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Matrices & Determinants — JEE Maths practice question

JEE Maths question with a full step-by-step solution.

Question
For θ=3π5\theta = \dfrac{3\pi}{5}, let B=[bij]B = [b_{ij}] be a square matrix of order 2 such that bij={cos⁡θ,i=jcos⁡(jπ2+θ),i>jsin⁡(jπ2−θ),i<jb_{ij} = \begin{cases} \cos\theta, & i = j \\ \cos\left(\frac{j\pi}{2}+\theta\right), & i > j \\ \sin\left(\frac{j\pi}{2}-\theta\right), & i < j \end{cases}. Then the trace of B5B^5 is:
A−1-1
B−2-2correct
C−3-3
D−4-4
Solution
Step 1: Determine the entries of BB
b11=b22=cos⁡θb_{11} = b_{22} = \cos\theta
b12 (i<j):sin⁡(2π2−θ)=sin⁡(π−θ)=sin⁡θb_{12}\ (i<j): \sin\left(\frac{2\pi}{2}-\theta\right) = \sin(\pi - \theta) = \sin\theta
b21 (i>j):cos⁡(π2+θ)=−sin⁡θb_{21}\ (i>j): \cos\left(\frac{\pi}{2}+\theta\right) = -\sin\theta
B=(cos⁡θsin⁡θ−sin⁡θcos⁡θ)B = \begin{pmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{pmatrix}
 Step 2: Use the rotation-matrix power formula
B5=(cos⁡5θsin⁡5θ−sin⁡5θcos⁡5θ)B^5 = \begin{pmatrix} \cos 5\theta & \sin 5\theta \\ -\sin 5\theta & \cos 5\theta \end{pmatrix}
 Step 3: Evaluate the trace
tr(B5)=2cos⁡5θ=2cos⁡(3π)=2(−1)=−2\text{tr}(B^5) = 2\cos 5\theta = 2\cos(3\pi) = 2(-1) = -2
 Answer: (2)
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