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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 = PQP⁻¹. Then which of the following is/are correct?

For x = 0, if R[1 a b] = 6[1 a b], then a + b = 5

There exists a real number x such that PQ = QP

For x = 1, there exists a unit vector αi^ + βj^ + γk^ for which R[α β γ] = [0 0 0]

det(R) = det([2 x x]
[0 4 0]
[x x 5]) + 8 for all x ∈ R

Solution:

Correct option is D.
For x = 0, if R[1 a b] = 6[1 a b], then a + b = 5
det(R) = det(P) × det(Q) × det(P⁻¹)
⇒ det(R) = det(Q) = 4(12 × x² ) = 48 − 4x²
Now [2 x x]
[0 4 0]
[x x 5] = 4(10 − x²) = 40 − 4x²
⇒ det(R) = det([2 x x]
[0 4 0]
[x x 5]) + 8 ∀ x ∈ R
At x = 1, det(Q) = 48 − 4 = 44 = det(R)
Because det(R) ≠ 0, so R[α β γ] = [0 0 0] ⇒ α = β = γ = 0, so αi^ + βj^ + γk^ is not a unit vector
P = [1 1 1]
[0 2 2]
[0 0 3], Q(x = 0) = [2 0 0]
[0 4 0]
[0 0 6]
R = PQP⁻¹ = [2 4 6]
[0 8 12]
[0 0 18]
1/6[6 −3 0]
[0 3 −2]
[0 0 2] = 1/6[12 6 4]
[0 24 8]
[0 0 36]
R = [2 1 2/3]
[0 4 4/3]
[0 0 6]
(R − 6I)[1 a b] = [−4 1 2/3 − 2]
[4/3 0]
[0 0 0][1 a b]
−4 + a + 2b/3 = 0
−2a + 4b/3 = 0
−8 + 8/3 = 0 ⇒ b = 3, a = 2
PQ = QP ⇒ a₁₂ for both are same = x + 4 + x = 2 + 2x + 0 ⇒ x ∈ No value exists