$If A = [\begin{array}{cc} \cos θ & - \sin θ \\ \sin θ & \cos θ \end{array}], then the matrix A^{- 50} when θ = \frac{π}{12}, is equal to$

Moderate

A

$[\begin{matrix} \frac{1}{2} & \frac{\sqrt{3}}{2} \\ - \frac{\sqrt{3}}{2} & \frac{1}{2} \end{matrix}]$
B

$[\begin{matrix} \frac{\sqrt{3}}{2} & - \frac{1}{2} \\ \frac{1}{2} & \frac{\sqrt{3}}{2} \end{matrix}]$
C

$[\begin{matrix} \frac{\sqrt{3}}{2} & \frac{1}{2} \\ - \frac{1}{2} & \frac{\sqrt{3}}{2} \end{matrix}]$
D

$[\begin{matrix} \frac{1}{2} & - \frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & \frac{1}{2} \end{matrix}]$

Solution

$We \underline{\underline{have}} A = [\begin{array}{cc} \cos θ & - \sin θ \\ \sin θ & \cos θ \end{array}]$
$∴ | A | = \cos^{2} θ + \sin^{2} θ = 1$
$And adj A = [\begin{array}{cc} \cos θ & \sin θ \\ - \sin θ & \cos θ \end{array}]$
$(∵ If A = [\begin{array}{ll} a b \\ c d \end{array}], then adj A = [\begin{array}{cc} d & - b \\ - c & a \end{array}])$
$\Rightarrow A^{- 1} = [\begin{matrix} \cos θ & \sin θ \\ - \sin θ & \cos θ \end{matrix}] (∵ A^{- 1} = \frac{a d j A}{| A |})$
$Note that, A^{- 50} = {(A^{- 1})}^{50}$
$Now, A^{- 2} = (A^{- 1}) (A^{- 1})$
$\begin{array}{l} \Rightarrow A^{- 2} = [\begin{array}{cc} \cos θ & \sin θ \\ - \sin θ & \cos θ \end{array}] [\begin{array}{cc} \cos θ & \sin θ \\ - \sin θ & \cos θ \end{array}] \\ = [\begin{array}{cc} \cos^{2} θ - \sin^{2} θ & \cos θ \sin θ + \sin θ \cos θ \\ - \cos θ \sin θ - \cos θ \sin θ & - \sin^{2} θ + \cos^{2} θ \end{array}] \\ = [\begin{array}{cc} \cos 2 θ & \sin 2 θ \\ - \sin 2 θ & \cos 2 θ \end{array}] \end{array}$
$Also, A^{- 3} = (A^{- 2}) (A^{- 1})$
$\begin{array}{l} A^{- 3} = [\begin{array}{cc} \cos 2 θ & \sin 2 θ \\ - \sin 2 θ & \cos 2 θ \end{array}] [\begin{array}{cc} \cos θ & \sin θ \\ - \sin θ & \cos θ \end{array}] \\ = [\begin{array}{cc} \cos 3 θ & \sin 3 θ \\ - \sin 3 θ & \cos 3 θ \end{array}] \end{array}$
$Similarly, A^{- 50} = [\begin{array}{cc} \cos 50 θ & \sin 50 θ \\ - \sin 50 θ & \cos 50 θ \end{array}]$
$\begin{array}{l} = [\begin{array}{cc} \cos \frac{25}{6} π & \sin \frac{25}{6} π \\ - \sin \frac{25}{6} π & \cos \frac{25}{6} π \end{array}] (when θ = \frac{π}{12}) \\ = [\begin{array}{cc} \cos \frac{π}{6} & \sin \frac{π}{6} \\ - \sin \frac{π}{6} & \cos \frac{π}{6} \end{array}] [\begin{array}{ll} ∵ \cos (\frac{25 π}{6}) = \cos (4 π + \frac{π}{6}) = \cos \frac{π}{6} \\ and \sin (\frac{25 π}{6}) = \sin (4 π + \frac{π}{6}) = \sin \frac{π}{6} \end{array}] \\ = [\begin{array}{cc} \frac{\sqrt{3}}{2} & \frac{1}{2} \\ \frac{- 1}{2} & \frac{\sqrt{3}}{2} \end{array}] \end{array}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

SCAN ME