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  • Explain Solution for RD Sharma Maths Class 12 Chapter 6 Adjoint and inverse of a Matrix Exercise very short answer type Question 23 maths textbook solution

Explain Solution for RD Sharma Maths Class 12 Chapter 6 Adjoint and inverse of a Matrix Exercise very short answer type Question 23 maths textbook solution

Answers (1)

VSQ : 23

Answer : $$ \left[\begin{array}{cc} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right]

Given : $$ \left[\begin{array}{cc} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]

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

Hint : Find the cofactor and then transpose it 

Solution : $$ \left[\begin{array}{cc} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]

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

$$ \begin{aligned} &c_{11}=\cos \theta, c_{12}=\sin \theta, c_{21}=-\sin \theta, c_{22}=\cos \theta \\ &|A|=\cos ^{2} \theta-\left(-\sin ^{2} \theta\right) \\ &=\cos ^{2} \theta+\sin ^{2} \theta \quad \end{aligned}                \left[\because \cos ^{2} \theta+\sin ^{2} \theta=1\right] \\

=1

A^{-1}=\frac{A d j|A|}{|A|}=\frac{1}{1}\left[\begin{array}{cc} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]

Now transpose it, we get

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

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