JEE Main201911 Jan 2019Evening ShiftMathematicsDeterminantsActual
If | array ccc a-b-c & 2 a & 2 a 2 b & b-c-a & 2 b 2 c & 2 c & c-a-b array |=(a+b+c)(x+a+b+c)², x 0 and a+b+c 0, then x is equal to
Options
- Aabc
- B-(a+b+c)
- C2(a+b+c)
- D-2(a+b+c)
Correct answer
D. -2(a+b+c)
Step-by-step solution
= | array ccc a-b-c & 2 a & 2 a 2 b & b-c-a & 2 b 2 c & 2 c & c-a-b array |R₁ R₁+R₂+R = | array ccc a+b+c & a+b+c & a+b+c 2 b & b-c-a & 2 b 2 c & 2 c & c-a-b array |=(a+b+c) | array ccc 1 & 1 & 1 2 b & b-c-a & 2 b 2 c & 2 c & c-a-b array | array l C₁ C₁-C_ y , C₂ C₂-C₃ =(a+b+c) | array ccc 0 & 0 & 1 0 & -b-c-a & 2 b c+a+b & c+a+b & c-a-b array | array =(a+b+c)(a+b+c)² Hence, x=-2(a+b+c)