Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa

Let \theta =\frac{\pi }{5} and A=\begin{bmatrix} \cos \; \theta & \sin \; \theta \\ -\sin \; \theta & \cos \; \theta \end{bmatrix}. If B=A+A^{4}, then det (B):
Option: 1 is one
Option: 2 Lies in (2,3)
Option: 3 is zero.  
Option: 4 Lies in (1,2).

Answers (1)

best_answer

\\A=\left[\begin{array}{cc} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right] \\ \\A^{2}=\left[\begin{array}{cc} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]\left[\begin{array}{cc} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right] \\ \\A^{2}=\left[\begin{array}{cc} \cos 2 \theta & \sin 2 \theta \\ -\sin 2 \theta & \cos 2 \theta \end{array}\right]

\\\mathrm{B}=\mathrm{A}+\mathrm{A}^{4} =\left[\begin{array}{cc} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]+\left[\begin{array}{cc} \cos 4 \theta & \sin 4 \theta \\ -\sin 4 \theta & \cos 4 \theta \end{array}\right] \\ \\\mathrm{B}=\left[\begin{array}{cc} (\cos \theta+\cos 4 \theta) & (\sin \theta+\sin 4 \theta) \\ -(\sin \theta+\sin 4 \theta) & (\cos \theta+\cos 4 \theta) \end{array}\right]

\\|\mathrm{B}|=\left(\cos \theta+\cos ^{4} \theta\right)^{2}+(\sin \theta+\sin 4 \theta)^{2} \\ \\|\mathrm{B}|=2+2 \cos 3 \theta, \text { when } \theta=\frac{\pi}{5} \\\\ |\mathrm{B}|=2+2 \cos \frac{3 \pi}{5}=2(1-\sin 18) \\ \\|\mathrm{B}|=2\left(1-\frac{\sqrt{5}-1}{4}\right)=2\left(\frac{5-\sqrt{5}}{4}\right)=\frac{5-\sqrt{5}}{2}

 

Posted by

himanshu.meshram

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main Result 2024 Live: JEE Mains session 2 results at jeemain.nta.ac.in soon; rank predictor
anam-khan

JEE Main 2024 session 2 final answer key out at jeemain.nta.ac.in; steps to check probable scores
anam-khan

JEE Main Result 2024 Live: Download session 2 final answer key at jeemain.nta.ac.in; rank predictor, cutoff

Latest Question