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Explain solution rd sharma class 12 chapter adjoint and inverse of a matrix exercise multiple choice question question 27

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\text { option (b) } A=\cos 2 \theta \& b=\sin 2 \theta

\text { Given: }\left[\begin{array}{cc} 1 & -\tan \theta \\ \tan \theta & 1 \end{array}\right]\left[\begin{array}{cc} 1 & \tan \theta \\ -\tan \theta & 1 \end{array}\right]^{-1}=\left[\begin{array}{cc} a & -b \\ b & a \end{array}\right]

Hint: use simple matrix multiplication.

Solution :

\begin{aligned} &\text { Inverse of }\left[\begin{array}{cc} 1 & \tan \theta \\ -\tan \theta & 1 \end{array}\right]^{-1}=\frac{1}{1+\tan ^{2} \theta}\left[\begin{array}{cc} 1 & -\tan \theta \\ \tan \theta & 1 \end{array}\right]\\ &\text { From the given part }\\ &\left[\begin{array}{cc} 1 & -\tan \theta \\ \tan \theta & 1 \end{array}\right] \frac{1}{1+\tan ^{2} \theta}\left[\begin{array}{cc} 1 & -\tan \theta \\ \tan \theta & 1 \end{array}\right]=\left[\begin{array}{cc} a & -b \\ b & a \end{array}\right]\\ &\frac{1}{1+\tan ^{2} \theta}\left[\begin{array}{cc} 1-\tan ^{2} \theta & -2 \tan \theta \\ 2 \tan \theta & 1-\tan ^{2} \theta \end{array}\right]=\left[\begin{array}{cc} a & -b \\ b & a \end{array}\right] \end{aligned}

\begin{aligned} &\left[\begin{array}{cc} \frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta} & \frac{-2 \tan \theta}{1+\tan ^{2} \theta} \\ \frac{2 \tan \theta}{1+\tan ^{2} \theta} & \frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta} \end{array}\right]=\left[\begin{array}{cc} a & -b \\ b & a \end{array}\right]\\ &\left[\begin{array}{cc} \cos 2 \theta & -\sin 2 \theta \\ \sin 2 \theta & \cos 2 \theta \end{array}\right]=\left[\begin{array}{cc} a & -b \\ b & a \end{array}\right]\\ &\text { On comparing, we get }\\ &A=\cos 2 \theta \ \& \ b=\sin 2 \theta \end{aligned}

Hence, option (b) is correct

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