JEE Advanced2019MathematicsDeterminantsActual
Let x ∈ R and let P = 1 1 1 0 2 2 0 0 3 , Q = 2 x x 0 4 0 x x 6 and R = P Q P - 1 . Then which of the following options is/are correct?
Options
- AFor x = 1 , there exists a unit vector α i ^ + β j ^ + γ k ^ for which R α β γ = 0 0 0
- BThere exists a real number x such that P Q = Q P
- Cdet R = det 2 x x 0 4 0 x x 5 + 8 , for all x ∈ R
- DFor x = 0 , if R 1 a b = 6 1 a b , then a + b = 5
Correct answer
C. det R = det 2 x x 0 4 0 x x 5 + 8 , for all x ∈ R
Step-by-step solution
P = 1 1 1 0 2 2 0 0 3 Q = 2 x x 0 4 0 x x 6 Option ( C ) Now R = P Q P - 1 det ⁡ R = det ⁡ R det ⁡ Q det ⁡ P - 1 det ⁡ R = det ⁡ Q det ⁡ P - 1 = 1 det ⁡ P det ⁡ R = det ⁡ 2 x x 0 4 0 x x 6 det ⁡ R = 48 - 4 x 2 now det ⁡ 2 x x 0 4 0 x x 5 = 40 - 4 x 2 det ⁡ R = det ⁡ 2 x x 0 4 0 x x 5 + 8 Option A R α β γ = 0 0 0 must have not trivial solution So det ⁡ R = 0 48 - 4 x 2 = 0 ⇒ x = ± 2 3 Option D R 1 a b