Q.2
If 𝑃 = ( − 𝑐𝑜𝑠𝛼 𝑠𝑖𝑛𝛼 𝑐𝑜𝑠𝛼 𝑠𝑖𝑛𝛼) and 𝑃 + 𝑃 𝑇 = I, the value of α (0 ≤ 𝛼 ≤ 𝜋⁄2) is
(A) 𝜋/2
(B) 𝜋/3
(C) 3𝜋/2
(D) 0
Introduction
In competitive exams like IIT JAM, BSc, BCA, and MSc entrance tests, matrix questions involving transpose and identity matrices are very common.
In this article, we solve a standard and important problem based on orthogonal matrices and trigonometric identities.
Question
If
P = \begin{pmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{pmatrix}
and P + PT = I
then find the value of α, where 0 ≤ α ≤ 2π
Options
- (A) π/2
- (B) π/3
- (C) 3π/2
- (D) 0
Step-by-Step Solution
Step 1: Find Transpose of P
Transpose means interchange rows and columns.
PT = \begin{pmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{pmatrix}
Step 2: Add P and Pᵀ
P + PT = \begin{pmatrix} \cos \alpha + \cos \alpha & \sin \alpha – \sin \alpha \\ -\sin \alpha + \sin \alpha & \cos \alpha + \cos \alpha \end{pmatrix}
= \begin{pmatrix} 2\cos \alpha & 0 \\ 0 & 2\cos \alpha \end{pmatrix}
Step 3: Compare with Identity Matrix
Identity matrix:
I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}
So, 2\cos α = 1 ⇒ \cos α = 1/2
Step 4: Find α
\cos α = 1/2 ⇒ α = π/3
This value lies in the given range 0 ≤ α ≤ 2π.
Correct Answer
✅ Option (B) π/3
Explanation of All Options
Option (A): π/2
❌ \cos(π/2) = 0 ⇒ P + PT = 0 ≠ I
Option (B): π/3
✅ \cos(π/3) = 1/2 ⇒ P + PT = I
Option (C): 3π/2
❌ Not in the given range. Also, \cos(3π/2) = 0
Option (D): 0
❌ \cos(0) = 1 ⇒ P + PT = 2I ≠ I
Final Conclusion
The matrix simplifies beautifully due to symmetry. Only Option (B) satisfies both the equation and the given range.
α = π/3