AP EAMCET202021 Sep 2020Evening ShiftMathematicsDeterminantsActual
For any (a, b, c R ), the determinant ( | array lll b c & b+c & 1 c a & c+a & 1 a b & a+b & 1 array | ) is equal to
Options
- A(a (b^2-c^2 )+b (c^2-a^2 )+c (a^2-b^2 ) )
- B(a(b-c)+b(c-a)+c(a-b) )
- C((a-b)(b-c)(c-a) )
- D(a b c )
Correct answer
C. ((a-b)(b-c)(c-a) )
Step-by-step solution
For any (a, b, c R ), the given determinant ( = | array lll b c & b+c & 1 c a & c+a & 1 a b & a+b & 1 array | ) On applying (R₂ R₂-R₁ ) and (R₃ R₃-R₁ ), we have ( gathered = | array ccc b c & b+c & 1 c(a-b) & a-b & 0 b(a-c) & a-c & 0 array | =(a-b)(a-c) | array ccc b c & b+c & 1 c & 1 & 0 b & 1 & 0 array | gathered ) On applying (R₃ R₃-R₂ ), we have ( aligned & =(a-b)(a-c) | array ccc b c & b+c & 1 c & 1 & 0 b-c & 0 & 0 array | & =(a-b)(a-c)[0-(b-c)]=(a-b)(b-c)(c-a) aligned ) Hence option (c) is correct.