AP EAMCET202119 Aug 2021Evening ShiftMathematicsDeterminantsActual
Let a , b , c be such that ( b + c ) ≠ 0 and a a + 1 a - 1 - b b + 1 b - 1 c c - 1 c + 1 + a + 1 b + 1 c - 1 a - 1 b - 1 c + 1 - 1 n + 2 a - 1 n - 1 b - 1 n c = 0 Then the value of n is
Options
- AZero
- BAny even integer
- CAny odd integer
- DAny integer
Correct answer
C. Any odd integer
Step-by-step solution
a a + 1 a - 1 - b b + 1 b - 1 c c - 1 c + 1 + a + 1 b + 1 c - 1 a - 1 b - 1 c + 1 - 1 n + 2 a - 1 n - 1 b - 1 n c = 0 ⇒ a a + 1 a - 1 - b b + 1 b - 1 c c - 1 c + 1 + - 1 n a + 1 b + 1 c - 1 a - 1 b - 1 c + 1 a - b c = 0 ⇒ a a + 1 a - 1 - b b + 1 b - 1 c c - 1 c + 1 + - 1 n a + 1 a - 1 a b + 1 b - 1 - b c - 1 c + 1 c = 0 ⇒ a a + 1 a - 1 - b b + 1 b - 1 c c - 1 c + 1 + - 1 n a a + 1 a - 1 - b b + 1 b - 1 c c - 1 c + 1 = 0 ⇒ ( 1 + - 1 n ) a a + 1 a - 1 - b b + 1 b - 1 c c - 1 c + 1 = 0 Hence, n