ] = a 2 + b 2 + c 2 a 1 b - c c + b 0 c - a - b 0 a + c - b [Operating C 2 → C 2 - C 1 , C 3 → C 3 - C 1 ] Expanding along C 1 , we get = a 2 + b 2 + c 2 a 1 - b c - ( - a - b ) ( a + c ) = a 2 + b 2 + c 2 a - b c + a 2 + a c + a b + b c = a 2 + b 2 + c 2 a + b + c Hence proved

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SENIOR SECONDARY SCHOOL MATHEMATICS FOR CLASS 12>Algebra>Chapter 6 - Determinants>EXERCISE

R S Aggarwal Solutions for Exercise 2: EXERCISE

Author:R S Aggarwal

R S Aggarwal Mathematics Solutions for Exercise - R S Aggarwal Solutions for Exercise 2: EXERCISE

Attempt the free practice questions from Exercise 2: EXERCISE with hints and solutions to strengthen your understanding. SENIOR SECONDARY SCHOOL MATHEMATICS FOR CLASS 12 solutions are prepared by Experienced Embibe Experts.