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# Let $n \ge 2$ be an integer$A = \left[ {\begin{array}{*{20}{c}}{\cos \left( {\dfrac{{2\pi }}{n}} \right)}&{\sin \left( {\dfrac{{2\pi }}{n}} \right)}&0\\{\sin \left( {\dfrac{{2\pi }}{n}} \right)}&{\cos \left( {\dfrac{{2\pi }}{n}} \right)}&0\\0&0&1\end{array}} \right]$ and $I$ is the identity matrix of order 3, then following of which is correctA.${A^n} = I$ and ${A^{n - 1}} \ne I$B.${A^m} \ne I$ for any positive integer $m$ C.$A$ is not invertibleD.${A^n} = 0$ for a positive integer $m$

Last updated date: 13th Jun 2024
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Hint: Here we will first find the square of the given matrix and then we will find its cube. Then we will see that it is forming a certain pattern. We will follow the same pattern to find the matrix raised to given power. We will simplify the matrix using trigonometric identities and find the correct answer.

The given matrix is $A = \left[ {\begin{array}{*{20}{c}}{\cos \left( {\dfrac{{2\pi }}{n}} \right)}&{\sin \left( {\dfrac{{2\pi }}{n}} \right)}&0\\{\sin \left( {\dfrac{{2\pi }}{n}} \right)}&{\cos \left( {\dfrac{{2\pi }}{n}} \right)}&0\\0&0&1\end{array}} \right]$.
Let $\dfrac{{2\pi }}{n} = x$ and we will substitute this value here.
$A = \left[ {\begin{array}{*{20}{c}}{\cos x}&{\sin x}&0\\{ - \sin x}&{\cos x}&0\\0&0&1\end{array}} \right]$
Now, we will find the product of matrices $A \times A$.
We can write $A \times A$ as ${A^2}$.
$A \times A = {A^2} = \left[ {\begin{array}{*{20}{c}}{\cos x}&{\sin x}&0\\{ - \sin x}&{\cos x}&0\\0&0&1\end{array}} \right] \times \left[ {\begin{array}{*{20}{c}}{\cos x}&{\sin x}&0\\{ - \sin x}&{\cos x}&0\\0&0&1\end{array}} \right]$
Now, we will multiply these matrices using the rule of multiplication of matrices. Therefore, we get
$\Rightarrow {A^2} = \left[ {\begin{array}{*{20}{c}}{\cos x \times \cos x + \sin x \times \left( { - \sin x} \right)}&{\cos x \times \sin x + \sin x \times \cos x}&0\\{\left( { - \sin x} \right) \times \cos x + \cos x \times \left( { - \sin x} \right)}&{\left( { - \sin x} \right) \times \sin x + \cos x \times \cos x}&0\\0&0&1\end{array}} \right]$
On further simplifying the terms, we get
$\Rightarrow {A^2} = \left[ {\begin{array}{*{20}{c}}{{{\cos }^2}x - {{\sin }^2}x}&{2 \cdot \cos x \cdot \sin x}&0\\{ - 2 \cdot \cos x \cdot \sin x}&{{{\cos }^2}x - {{\sin }^2}x}&0\\0&0&1\end{array}} \right]$
Now using the trigonometric identities $2\sin x\cos x = \sin 2x$and ${\cos ^2}x - {\sin ^2}x = \cos 2x$ in the above matrix, we get.
$\Rightarrow {A^2} = \left[ {\begin{array}{*{20}{c}}{\cos 2x}&{\sin 2x}&0\\{ - \sin 2x}&{\cos 2x}&0\\0&0&1\end{array}} \right]$
Similarly, we will find the product of matrices ${A^2} \times A$ .
We can write ${A^2} \times A$ as ${A^3}$.
${A^2} \times A = {A^3} = \left[ {\begin{array}{*{20}{c}}{\cos 2x}&{\sin 2x}&0\\{ - \sin 2x}&{\cos 2x}&0\\0&0&1\end{array}} \right] \times \left[ {\begin{array}{*{20}{c}}{\cos x}&{\sin x}&0\\{ - \sin x}&{\cos x}&0\\0&0&1\end{array}} \right]$
Now, we will multiply these matrices using the rule of multiplication of matrices.
$\Rightarrow {A^2} = \left[ {\begin{array}{*{20}{c}}{\cos 2x \times \cos x + \sin 2x \times \left( { - \sin x} \right)}&{\cos 2x \times \sin x + \sin 2x \times \cos x}&0\\{\left( { - \sin 2x} \right) \times \cos x + \cos 2x \times \left( { - \sin x} \right)}&{\left( { - \sin 2x} \right) \times \sin x + \cos 2x \times \cos x}&0\\0&0&1\end{array}} \right]$

Now using the trigonometric identities $\sin A\cos B + \cos A\sin B = \sin \left( {A + B} \right)$ and $\cos A\cos B - \sin A\sin B = \cos \left( {A + B} \right)$ in the above matrix, we get
$\Rightarrow {A^3} = \left[ {\begin{array}{*{20}{c}}{\cos 3x}&{\sin 3x}&0\\{ - \sin 3x}&{\cos 3x}&0\\0&0&1\end{array}} \right]$
We can see that it is following a certain pattern as shown below:
${A^n} = \left[ {\begin{array}{*{20}{c}}{\cos nx}&{\sin nx}&0\\{ - \sin nx}&{\cos nx}&0\\0&0&1\end{array}} \right]$
Now, we will substitute the value $\dfrac{{2\pi }}{n} = x$ in the above matrix. Therefore, we get
$\Rightarrow {A^n} = \left[ {\begin{array}{*{20}{c}}{\cos n \cdot \dfrac{{2\pi }}{n}}&{\sin n \cdot \dfrac{{2\pi }}{n}}&0\\{ - \sin n \cdot \dfrac{{2\pi }}{n}}&{\cos n \cdot \dfrac{{2\pi }}{n}}&0\\0&0&1\end{array}} \right]$
On further multiplying the terms, we get
$\Rightarrow {A^n} = \left[ {\begin{array}{*{20}{c}}{\cos 2\pi }&{\sin 2\pi }&0\\{ - \sin 2\pi }&{\cos 2\pi }&0\\0&0&1\end{array}} \right]$

Now, substituting $\sin 2\pi = 0$ and $\cos 2\pi = 1$ in the above matrix, we get
$\Rightarrow {A^n} = \left[ {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{array}} \right]$
We know that $I = \left[ {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{array}} \right]$
Therefore, we have
$\begin{array}{l} \Rightarrow {A^n} = \left[ {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{array}} \right] = I\\ \Rightarrow {A^n} = I\end{array}$
Also, ${A^{n - 1}} \ne I$
Hence, the correct option is option A.

Note: To solve this question, we need to know the meaning or definition of the trigonometric identities. Trigonometric identities are defined as the equalities which involve the trigonometric functions. They are always true for every value of the occurring variables for which both sides of the equality are defined. We need to remember that all the trigonometric identities are periodic in nature. They repeat their values after a certain interval. These intervals are a multiple of $2\pi$.