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  • Please Solve R D Sharma Class 12 Chapter 4 Algebra of Matrices Exercise MCQs Question 6 Maths Textbook Solution

Please Solve R.D.Sharma Class 12 Chapter 4 Algebra of Matrices Exercise MCQs Question 6 Maths Textbook Solution.

Answers (1)

Answer: The correct option is (D), k = 7

Given: \left[\begin{array}{cc} \cos \frac{2 \pi}{7} & -\sin \frac{2 \pi}{7} \\ \sin \frac{2 \pi}{7} & \cos \frac{2 \pi}{7} \end{array}\right]^{\hat{k}}=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]

Solution:

Let A=\left[\begin{array}{cc} \cos \frac{2 \pi}{7} & -\sin \frac{2 \pi}{7} \\ \sin \frac{2 \pi}{7} & \cos \frac{2 \pi}{7} \end{array}\right]

Then \begin{aligned} |\mathrm{A}| &=\left(\cos \frac{2 \pi}{7}\right)\left(\cos \frac{2 \pi}{7}\right)-\left(\sin \frac{2 \pi}{7}\right)\left(-\sin \frac{2 \pi}{7}\right) \\ &=\left(\cos ^{2} \frac{2 \pi}{7}\right)+\left(\sin ^{2} \frac{2 \pi}{7}\right) \\ &=1 \end{aligned}

I=1

I^{k}=1                                                                                                                            {k cab be anything}

\text { Let } \theta=\frac{2 \pi}{7}

\begin{aligned} \mathrm{A}^{2} &=\left[\begin{array}{cc} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right] \times\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 \theta \cos \theta-\sin \theta \cos \theta \\ \sin \theta \cos \theta+\sin \theta \cos \theta & \cos ^{2} \theta-\sin ^{2} \theta \end{array}\right] \\ &=\left[\begin{array}{cc} \cos ^{2} \theta-\sin ^{2} \theta & -2 \sin \theta \cos \theta \\ 2 \sin \theta \cos \theta & \cos ^{2} \theta-\sin ^{2} \theta \end{array}\right] \\ &=\left[\begin{array}{cc} \cos 2 \theta & -\sin 2 \theta \\ \sin 2 \theta & \cos 2 \theta \end{array}\right] \quad\left[\because \cos ^{2} \theta-\sin ^{2} \theta=\cos 2 \theta, 2 \sin \theta \cos \theta=\sin 2 \theta\right] \end{aligned}

Now,\begin{aligned} \mathrm{A}^{4} &=\left[\begin{array}{cc} \cos 2 \theta & -\sin 2 \theta \\ \sin 2 \theta & \cos 2 \theta \end{array}\right] \times\left[\begin{array}{cc} \cos 2 \theta & -\sin 2 \theta \\ \sin 2 \theta & \cos 2 \theta \end{array}\right] \\ &=\left[\begin{array}{cc} \cos 4 \theta & -\sin 4 \theta \\ \sin 4 \theta & \cos 4 \theta \end{array}\right] \end{aligned}

Similarly,A^{7}=\left[\begin{array}{cc} \cos 7 \theta & -\sin 7 \theta \\ \sin 7 \theta & \cos 7 \theta \end{array}\right]

Hence, \theta=\frac{2 \pi}{7}

Now, multiplying \cos \Thetaand \sin \Theta to LHS and RHS,

Then,

\begin{aligned} &\cos 7 \theta=\cos 2 \pi=1 \\ &\sin 7 \theta=\sin 2 \pi=0 \\ &{\left[\begin{array}{cc} \cos 7 \theta & -\sin 7 \theta \\ \sin 7 \theta & \cos 7 \theta \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]} \\ &\mathrm{A}^{7}=1 \end{aligned}

Hence,k=7

So, the option (D) is correct.

 

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