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Need solution for RD Sharma Maths Class 12 Chapter Algebra of Matrices Excercise 4.4 Question 8

Answers (1)

Answer:A^{T}A=I_{2}

Given:=\begin{bmatrix} \cos \alpha &-\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix}

Hint: I_{2} refers to an identity matrix with two rows and two columns.

Solution:

               A^{T}A=I_{2}

Consider: LHS=A^{T}A

               =\begin{bmatrix} \cos \alpha &-\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix}\begin{bmatrix} \cos \alpha &\sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix}

               =\begin{bmatrix} \cos \alpha &-\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix}\begin{bmatrix} \cos \alpha &\sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix}

               \begin{bmatrix} \cos ^{2 }\alpha +\sin^{2} \alpha &\cos \alpha \sin \alpha -\sin \alpha \cos \alpha \\ \sin \alpha \cos \alpha -\cos \alpha \sin \alpha & \sin ^{2}\alpha +\cos ^{2}\alpha \end{bmatrix}

               = \begin{bmatrix} 1 & 0\\ 0& 1 \end{bmatrix}=RHS

∴LHS=RHS

Hence,A^{T}A=I_{2} is proved.

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