Question

If we are given a matrix as $A=\left[ \begin{matrix} \cos \theta & -\sin \theta \\\ \sin \theta & \cos \theta \\\ \end{matrix} \right]$, then the matrix ${{A}^{-50}}$ when $\theta =\dfrac{\pi }{12}$, is equal to
A. $\left[ \begin{matrix} \dfrac{\sqrt{3}}{2} & \dfrac{1}{2} \\\ -\dfrac{1}{2} & \dfrac{\sqrt{3}}{2} \\\ \end{matrix} \right]$
B. $\left[ \begin{matrix} \dfrac{1}{2} & \dfrac{\sqrt{3}}{2} \\\ -\dfrac{\sqrt{3}}{2} & \dfrac{1}{2} \\\ \end{matrix} \right]$
C. $\left[ \begin{matrix} \dfrac{1}{2} & -\dfrac{\sqrt{3}}{2} \\\ \dfrac{\sqrt{3}}{2} & \dfrac{1}{2} \\\ \end{matrix} \right]$
D. $\left[ \begin{matrix} \dfrac{\sqrt{3}}{2} & -\dfrac{1}{2} \\\ \dfrac{1}{2} & \dfrac{\sqrt{3}}{2} \\\ \end{matrix} \right]$

Answer 1

First we will calculate the inverse of the given matrix $A$ by calculating the $\left| A \right|$ and $adj\left( A \right)$. Now we will calculate the values of ${{A}^{-2}}$, ${{A}^{-3}}$,… by using the value of inverse of matrix $A$ i.e. ${{A}^{-1}}$. From the values of ${{A}^{-2}}$, ${{A}^{-3}}$,… we will write the value of ${{A}^{-n}}$. From the value of ${{A}^{-n}}$ we will calculate the required value ${{A}^{-50}}$.

