Matrices & DeterminantsmediumFree

Matrices & Determinants: Let Square Matrix Order Cases Cases Trace

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)
Still stuck on this question?Ask your doubt on WhatsApp
Similar questions
Matrices & Determinants · easy
If AA is a 3×33 \times 3 matrix and det⁡A=5\det A = 5, then det⁡(Adj(Adj A))\det\big(\text{Adj}(\text{Adj}\,A)\big) equals:
Matrices & Determinants · easy
The equation (12213434k)(xyz)=(000)\begin{pmatrix} 1 & 2 & 2 \\ 1 & 3 & 4 \\ 3 & 4 & k \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} has a solution (x,y,z)(x, y, z) besides (0,0,0)(0, 0, 0). The value of kk equals:
Matrices & Determinants · easy
AA is an involutory matrix given by A=(01−14−343−34)A = \begin{pmatrix}0 & 1 & -1\\ 4 & -3 & 4\\ 3 & -3 & 4\end{pmatrix}, then the inverse of A2\dfrac{A}{2} will be:
Matrices & Determinants · easy
A square matrix PP satisfies P2=I−PP^2 = I - P, where II is the identity matrix. If Pn=5I−8PP^n = 5I - 8P, then nn is:

Solve more, learn faster

Sign up free to solve more JEE Maths questions and explore doMath — timed drills, mastery sprints, bookmarks, and chapter-wise progress tracking.

Sign up freeExplore doMath