Question

If $A=\left[ \begin{matrix} \cos \theta & -\sin \theta \\\ \sin \theta & \cos \theta \\\ \end{matrix} \right]$, 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

Hint: First try to find ${{A}^{-1}}$ using formula if $A=\left[ \begin{matrix} a & b \\\ c & d \\\ \end{matrix} \right]$ then $adj(A)=\left[ \begin{matrix} d & -b \\\ -c & a \\\ \end{matrix} \right]$ and
${{A}^{-1}}=\dfrac{1}{\left( ad-bc \right)}adj(A)$ after that find ${{A}^{-2}}$ , then ${{A}^{-3}}$. Then find any pattern they are following after generalize the value of ${{A}^{-50}}$ using identities like $\cos \left( 4\pi +\theta \right)=\cos \theta$ and
$\sin \left( 4\pi +\theta \right)=\sin \theta$ to get the desired result.

Complete step-by-step answer:
In the equation we are given,

\cos \theta & -\sin \theta \\\ \sin \theta & \cos \theta \\\ \end{matrix} \right]$$ Now we will first try to find ${{A}^{-1}}$ . We will use the formula as, ${{A}^{-1}}=\dfrac{1}{\left| A \right|}adj(A)$ Here $|A|$ , means determinant value of A which is done by formula, if $A=\left[ \begin{matrix} a & b \\\ c & d \\\ \end{matrix} \right]$, then $$\left| A \right|=ad-bc$$ So, if $A=\left[ \begin{matrix} \cos \theta & -\sin \theta \\\ \sin \theta & \cos \theta \\\ \end{matrix} \right]$ then $\begin{aligned} & \left| A \right|={{\cos }^{2}}\theta -\left( -{{\sin }^{2}}\theta \right) \\\ & \left| A \right|=1..........(i) \\\ \end{aligned}$ Now adj(A) means adjoint matrix of A, by using the formula which is, if $A=\left[ \begin{matrix} a & b \\\ c & d \\\ \end{matrix} \right]$, then adj. $A=\left[ \begin{matrix} d & -b \\\ -c & a \\\ \end{matrix} \right]$ So, if $A=\left[ \begin{matrix} \cos \theta & -\sin \theta \\\ \sin \theta & \cos \theta \\\ \end{matrix} \right]$ Then, $adj(A)=\left[ \begin{matrix} \cos \theta & \sin \theta \\\ -\sin \theta & \cos \theta \\\ \end{matrix} \right].........(ii)$ Substituting the values from equation (i) and (ii), we get $${{A}^{-1}}=\dfrac{1}{1}\left[ \begin{matrix} \cos \theta & \sin \theta \\\ -\sin \theta & \cos \theta \\\ \end{matrix} \right]$$ $${{A}^{-1}}=\left[ \begin{matrix} \cos \theta & \sin \theta \\\ -\sin \theta & \cos \theta \\\ \end{matrix} \right]$$ Now we will find ${{A}^{-2}}$ which is ${{A}^{-1}}\times {{A}^{-1}}$ which can be found by using formula, If $${{A}^{-1}}=\left[ \begin{matrix} a & b \\\ c & d \\\ \end{matrix} \right]$$ then, $${{A}^{-2}}=\left[ \begin{matrix} a & b \\\ c & d \\\ \end{matrix} \right]\times \left[ \begin{matrix} a & b \\\ c & d \\\ \end{matrix} \right]=\left[ \begin{matrix} {{a}^{2}}+bc & ab+bd \\\ ac+cd & bc+{{d}^{2}} \\\ \end{matrix} \right]$$ So, if $${{A}^{-1}}=\left[ \begin{matrix} \cos \theta & \sin \theta \\\ -\sin \theta & \cos \theta \\\ \end{matrix} \right]$$ then $$\begin{aligned} & {{A}^{-2}}=\left[ \begin{matrix} \cos \theta & \sin \theta \\\ -\sin \theta & \cos \theta \\\ \end{matrix} \right]\times \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 & 2cos\theta \sin \theta \\\ -2\sin \theta \cos \theta & co{{s}^{2}}\theta -{{\sin }^{2}}\theta \\\ \end{matrix} \right] \\\ \end{aligned}$$ Now by applying identities, we can write, ${{\cos }^{2}}\theta -{{\sin }^{2}}\theta =\cos 2\theta ,2\sin \theta \cos \theta =\sin 2\theta $ , so we can write as, ${{A}^{-2}}=\left[ \begin{matrix} \cos 2\theta & \sin 2\theta \\\ -\sin 2\theta & \cos 2\theta \\\ \end{matrix} \right]$ Now to find ${{A}^{-3}}$ we will write as, $$\begin{aligned} & {{A}^{-3}}={{A}^{-2}}\times {{A}^{-1}} \\\ & {{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] \\\ & {{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 \\\ -\sin 2\theta \cos \theta -\cos 2\theta \sin \theta & -\sin 2\theta \sin \theta +\cos 2\theta \cos \theta \\\ \end{matrix} \right] \\\ & {{A}^{-3}}=\left[ \begin{matrix} \cos 2\theta \cos \theta -\sin 2\theta \sin \theta & \sin 2\theta \cos \theta +\cos 2\theta \sin \theta \\\ -(\sin 2\theta \cos \theta +\cos 2\theta \sin \theta ) & \cos 2\theta \cos \theta -\sin 2\theta \sin \theta \\\ \end{matrix} \right] \\\ \end{aligned}$$ Now we will use the identities, $\sin (A+B)=\sin A\cos B+\cos A\sin B,\cos (A+B)=\cos A\cos B-\sin A\sin B$, so the above expression can be written as, $$\begin{aligned} & {{A}^{-3}}=\left[ \begin{matrix} \cos (2\theta +\theta ) & \sin (2\theta +\theta ) \\\ -\sin (2\theta +\theta ) & \cos (2\theta +\theta ) \\\ \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}$$ Like this following the pattern we can say that, $${{A}^{-n}}=\left[ \begin{matrix} \cos n\theta & \sin n\theta \\\ -\sin n\theta & \cos n\theta \\\ \end{matrix} \right]$$ So, the value of n = 50, we get, $${{A}^{-50}}=\left[ \begin{matrix} \cos 50\theta & \sin 50\theta \\\ -\sin 50\theta & \cos 50\theta \\\ \end{matrix} \right]$$ Now for the value of $$\theta =\dfrac{\pi }{12}$$ the value of $${{A}^{-50}}\text{ }$$ $$\begin{aligned} & {{A}^{-50}}\text{ = }\left[ \begin{matrix} \cos \left( \dfrac{50}{12}\pi \right) & \sin \left( \dfrac{50}{12}\pi \right) \\\ -\sin \left( \dfrac{50\pi }{12} \right) & \cos \left( \dfrac{50}{12}\pi \right) \\\ \end{matrix} \right] \\\ & \Rightarrow {{A}^{-50}}\text{ = }\left[ \begin{matrix} \cos \left( \dfrac{25}{6}\pi \right) & \sin \left( \dfrac{25}{6}\pi \right) \\\ -\sin \left( \dfrac{25}{6}\pi \right) & \cos \left( \dfrac{25}{6}\pi \right) \\\ \end{matrix} \right] \\\ \end{aligned}$$ This can be written as, $${{A}^{-50}}\text{ = }\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]$$ Now we will use the identity which is, $$\sin \left( 4\pi +\theta \right)=\sin \theta ,\cos \left( 4\pi +\theta \right)=\cos \theta $$, so above equation can be written as, $${{A}^{-50}}=\left[ \begin{matrix} \cos \dfrac{\pi }{6} & \sin \dfrac{\pi }{6} \\\ -\sin \dfrac{\pi }{6} & \cos \dfrac{\pi }{6} \\\ \end{matrix} \right]$$ Now we know that $\cos \dfrac{\pi }{6}=\dfrac{\sqrt{3}}{2}$ and $$\sin \dfrac{\pi }{6}=\dfrac{1}{2}$$, substituting the values, we get $${{A}^{-50}}=\left[ \begin{matrix} \dfrac{\sqrt{3}}{2} & \dfrac{1}{2} \\\ \dfrac{-1}{2} & \dfrac{\sqrt{3}}{2} \\\ \end{matrix} \right]$$ So the correct option is (a). Note: At first try to understand what is and how to find the inverse of matrix A. Then understand the product of two matrices as the sum contains lots of simplification. Be careful about them to avoid any errors or mistakes. Students often make mistake when getting ${{A}^{-3}}$, they use $${{A}^{-2}}=\left[ \begin{matrix} {{\cos }^{2}}\theta -{{\sin }^{2}}\theta & 2cos\theta \sin \theta \\\ -2\sin \theta \cos \theta & co{{s}^{2}}\theta -{{\sin }^{2}}\theta \\\ \end{matrix} \right]$$, directly without simplifying this, in this way the solution can get complicated.