#### Provide solution for RD Sharma maths Class 12 Chapter 5 Determinants Exercise VSQ Question 52 maths textbook solution.

Hint: Here we use basic concept of determinant of matrix

Given: $A=\left[\begin{array}{cc} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right] \text { Find }\left|A^{n}\right|=?$

Solution :

\begin{aligned} &\left|A^{n}\right| \text { let } \mathrm{n}=1\\ &A^{1}=\left[\begin{array}{cc} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]\\ &\rightarrow \text { Let's take } n=2\\ &A^{2}=A \times A=\left[\begin{array}{cc} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right] \times\left[\begin{array}{cc} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]\\ &A^{2}=\left[\begin{array}{ccc} \cos 2 \theta & \sin 2 & \theta \\ -\sin 2 \theta & \cos 2 \theta \end{array}\right] \end{aligned}

$\rightarrow$ Let's take n = 3

\begin{aligned} A^{3} &=A^{2} \times A=\left[\begin{array}{ccc} \cos 2 \theta & \sin 2 \theta \\ -\sin 2 \theta & \cos 2 \theta \end{array}\right] \times\left[\begin{array}{cc} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right] \\ A^{3} &=\left[\begin{array}{cc} \cos 3 \theta & \sin 3 \theta \\ -\sin 3 \theta & \cos 3 \theta \end{array}\right] \end{aligned}

$\rightarrow$ So, we can clearly see that

$A^{n}=\left[\begin{array}{cc} \cos n \theta & \sin n \theta \\ -\sin n \theta & \cos n \theta \end{array}\right]$

$\rightarrow$ So, Let's find $\left|A^{n}\right|$

\begin{aligned} &\left|A^{n}\right|=[\cos n \theta \times \cos n \theta]-[\sin n \theta \times-\sin n \theta] \\ &=\cos ^{2} n \theta+\sin ^{2} n \theta \\ &\left|A^{n}\right|=1 \end{aligned}