Complete step-by-step solution
Given that, $A=\left[ \begin{matrix} \cos \theta & -\sin \theta \\\ \sin \theta & \cos \theta \\\ \end{matrix} \right]$
Determinant of the matrix $A$ is given by
$\begin{aligned} & \left| A \right|=\left| \begin{matrix} \cos \theta & -\sin \theta \\\ \sin \theta & \cos \theta \\\ \end{matrix} \right| \\\ & \Rightarrow \left| A \right|=\cos \theta \left( \cos \theta \right)-\left( -\sin \theta \right)\left( \sin \theta \right) \\\ & \Rightarrow \left| A \right|={{\cos }^{2}}\theta +{{\sin }^{2}}\theta \\\ \end{aligned}$
We have the trigonometric identity ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$.
$\therefore \left| A \right|=1$
We know that adjoint matrix of $x=\left[ \begin{matrix} a & c \\\ b & d \\\ \end{matrix} \right]$ can be written as $adj\left( x \right)=\left[ \begin{matrix} d & -c \\\ -b & a \\\ \end{matrix} \right]$.
$\therefore adj\left( A \right)=\left[ \begin{matrix} \cos \theta & \sin \theta \\\ -\sin \theta & \cos \theta \\\ \end{matrix} \right]$
Now the inverse of the matrix $A$ is given by
$\begin{aligned} & {{A}^{-1}}=\dfrac{1}{\left| A \right|}adj\left( A \right) \\\ & \Rightarrow {{A}^{-1}}=\dfrac{1}{1}\left[ \begin{matrix} \cos \theta & \sin \theta \\\ -\sin \theta & \cos \theta \\\ \end{matrix} \right] \\\ & \Rightarrow {{A}^{-1}}=\left[ \begin{matrix} \cos \theta & \sin \theta \\\ -\sin \theta & \cos \theta \\\ \end{matrix} \right] \\\ \end{aligned}$
Now the value of ${{A}^{-2}}$ can be calculated as
$\begin{aligned} & {{A}^{-2}}={{\left( {{A}^{-1}} \right)}^{2}} \\\ & \Rightarrow {{A}^{-2}}={{A}^{-1}}.{{A}^{-1}} \\\ \end{aligned}$
Substituting ${{A}^{-1}}=\left[ \begin{matrix} \cos \theta & \sin \theta \\\ -\sin \theta & \cos \theta \\\ \end{matrix} \right]$ in the above equation, then we will get
$\begin{aligned} & {{A}^{-2}}=\left[ \begin{matrix} \cos \theta & \sin \theta \\\ -\sin \theta & \cos \theta \\\ \end{matrix} \right].\left[ \begin{matrix} \cos \theta & \sin \theta \\\ -\sin \theta & \cos \theta \\\ \end{matrix} \right] \\\ & \Rightarrow {{A}^{-2}}=\left[ \begin{matrix} \cos \theta \left( \cos \theta \right)+\sin \theta \left( -\sin \theta \right) & \cos \theta \left( \sin \theta \right)+\sin \theta \left( \cos \theta \right) \\\ -\sin \theta \left( \cos \theta \right)+\cos \theta \left( -\sin \theta \right) & -\sin \theta \left( \sin \theta \right)+\cos \theta \left( \cos \theta \right) \\\ \end{matrix} \right] \\\ & \Rightarrow {{A}^{-2}}=\left[ \begin{matrix} {{\cos }^{2}}\theta -{{\sin }^{2}}\theta & 2\sin \theta \cos \theta \\\ -2\sin \theta \cos \theta & {{\cos }^{2}}\theta -{{\sin }^{2}}\theta \\\ \end{matrix} \right] \\\ \end{aligned}$
We have the trigonometric formulas as ${{\cos }^{2}}\theta -{{\sin }^{2}}\theta =\cos 2\theta$ and $2\sin \theta \cos \theta =\sin 2\theta$, now the value of ${{A}^{-2}}$ will be
${{A}^{-2}}=\left[ \begin{matrix} \cos 2\theta & \sin 2\theta \\\ -\sin 2\theta & \cos 2\theta \\\ \end{matrix} \right]....\left( \text{i} \right)$
Now we will calculate the value of ${{A}^{-3}}$ by multiplying ${{A}^{-2}}$ with ${{A}^{-1}}$, then we will get
$\begin{aligned} & {{A}^{-3}}={{A}^{-2}}.{{A}^{-1}} \\\ & \Rightarrow {{A}^{-3}}=\left[ \begin{matrix} \cos 2\theta & \sin 2\theta \\\ -\sin 2\theta & \cos 2\theta \\\ \end{matrix} \right].\left[ \begin{matrix} \cos \theta & \sin \theta \\\ -\sin \theta & \cos \theta \\\ \end{matrix} \right] \\\ & \Rightarrow {{A}^{-3}}=\left[ \begin{matrix} \cos 2\theta \cos \theta -\sin 2\theta .\sin \theta & \cos 2\theta .\sin \theta +\sin 2\theta .\cos \theta \\\ -\left( \sin 2\theta .\cos \theta +\cos 2\theta .\sin \theta \right) & \cos 2\theta .\cos \theta -\sin 2\theta .\sin \theta \\\ \end{matrix} \right] \\\ \end{aligned}$
We have trigonometric formulas as $\sin \left( A+B \right)=\sin A.\cos B+\cos A.\sin B$ and $\cos \left( A+B \right)=\cos A.\cos B-\sin A.\sin B$, then we will get
$\begin{aligned} & \Rightarrow {{A}^{-3}}=\left[ \begin{matrix} \cos \left( 2\theta +\theta \right) & \sin \left( 2\theta +\theta \right) \\\ -\sin \left( 2\theta +\theta \right) & \cos \left( 2\theta +\theta \right) \\\ \end{matrix} \right] \\\ & \Rightarrow {{A}^{-3}}=\left[ \begin{matrix} \cos 3\theta & \sin 3\theta \\\ -\sin 3\theta & \cos 3\theta \\\ \end{matrix} \right] \\\ \end{aligned}$
We have the values of ${{A}^{-1}}$, ${{A}^{-2}}$, ${{A}^{-3}}$ as
${{A}^{-1}}=\left[ \begin{matrix} \cos \theta & \sin \theta \\\ -\sin \theta & \cos \theta \\\ \end{matrix} \right]$
${{A}^{-2}}=\left[ \begin{matrix} \cos 2\theta & \sin 2\theta \\\ -\sin 2\theta & \cos 2\theta \\\ \end{matrix} \right]$
${{A}^{-3}}=\left[ \begin{matrix} \cos 3\theta & \sin 3\theta \\\ -\sin 3\theta & \cos 3\theta \\\ \end{matrix} \right]$
Similarly, we can write that ${{A}^{-n}}=\left[ \begin{matrix} \cos n\theta & \sin n\theta \\\ -\sin n\theta & \cos n\theta \\\ \end{matrix} \right]$
Hence the value of ${{A}^{-50}}$ is
${{A}^{-50}}=\left[ \begin{matrix} \cos 50\theta & \sin 50\theta \\\ -\sin 50\theta & \cos 50\theta \\\ \end{matrix} \right]$
Given that $\theta =\dfrac{\pi }{12}$, then the value of ${{A}^{-50}}$ will be
$\begin{aligned} & {{A}^{-50}}=\left[ \begin{matrix} \cos 50\times \dfrac{\pi }{12} & \sin 50\times \dfrac{\pi }{12} \\\ -\sin 50\times \dfrac{\pi }{12} & \cos 50\times \dfrac{\pi }{12} \\\ \end{matrix} \right] \\\ & \Rightarrow {{A}^{-50}}=\left[ \begin{matrix} \cos \left( 4\pi +\dfrac{\pi }{6} \right) & \sin \left( 4\pi +\dfrac{\pi }{6} \right) \\\ -\sin \left( 4\pi +\dfrac{\pi }{6} \right) & \cos \left( 4\pi +\dfrac{\pi }{6} \right) \\\ \end{matrix} \right] \\\ \end{aligned}$
We know that $\cos \left( n\pi +\theta \right)=\cos \theta$ and $\sin \left( n\pi +\theta \right)=\sin \theta$, then we will get
${{A}^{-50}}=\left[ \begin{matrix} \dfrac{\sqrt{3}}{2} & \dfrac{1}{2} \\\ -\dfrac{1}{2} & \dfrac{\sqrt{3}}{2} \\\ \end{matrix} \right]$
$\therefore$ Option – A is the correct option.

Note: For this kind of problem students should know how to multiply matrices. Because the whole problem completely depends on the values of ${{A}^{-1}}$, ${{A}^{-2}}$, ${{A}^{-3}}$ and it is also important to know how to calculate the values trigonometric ratios for a given angle. If you do everything correct up to the calculation of ${{A}^{-n}}$ and make mistakes in the calculation of trigonometric ratios at given $\theta$ you won’t get the correct answer.