Matrices and Determinants 3 Question 3
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4. If $A=\begin{array}{cc}\cos \theta & -\sin \theta \ \sin \theta & \cos \theta\end{array}$, then the matrix $A^{-50}$ when $\theta=\frac{\pi}{12}$, is equal to
======= ####4. If $A=\begin{bmatrix}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta\end{bmatrix}$, then the matrix $A^{-50}$ when $\theta=\frac{\pi}{12}$, is equal to
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(2019 Main, 9 Jan I)
(a) $\begin{bmatrix}\frac{1}{2} & \frac{\sqrt{3}}{2} \\ -\frac{\sqrt{3}}{2} & \frac{1}{2}\end{bmatrix}$
(b) $\begin{bmatrix}\frac{\sqrt{3}}{2} & -\frac{1}{2} \\ \frac{1}{2} & \frac{\sqrt{3}}{2}\end{bmatrix}$
(c) $\begin{bmatrix}\frac{\sqrt{3}}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{\sqrt{3}}{2}\end{bmatrix}$
(d) $\begin{bmatrix}\frac{1}{2} & -\frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & \frac{1}{2}\end{bmatrix}$
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Answer:
Correct Answer: 4. (c)
Solution:
- We have, $A=\begin{bmatrix}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta\end{bmatrix}$
$ \therefore \quad|A|=\cos ^{2} \theta+\sin ^{2} \theta=1 $
and $\quad \operatorname{adj} A=\begin{bmatrix}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{bmatrix}$
$\because.$ If $A=\begin{bmatrix}a & b \\ c & d\end{bmatrix}$, then $\operatorname{adj} A=\begin{bmatrix}d & -b \\ -c & a\end{bmatrix}$
$\Rightarrow \quad A^{-1}=\begin{bmatrix}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{bmatrix} \quad \because A^{-1}=\frac{\operatorname{adj} A}{|A|}$
Note that, $A^{-50}=\left(A^{-1}\right)^{50}$
Now, $\quad A^{-2}=\left(A^{-1}\right)\left(A^{-1}\right)$
$\Rightarrow A^{-2}=\begin{bmatrix}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{bmatrix}$ $\begin{bmatrix}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{bmatrix}$
= $\begin{bmatrix} \cos ^{2} \theta-\sin ^{2} \theta & \cos \theta \sin \theta+\sin \theta \cos \theta\\ -\cos \theta \sin \theta -\cos \theta \sin \theta &-\sin ^{2} \theta+\cos ^{2} \theta \end{bmatrix}$
= $\begin{bmatrix} \cos 2 \theta &\sin 2 \theta\\ -\sin 2 \theta & \cos 2 \theta \end{bmatrix}$
Also, $A^{-3}=\left(A^{-2}\right)\left(A^{-1}\right)$
$\begin{aligned} & A^{-3}=\left[\begin{array}{cc} \cos 2 \theta & \sin 2 \theta \\ -\sin 2 \theta & \cos 2 \theta \end{array}\right]\left[\begin{array}{cc} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right] \\ & =\left[\begin{array}{cc} \cos 3 \theta & \sin 3 \theta \\ -\sin 3 \theta & \cos 3 \theta \end{array}\right] \\ & \text { Similarly, } A^{-50}=\left[\begin{array}{cc} \cos 50 \theta & \sin 50 \theta \\ -\sin 50 \theta & \cos 50 \theta \end{array}\right] \\ & =\left[\begin{array}{cc} \cos \frac{25}{6} \pi & \sin \frac{25}{6} \pi \\ -\sin \frac{25}{6} \pi & \cos \frac{25}{6} \pi \end{array}\right] \\ & \left(\text { when } \theta=\frac{\pi}{12}\right) \\ & =\left[\begin{array}{cc} \cos \frac{\pi}{6} & \sin \frac{\pi}{6} \\ -\sin \frac{\pi}{6} & \cos \frac{\pi}{6} \end{array}\right]\left[\begin{array}{l} \because \cos \left(\frac{25 \pi}{6}\right)=\cos \left(4 \pi+\frac{\pi}{6}\right)=\cos \frac{\pi}{6} \\ \text { and } \sin \left(\frac{25 \pi}{6}\right)=\sin \left(4 \pi+\frac{\pi}{6}\right)=\sin \frac{\pi}{6} \end{array}\right] \\ & =\left[\begin{array}{cc} \frac{\sqrt{3}}{2} & \frac{1}{2} \\ \frac{-1}{2} & \frac{\sqrt{3}}{2} \end{array}\right] \\ & \end{aligned}$
