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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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Evaluate the determinant $ \begin{bmatrix} cos \theta & -sin \theta \\ sin \theta & cos \theta \end{bmatrix} $

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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 meena.p
 

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