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  • Please Solve R.D. Sharma Class 12 Chapter 4 Algebra of Matrices Exercise Multiple choice Questions Question 6 Maths Textbook Solution.

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

Answers (1)

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

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

Ik = 1                                                                                                              {k can be anything}

\begin{aligned} &\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] &\left[\because \cos ^{2} \theta-\sin ^{2} \theta=\cos 2 \theta, 2 \sin \theta \cos \theta=\sin 2 \theta\right] \end{aligned} \end{aligned}

Now,\begin{aligned} 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}

7 \theta=2 \pi

Now, multiplying \cos \Theta and \sin \Thetato LHS and RHS,

Then, \begin{aligned} &\cos 7 \theta=\cos 2 \pi=1 \\ &\sin 7 \theta=\sin 2 \pi=0 \\ &{\left[\begin{array}{ll} \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.

Posted by

infoexpert21

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