If $A=\left[ \begin{matrix}
   \cos 2\theta & -\sin 2\theta \\
   \sin 2\theta & \cos 2\theta \\
\end{matrix} \right]$ and $A+{{A}^{T}}=I$ then find the value of $\theta $. \[\]

\[\] Hint: We find ${{A}^{T}}$ by replacing the entices as ${{a}_{ji}}={{a}_{ij}}$. We put $A,{{A}^{T}}$ and the second order identity matrix $I=\left[ \begin{matrix}
   1 & 0 \\
   0 & 1 \\
\end{matrix} \right]$ in the given equation. We add $A,{{A}^{T}}$. We use the condition of equality and equate respective elements from both matrices at left and right hand side. We get an equation in cosine which we solve to the value of $\theta $. \[\]

Complete step by step answer:
We know the transpose of a matrix $A$ with $m$ rows and $n$ columns (order $m\times n$) is matrix denoted as ${{A}^{T}}$ with $n$ rows and $m$ column(order $n\times m$). If the entries of A are denoted as ${{a}_{ij}}$ where $i=1,2,...,m$ is the row position of the entry and $j=1,2,...,n$ is the column position of the entry then entries of ${{A}^{T}}$ are given as ${{a}_{ji}}$ with $j$ as the row position and $i$ as column position. The identity matrix $I$ is the square matrix with say number of rows and columns $n$ (order $n\times n$) then the entries of $I$ is ${{a}_{ij}}=1$ if $i=j$ otherwise ${{a}_{ij}}=0$. Two matrices. The matrices $A$ and $B$ with entries ${{b}_{ij}}$are equal if and only if they are of same order and ${{a}_{ij}}={{b}_{ij}}$. Two matrices $A$ and $B$ can be added when they are of same order and the entries of sum matrix is ${{a}_{ij}}+{{b}_{ij}}$\[\]
The given matrix is square matrix of order $2\times 2$ which is $A=\left[ \begin{matrix}
   \cos 2\theta & -\sin 2\theta \\
   \sin 2\theta & \cos 2\theta \\
\end{matrix} \right]$ . Then its transpose is ${{A}^{T}}=\left[ \begin{matrix}
   \cos 2\theta & \sin 2\theta \\
   -\sin 2\theta & \cos 2\theta \\
\end{matrix} \right]$ . The identity matrix of order $2\times 2$ is $\left[ \begin{matrix}
   1 & 0 \\
   0 & 1 \\
\end{matrix} \right]$. We put these matrices in given equation of matrices which is,
\[\begin{align}
  & A+{{A}^{T}}=I \\
 & \Rightarrow \left[ \begin{matrix}
   \cos 2\theta & -\sin 2\theta \\
   \sin 2\theta & \cos 2\theta \\
\end{matrix} \right]+\left[ \begin{matrix}
   \cos 2\theta & \sin 2\theta \\
   -\sin 2\theta & \cos 2\theta \\
\end{matrix} \right]=\left[ \begin{matrix}
   1 & 0 \\
   0 & 1 \\
\end{matrix} \right] \\
\end{align}\]
We add the respective elements and get
\[\Rightarrow \left[ \begin{matrix}
   2\cos 2\theta & 0 \\
   0 & 2\cos 2\theta \\
\end{matrix} \right]=\left[ \begin{matrix}
   1 & 0 \\
   0 & 1 \\
\end{matrix} \right]\]
We use the condition of equality between matrices and equate respective elements to get the equation
\[\begin{align}
  & 2\cos 2\theta =1 \\
 & \Rightarrow \cos 2\theta =\dfrac{1}{2} \\
 & \Rightarrow \cos 2\theta =\cos \left( \dfrac{\pi }{3} \right) \\
\end{align}\]
We know that the solutions of the equation $\cos x=\cos \alpha $ are $x=2n\pi \pm \alpha $ where $n$ is an integer. Here we have $x=2\theta ,\alpha =\dfrac{\pi }{3}$ in the equation $\cos 2\theta =1$. So we have,
\[\begin{align}
  & \cos 2\theta =\cos \left( \dfrac{\pi }{3} \right) \\
 & \Rightarrow 2\theta =2n\pi \pm \dfrac{\pi }{3} \\
 & \Rightarrow \theta =n\pi \pm \dfrac{\pi }{3} \\
\end{align}\]
So the values of $\theta $ are $n\pi \pm \dfrac{\pi }{3}$ where $n$ is any integer. \[\]

Note:
We note that we could add and compare matrices because they were of the same order. We also note the determinant values of $A,{{A}^{T}}$ are equal. We should care of the confusion between solution of $\cos x=\cos \alpha $ from $\sin x=\sin \alpha $ which has solutions $x=n\pi +{{\left( -1 \right)}^{n}}\alpha.$