Question
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?
JEE Advanced
Mathematics
Determinants
2019
JEE Advanced 2019 (Paper 2)
JEE Advanced Mathematics Question (2019) — Solution
Options
- A. For x = 1 , there exists a unit vector α i ^ + β j ^ + γ k ^ for which R α β γ = 0 0 0
- B. There exists a real number x such that P Q = Q P
- C. det R = det 2 x x 0 4 0 x x 5 + 8 , for all x ∈ R
- D. For x = 0 , if R 1 a b = 6 1 a b , then a + b = 5
Answer
D. For x = 0 , if R 1 a b = 6 1 a b , then a + b = 5
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 = 6 6 a 6 b P Q P - 1 1 a b = 6 6 a 6 b …(i) P - 1 = 1 6 6 - 3 0 0 3 - 2 0 0 2 Putting P ,   Q ,   P - 1 in equation (i) 1 6 12 6 4 0 24 8 0 0 36 1 a b = 6 6 a 6 b ⇒ 12 + 6 a + 4 b = 36 24 a + 8 b = 36 a a = 2 and b = 3 a + b = 5 Option B P Q = Q P P Q = Q P ⇒ P Q P - 1 = Q R = Q Not possible for any value of x .
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