Question

$\sf{If \: A~ = \: \bigg[ \begin{matrix} \theta& - \sin \theta \\ sin \theta& cos \theta \end{matrix} \bigg],} \: find \: the \: values$
$\sf{}of \: \theta \: satisfying \: the \: equation \:$
${A}^{T} + A \: \: \: I_2.$
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Answer 1

Please refer the attachment.

Answer 2

Appropriate Question :-

If ${\sf A = \bigg[ \begin{matrix} \theta& - \sin \theta \\ \sin \theta& \cos \theta \end{matrix} \bigg] }$

Then , find the value of $\theta$ . Which satisfies the equation ${\sf A^{T} + A = I_{2}}$. Where ${\sf I_2 }$ is the identity matrix .

Solution :-

Before starting the answer , let's recall ;

Transpose of a matrix is given by interchanging row and columns of the matrix and is written as $\bf A^{T}$ . And Identity matrix is basically a diagonal matrix . In addition of matrices with same order , we add the corresponding elements and form a matrix of same order and if it is in a equation we can equate corresponding elements too . Also , identity matrix is given by :-

${\sf \quad \leadsto \quad \bigg[ \begin{matrix} 1& 0 \\ 0& 1 \end{matrix} \bigg] }$

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Now , Consider ;

• ${\sf A = \bigg[ \begin{matrix} \theta& - \sin \theta \\ \sin \theta& \cos \theta \end{matrix} \bigg] }$

Now , By interchanging row and columns we have the transpose of A as follows ;

${\quad \leadsto \quad \sf A^{T} = \bigg[ \begin{matrix} \theta& \sin \theta \\ - \sin \theta& \cos \theta \end{matrix} \bigg] }$

Now , According to the question ;

${\quad \leadsto \quad \sf A^{T} + A = I_{2}}$

Here , the Identity matrix is of 2 × 2 order as it is ${\sf I_2}$

${ : \implies \quad \sf \sf \bigg[ \begin{matrix} \theta& - \sin \theta \\ \sin \theta& \cos \theta \end{matrix} \bigg] + \sf \bigg[ \begin{matrix} \theta& \sin \theta \\ - \sin \theta& \cos \theta \end{matrix} \bigg] = \sf \bigg[ \begin{matrix} 1& 0 \\ 0& 1 \end{matrix} \bigg] }$

${ : \implies \quad \sf \bigg[ \begin{matrix} \theta + \theta& - \sin \theta + \sin \theta \\ \sin \theta - \sin \theta & \cos \theta + \cos \theta \end{matrix} \bigg] = \sf \bigg[ \begin{matrix} 1& 0 \\ 0& 1 \end{matrix} \bigg] }$

${ : \implies \quad \sf \bigg[ \begin{matrix} 2 \theta& - \cancel{\sin \theta} + \cancel{\sin \theta} \\ \cancel{\sin \theta} - \cancel{\sin \theta} & 2 \cos \theta \end{matrix} \bigg] = \sf \bigg[ \begin{matrix} 1& 0 \\ 0& 1 \end{matrix} \bigg] }$

${ : \implies \quad \sf \bigg[ \begin{matrix} 2 \theta& 0\\ 0 & 2 \cos \theta \end{matrix} \bigg] = \sf \bigg[ \begin{matrix} 1& 0 \\ 0& 1 \end{matrix} \bigg] }$

Equating corresponding elements we have ;

${ : \implies \quad \sf 2\theta = 1 \quad \qquad or \quad \qquad 2 \cos \theta = 1 }$

${ : \implies \quad \sf \theta = \dfrac{1}{2} \quad \qquad or \quad \qquad \cos \theta = \dfrac{1}{2} }$

${ : \implies \quad \sf \theta = \dfrac{1}{2} \quad \qquad or \quad \qquad \cos \theta = \cos \bigg( \dfrac{\pi}{3} \bigg) }$

${ : \implies \quad \quad \bf \theta = \dfrac{1}{2} \quad \qquad or \quad \qquad \theta = \dfrac{\pi}{3} }$

But for $\bf \theta = \cfrac{1}{2}$ the whole matrix willn't equal to the required .So writing here $\cos \theta$ as principal solution we may write it as ;

${ : \implies \quad \therefore \quad \bf \theta = 2n \pi \pm \dfrac{\pi}{3} \quad \forall \quad n \: \in \mathbb Z }$

Henceforth , for ${ \bf \theta = 2n \pi \pm \dfrac{\pi}{3} \quad \forall \quad n \: \in \mathbb Z }$ the above equation will be satisfied :D

Used Concepts :-

• $\sf \cos \bigg( \dfrac{\pi}{3}\bigg) = \dfrac{1}{2}$