Verified Solution • 2025–26 Session moderate

QJEEM23Ma61b0: JEE Main 2023 Mathematics - Class 12 JEEM23Ma61b0 Solution: If A = 1 2 [ 1 3 − 3 1 ] , then :

If A=12[13−31], then :

Multiple Choice Options

  • A A30−A25=2I\mathrm{A^{30}-A^{25}=2I}
  • B A30+A25−A=I\mathrm{A^{30}+A^{25}-A=I}
  • C A30=A25\mathrm{A^{30}=A^{25}}
  • D A30+A25+A=I\mathrm{A^{30}+A^{25}+A=I}

Step-by-Step Verified Solution

A=12[13−31]

Let θ=π3\theta=\frac{\pi}{3}

A2amp;=[cos⁡θamp;sin⁡θ−sin⁡θamp;cos⁡θ][cos⁡θamp;sin⁡θ−sin⁡θamp;cos⁡θ]amp;=[cos⁡2θamp;sin⁡2θ−sin⁡2θamp;cos⁡2θ]A3amp;=[cos⁡2θamp;sin⁡2θ−sin⁡2θamp;cos⁡2θ][cos⁡θamp;sin⁡θ−sin⁡θamp;cos⁡θ]amp;=[cos⁡3θamp;sin⁡3θ−sin⁡3θamp;cos⁡3θ]\begin{aligned}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] \\\\& =\left[\begin{array}{cc}\cos 2 \theta & \sin 2 \theta \\\\-\sin 2 \theta & \cos 2 \theta\end{array}\right] \\\\A^3 & =\left[\begin{array}{cc}\cos 2 \theta & \sin 2 \theta \\\\-\sin 2 \theta & \cos 2 \theta\end{array}\right]\left[\begin{array}{cc}\cos \theta & \sin \theta \\\\-\sin \theta & \cos \theta\end{array}\right] \\\\& =\left[\begin{array}{cc}\cos 3 \theta & \sin 3 \theta \\\\-\sin 3 \theta & \cos 3 \theta\end{array}\right]\end{aligned}

amp;∴A30=[cos⁡30θamp;sin⁡30θ−sin⁡30θamp;cos⁡30θ]=[1amp;00amp;1]amp;A25=[cos⁡25θamp;sin⁡25θ−sin⁡25θamp;cos⁡25θ]=12[1amp;3−3amp;1]=Aamp;∴A30+A25−A=I\begin{aligned} & \therefore \quad A^{30}=\left[\begin{array}{cc}\cos 30 \theta & \sin 30 \theta \\\\ -\sin 30 \theta & \cos 30 \theta\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] \\\\ & A^{25}=\left[\begin{array}{cc}\cos 25 \theta & \sin 25 \theta \\\\ -\sin 25 \theta & \cos 25 \theta\end{array}\right]=\frac{1}{2}\left[\begin{array}{cc}1 & \sqrt{3} \\ -\sqrt{3} & 1\end{array}\right]=A \\\\ & \therefore \quad A^{30}+A^{25}-A=I\end{aligned}

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring