Question

IF A= [ COS THETA -SIN THETA]
[ SIN THETA. COS THETA]
THEN FIND THE VALUE OF THETA SATISFYING THE EQUATION A^T+A=I2

Answer 1

$\textbf{Given:}$

$A=\left(\begin{array}{cc}cos\,\theta&-sin\,\theta\\sin\,\theta&cos\,\theta\end{array}\right)$

$\text{and A^T+A=I_2 }$

$\textbf{To find:}$

$\text{The value of \theta satisfying the given equation}$

$\textbf{Solution:}$

$\text{Consider,}$

$A=\left(\begin{array}{cc}cos\,\theta&-sin\,\theta\\sin\,\theta&cos\,\theta\end{array}\right)$

$A^T=\left(\begin{array}{cc}cos\,\theta&sin\,\theta\\-sin\,\theta&cos\,\theta\end{array}\right)$

$\text{Now, A^T+A=I_2 }$

$\implies\left(\begin{array}{cc}cos\,\theta&sin\,\theta\\-sin\,\theta&cos\,\theta\end{array}\right)+\left(\begin{array}{cc}cos\,\theta&-sin\,\theta\\sin\,\theta&cos\,\theta\end{array}\right)=\left(\begin{array}{cc}1&0\\0&1\end{array}\right)$

$\implies\left(\begin{array}{cc}cos\,\theta+cos\,\theta&sin\,\theta-sin\,\theta\\-sin\,\theta+sin\,\theta&cos\,\theta+cos\,\theta\end{array}\right)=\left(\begin{array}{cc}1&0\\0&1\end{array}\right)$

$\implies\left(\begin{array}{cc}2\,cos\,\theta&0\\0&cos\,\theta\end{array}\right)=\left(\begin{array}{cc}1&0\\0&1\end{array}\right)$

$\text{Equating the corresponding elements on bothsides, we get}$

$2\,cos\,\theta=1$

$\implies\,cos\,\theta=\dfrac{1}{2}$

$\implies\bf\,\theta=\dfrac{\pi}{3}$

$\therefore\textbf{The value of \bf\theta is \bf\dfrac{\pi}{3} }$

Find more:

$\text{A square matrix A is said to be}\;\textbf{Symmetric}\;\text{if}\;A=-A^{T}$

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Answer 2

Answer:

Holaaaà!!!!

Step-by-step explanation:

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