NTA Abhyas JEE Main2020MathematicsDeterminantsPractice
The value of the determinant ∆ = 13 + 3 2 5 5 15 + 26 5 10 3 + 65 15 5 is equal to
Options
- A15 2 - 25 3
- B25 3 - 15 2
- C3 5
- D- 15 2 + 7 3
Correct answer
A. 15 2 - 25 3
Step-by-step solution
Taking 5 common from C 2 and C 3 , we get ∆ = 5 2 13 + 3 2 1 15 + 26 5 2 3 + 65 3 5 Applying C 1 → C 1 - 13 C 3 - 3 C 2 , we get ∆ = 5 - 3 2 1 0 5 2 0 3 5 = 5 - 3 5 - 6 = 5 18 - 25 3 = 15 2 - 25 3