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Question 119 of 121

Q.Find the inverse of [cos⁡θ−sin⁡θ0sin⁡θcos⁡θ0001]\begin{bmatrix}\cos\theta & -\sin\theta & 0\\ \sin\theta & \cos\theta & 0\\ 0 & 0 & 1\end{bmatrix} by elementary row transformations.

Maharashtra MsbshseMaharashtra HSC (MSBSHSE) Board 2025Subjective· 4mImportance★★★★★
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Augment AA with the identity matrix and reduce the left block to II using elementary row operations.

A=[cos⁡θ−sin⁡θ0sin⁡θcos⁡θ0001]A=\begin{bmatrix}\cos\theta & -\sin\theta & 0\\ \sin\theta & \cos\theta & 0\\ 0 & 0 & 1\end{bmatrix}

Write [A∣I][A\mid I]:

[cos⁡θ−sin⁡θ0100sin⁡θcos⁡θ0010001001]\left[\begin{array}{ccc|ccc}\cos\theta & -\sin\theta & 0 & 1 & 0 & 0\\ \sin\theta & \cos\theta & 0 & 0 & 1 & 0\\ 0 & 0 & 1 & 0 & 0 & 1\end{array}\right]

Apply R1→(cos⁡θ)R1+(sin⁡θ)R2R_1\to(\cos\theta)R_1+(\sin\theta)R_2 and R2→−(sin⁡θ)R1old+(cos⁡θ)R2R_2\to-(\sin\theta)R_1^{\text{old}}+(\cos\theta)R_2 (both using the original R1,R2R_1,R_2):

New R1R_1: (cos⁡2θ+sin⁡2θ, −sin⁡θcos⁡θ+sin⁡θcos⁡θ, 0∣cos⁡θ, sin⁡θ, 0)=(1,0,0∣cos⁡θ,sin⁡θ,0)(\cos^2\theta+\sin^2\theta,\ -\sin\theta\cos\theta+\sin\theta\cos\theta,\ 0 \mid \cos\theta,\ \sin\theta,\ 0) = (1,0,0\mid\cos\theta,\sin\theta,0)

New R2R_2: (−sin⁡θcos⁡θ+sin⁡θcos⁡θ, sin⁡2θ+cos⁡2θ, 0∣−sin⁡θ, cos⁡θ, 0)=(0,1,0∣−sin⁡θ,cos⁡θ,0)(-\sin\theta\cos\theta+\sin\theta\cos\theta,\ \sin^2\theta+\cos^2\theta,\ 0\mid -\sin\theta,\ \cos\theta,\ 0)=(0,1,0\mid-\sin\theta,\cos\theta,0)

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