# Evaluate the determinant $\begin{bmatrix} cos \theta & -sin \theta \\ sin \theta & cos \theta \end{bmatrix}$

## 1 Answer

Toolbox:
• The value of the determinant value of a $2 \times 2$ square matrix is $|A|=a_{11} \times a_{22}- a_{21} \times a_{12}$
• $cos \large\frac{\pi}{3}=\frac{1}{2}$
Given $A=\begin{bmatrix} cos \theta & -sin \theta \\ sin \theta & cos \theta \end{bmatrix}$
Let transpose of A be A'
$A'=\begin{bmatrix} cos \theta & sin \theta \\ -sin \theta & cos \theta \end{bmatrix}$
(Transpose of a matrix A can be obtained by interchanging the rows and columns of matrix A)
$A+A'=\begin{bmatrix} cos \theta & -sin \theta \\ sin \theta & cos \theta \end{bmatrix} + \begin{bmatrix} cos \theta & sin \theta \\ -sin \theta & cos \theta \end{bmatrix}$
$= \begin{bmatrix} cos \theta +cos \theta & -sin \theta+\sin \theta \\ sin \theta-\sin \theta & cos \theta+cos \theta \end{bmatrix}$
$=\begin{bmatrix} 2 cos \theta & 0 \\ 0 & 2cos \theta \end{bmatrix}$
Now taking $2 cos \theta$ as the common factor we get
$A+A'=2cos \theta \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$
it is also given $A+A'=I$
Therefore $2cos \theta \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} =\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$
$=> 2 \cos \theta=1$
Therefore $\cos \theta=\frac{1}{2}$
$=> \theta=2 n \pi \pm \large\frac{\pi}{3},n \in z$
answered Apr 8, 2013 by

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