Solve x a a m m m b x b = 0

Given: x a a m m m b x b = 0 Applying C 2 → C 2 − C 1 , C 3 → C 3 − C 1 ⇒ x a − x a − x m 0 0 b x − 6 0 = 0 Expanding along R 2 , we get: ⇒ − m [ − ( a − x ) ( x − b ) ] + 0 − 0 = 0 ⇒ m ( a − x ) ( x − b ) = 0 ⇒ m = 0 , x − b = 0 , a − x = 0 ⇒ x = b , x = a

Solve 7 6 x 2 x 2 x 3 7 = 0

Given: 7 6 x 2 x 2 x 3 7 = 0 Applying R 1 → R 1 + R 2 + R 3 ⇒ 9 + x 9 + x 9 + x 2 x 2 x 3 7 = 0 Taking 9 + x common from R 1 , we get: ⇒ ( 9 + x ) 1 1 1 2 x 2 x 3 7 = 0 Applying C 2 → C 2 − C 1 , C 3 → C 3 − C 1 ⇒ ( 9 + x ) 1 0 0 2 x − 2 0 x 3 − x 7 − x = 0 Expanding along R 1 , we get: ( 9 + x ) [ ( x − 2 ) ( 7 − x ) − 0 ] − 0 + 0 = 0 ⇒ ( 9 + x ) ( x − 2 ) ( 7 − x ) = 0 ⇒ x = 2 , x = 7 , x = − 9 .

Solve 0 x − a x − b x + a 0 x − c x + b x + c 0 = 0 .

Given: 0 x − a x − b x + a 0 x − c x + b x + c 0 = 0 ⇒ 0 0 x - c x + c 0 - x - a x + a x - c x + b 0 + x - b x + a 0 x + b x + c = 0 ⇒ 0 - x - a - x - c x + b + x - b x + a x + c = 0 ⇒ x - a x - c x + b + x - b x + a x + c = 0 ⇒ x - a x 2 + b x - c x - b c + x - b x 2 + c x + a x + a c = 0 ⇒ x 3 + b x 2 - c x 2 - b c x - a x 2 - a b x + a c x + a b c + x 3 + c x 2 + a x 2 + a c x - b x 2 - b c x - a b x - a b c = 0 ⇒ x 3 - b c x - a x 2 - a b x + a c x + x 3 + a x 2 + a c x - b c x - a b x = 0 ⇒ 2 x 3 - 2 b c x - 2 a b x + 2 a c x = 0 ⇒ x 3 - b c x - a b x + a c x = 0 ⇒ x x 2 - b c - a b + a c = 0 ⇒ x = 0 or x 2 - b c - a b + a c = 0 ⇒ x = 0 or x = ± b c + a b - a c

Solve x + a b c b x + c a c a x + b = 0

Given, x + a b c b x + c a c a x + b = 0 Applying C 1 → C 1 + C 2 + C 3 ⇒ x + a + b + c b c x + a + b + c x + c a x + a + b + c a x + b = 0 Taking x + a + b + c common from C 1 , we get: ⇒ x + a + b + c 1 b c 1 x + c a 1 a x + b = 0 Applying R 2 → R 2 - R 1 , R 3 → R 3 - R 1 ⇒ x + a + b + c 1 b c 0 x + c - b a - c 0 a - b x + b - c = 0 Expanding along C 1 , we get: ⇒ x + a + b + c × 1 x + c - b a - c a - b x + b - c = 0 ⇒ x + a + b + c x + c - b x + b - c - a - c a - b = 0 ⇒ x + a + b + c x 2 + b x - c x + c x + b c - c 2 - b x - b 2 + b c - a 2 - a b - a c + b c = 0 ⇒ x + a + b + c x 2 + b c - c 2 - b 2 + b c - a 2 + a b + a c - b c = 0 ⇒ x + a + b + c x 2 - c 2 - b 2 + b c - a 2 + a b + a c = 0 ⇒ ( x + a + b + c ) = 0 or x 2 − c 2 − b 2 + b c − a 2 + a b + a c = 0 ⇒ x = - a + b + c or x 2 = c 2 + b 2 + a 2 - a b - a c - b c ⇒ x = - a + b + c or x = ± c 2 + b 2 + a 2 - a b - a c - b c

Solve 1 + x 1 1 1 1 + x 1 1 1 1 + x = 0

Given: 1 + x 1 1 1 1 + x 1 1 1 1 + x = 0 Applying R 1 → R 1 + R 2 + R 3 ⇒ 3 + x 3 + x 3 + x 1 1 + x 1 1 1 1 + x = 0 Taking out 3 + x common from R 1 , we get: ⇒ 3 + x 1 1 1 1 1 + x 1 1 1 1 + x = 0 Applying R 2 → R 2 - R 1 , R 3 → R 3 - R 1 ⇒ ( 3 + x ) 1 1 1 0 x 0 0 0 x = 0 ⇒ 3 + x × 1 x 0 0 x = 0 [Expanding along C 1 ] ⇒ 3 + x × x 2 = 0 ⇒ 3 + x = 0 or x 2 = 0 ⇒ x = 0 , - 3

Solve 1 4 20 1 − 2 5 1 2 x 5 x 2 = 0

Given: 1 4 20 1 − 2 5 1 2 x 5 x 2 = 0 Applying R 2 → R 2 - R 1 , R 3 → R 3 - R 1 1 4 20 0 - 6 - 15 0 2 x - 4 5 x 2 - 20 = 0 Expanding along C 1 , we get: 1 - 6 - 15 2 x - 4 5 x 2 - 20 = 0 ⇒ - 6 5 x 2 − 20 + 15 2 x − 4 = 0 ⇒ - 30 x 2 + 120 + 30 x − 60 = 0 ⇒ - 30 x 2 + 30 x + 60 = 0 ⇒ x 2 - x - 2 = 0 ⇒ x 2 - 2 x + x - 2 = 0 ⇒ x x - 2 + 1 x - 2 = 0 ⇒ x + 1 x - 2 = 0 ⇒ x + 1 = 0 or x - 2 = 0 ⇒ x = - 1 , 2