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Please Solve RD Sharma Class 12 Chapter 4 Algebra of Matrices Exercise Very Short Asnwer Question 8 Maths Textbook Solution.

Answers (1)

Answer: x=\frac{\pi}{3}

Hint: Here we use basic of transpose matrix

Given: A=\begin{bmatrix} \cos x & \sin x\\ -\sin x & \cos x \end{bmatrix}A+A^{T}=I

Solution:

A+A^{T}=I

\begin{bmatrix} \cos x & \sin x\\ -\sin x & \cos x \end{bmatrix}+\begin{bmatrix} \cos x & -sin x\\ \sin x & \cos x \end{bmatrix}=\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}

\begin{bmatrix} \cos x+\cos x & \sin x-\sin x\\ -\sin x+\sin x & \cos x+\cos x \end{bmatrix}=\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}

\begin{bmatrix} 2\cos x & 0\\ 0&2 \cos x \end{bmatrix}=\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}

So,\cos x=\frac{1}{2}

x=\frac{\pi}{3}

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