Evaluate, Δ = b + c a a b c + a b c c a + b .

Given, Δ = b + c a a b c + a b c c a + b Expanding the given determinant along first row, Δ = b + c a a b c + a b c c a + b = b + c c + a a + b - b c - a b a + b - b c + a b c - c c + a = b + c a c + a 2 + b c + a b - b c - a a b + b 2 - b c + a b c - c 2 - a c = b + c a c + a 2 + a b - a 2 b - a b 2 + a b c + a b c - a c 2 - a 2 c = a b c + a 2 b + a b 2 + a c 2 + a 2 c + a b c - a 2 b - a b 2 + a b c + a b c - a c 2 - a 2 c = 4 a b c

If A = 1 - 1 5 3 2 - 4 7 8 9 , then find | A | .

A = 1 - 1 5 3 2 - 4 7 8 9 Hence, determinant will be : A = 1 18 + 32 + 1 27 + 28 + 5 24 - 14 = 50 + 55 + 50 = 155

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 3 - x - 6 3 - 6 3 - x 3 3 3 - 6 - x = 0

Given, 3 - x - 6 3 - 6 3 - x 3 3 3 - 6 - x = 0 Apply, C 1 → C 1 + C 2 + C 3 ⇒ - x - 6 3 - x 3 - x 3 - x 3 - 6 - x = 0 Apply, R 2 → R 2 - R 1 , R 3 → R 3 - R 1 ⇒ - x - 6 3 0 9 - x 0 0 9 - 9 - x = 0 ⇒ - x 9 - x - 9 - x = 0 ⇒ x = 0 , 9 , - 9

If x , y , z are all different and if x x 2 1 + x 3 y y 2 1 + y 3 z z 2 1 + z 3 = 0 , prove that x y z = - 1 .

Given, x x 2 1 + x 3 y y 2 1 + y 3 z z 2 1 + z 3 = 0 It can be written as, ⇒ x x 2 1 y y 2 1 z z 2 1 + x x 2 x 3 y y 2 y 3 z z 2 z 3 = 0 Changing row/column changes the sign of the determinant. ⇒ - 1 2 1 1 1 x x 2 y y 2 z z 2 + x y z 1 1 1 x x 2 y y 2 z z 2 = 0 ⇒ 1 + x y z 1 1 1 x x 2 y y 2 z z 2 = 0 ∴ x y z = - 1

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

Given, ( x - a ) 2 b 2 c 2 a 2 ( x - b ) 2 c 2 a 2 b 2 ( x - c ) 2 ⇒ x 2 + a 2 - 2 a x b 2 c 2 a 2 x 2 + b 2 - 2 b x c 2 a 2 b 2 x 2 + c 2 - 2 c x R 2 → R 2 - R 1 , R 3 → R 3 - R 1 ⇒ x 2 + a 2 - 2 a x b 2 c 2 - x x - 2 a x x - 2 b 0 - x x - 2 a 0 x x - 2 c Taking - x common from R 2 & R 3 ⇒ - x 2 x 2 + a 2 - 2 a x b 2 c 2 x - 2 a - x - 2 b 0 x - 2 a 0 - x - 2 c ⇒ - x 2 x 2 + a 2 - 2 a x x - 2 b x - 2 c + b 2 x - 2 a x - 2 c + c 2 x - 2 a x - 2 b ⇒ x 2 x - 2 a x - 2 b x - 2 c x + a 2 x - 2 a + b 2 x - 2 b + c 2 x - 2 c

If p + q + r = 0 , prove that p a q b r c q c r a p b r b p c q a = - p q r a b c b c a c a b .

Given: p + q + r = 0 . . . . i Now, L . H . S . = p a q b r c q c r a p b r b p c q a Expanding along R 1 , we have = p a q r a 2 - p 2 b c - q b q 2 a c - p r b 2 + r c p q c 2 - r 2 a b = p q r a 3 + b 3 + c 3 - a b c p 3 + q 3 + r 3 . . . ( 1 ) Since p + q + r = 0 ⇒ p 3 + q 3 + r 3 = 3 p q r , hence from ( 1 ) , we have L . H . S . = p q r a 3 + b 3 + c 3 - a b c 3 p q r ⇒ L . H . S . = p q r a 3 + b 3 + c 3 - 3 a b c . . . ( 2 ) Now, a b c b c a c a b Applying R 1 → R 1 + R 2 + R 3 , we get = a + b + c b + c + a c + a + b b c a c a b Taking out ( a + b + c ) common from first row, we get = a + b + c 1 1 1 b c a c a b Applying C 2 → C 2 - C 1 and C 3 → C 3 - C 1 = a + b + c 1 0 0 b c - b a - b c a - c b - c Expanding along first row, we get = a + b + c c - b a - b a - c b - c = a + b + c ( c - b ) ( b - c ) - ( a - b ) ( a - c ) = a + b + c c b - c 2 - b 2 + b c - a 2 + a c + a b - b c = a + b + c a b + b c + c a - a 2 - b 2 - c 2 = - a 3 + b 3 + c 3 - 3 a b c . . . . . ( 3 ) ∵ l 3 + m 3 + n 3 - 3 l m n = ( l + m + n ) ( l 2 + m 2 + n 2 - l m - m n - n l ) From ( 2 ) and ( 3 ) , we get p a q b r c q c r a p b r b p c q a = - p q r a b c b c a c a b Hence, proved.

Find the area of the triangle whose vertices are ( 3 , 8 ) , ( - 4 , 2 ) and ( 5 , 1 ) .

The area of triangle with vertices ( 3 , 8 ) , ( - 4 , 2 ) , 5 , 1 is given as follows Δ = 1 2 3 8 1 − 4 2 1 5 1 1 Expanding the determinant along C 1 , we have = 1 2 [ 3 ( 2 - 1 ) - 8 ( - 4 - 5 ) + 1 ( - 4 - 10 ) ] = 1 2 ( 3 + 72 - 14 ) = 61 2