# If $$A = \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta &\cos\theta \end{bmatrix}$$ and $A + A'= I$, then find the value of $$\theta$$.

Toolbox:
• If A_{i,j} be a matrix m*n matrix , then the matrix obtained by interchanging the rows and column of A is called as transpose of A.
• The sum / difference $A(+/-)B$ of two $m$-by-$n$ matrices $A$ and $B$ is calculated entrywise: $(A (+/-) B)_{i,j} = A_{i,j} +/- B_{i,j}$ where 1 $\leq$ i $\leq$ m and 1 $\leq$j $\leq$ n.
• An identity matrix or unit matrix of size n is the n × n square matrix with ones on the main diagonal and zeros elsewhere. An identity matrix of order 2, $I_{2}= \begin{bmatrix} 1 &0 \\ 0&1 \end{bmatrix}$
• If the order of 2 matrices are equal, their corresponding elements are equal, i.e, if $A_{ij}=B_{ij}$, then any element $a_{ij}$ in matrix A is equal to corresponding element $b_{ij}$ in matrix B.
• We can then match the corresponding elements and solve the resulting equations to find the values of the unknown variables.
Step1:
Given:
$A= \begin{bmatrix} cos\theta & -sin\theta \\ sin\theta & cos\theta \end{bmatrix}$
Transpose can be obtained by changing the rows and column.
$A'= \begin{bmatrix} cos\theta & sin\theta \\ -sin\theta & cos\theta \end{bmatrix}$
$A+A'= \begin{bmatrix} cos\theta & -sin\theta \\ sin\theta & cos\theta \end{bmatrix}+ \begin{bmatrix} cos\theta & sin\theta \\ -sin\theta & cos\theta \end{bmatrix}$
$\qquad\;\;\;= \begin{bmatrix} cos\theta + cos\theta&-sin\theta+sin\theta \\ sin\theta-sin\theta & cos\theta+cos\theta \end{bmatrix}$
$\qquad\;\;\;=\begin{bmatrix}2cos\theta &0\\0 & 2cos\theta\end{bmatrix}$
Step2:
Given:
A+A'=I
$\begin{bmatrix}2cos\theta &0\\0 & 2cos\theta\end{bmatrix}=\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}$
The given two matrices are equal ,hence their corresponding elements should be equal.
$\Rightarrow 2cos\theta=1$
$cos\theta=\frac{1}{2}$
$\theta=cos^{-1}(\frac{1}{2})$
$\theta=\Large \frac{\pi}{3}$

edited Dec 24, 2013