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Find the value of $\alpha$ if $A=\begin{bmatrix}\cos\alpha & \sin\alpha\\-\sin\alpha & \cos\alpha\end{bmatrix}$ and $\;A^{-1}=A'$

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  • 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.
  • If A is an m-by-n matrix and B is an n-by-p matrix, then their matrix product AB is the m-by-p matrix whose entries are given by dot product of the corresponding row of A and the corresponding column of B: $\begin{bmatrix}AB\end{bmatrix}_{i,j} = A_{i,1}B_{1,j} + A_{i,2}B_{2,j} + A_{i,3}B_{3,j} ... A_{i,n}B_{n,j}$
Given $A^{-1}=A'$
$A^1=\begin{bmatrix}cos\alpha & -sin\alpha\\sin\alpha & cos\alpha\end{bmatrix}$
Let $A'$ be a matrix.
Hence $A.A^T=\begin{bmatrix}cos\alpha & sin\alpha\\-sin\alpha & cos\alpha\end{bmatrix}\begin{bmatrix}cos\alpha & -sin\alpha\\sin\alpha & cos\alpha\end{bmatrix}$
$\;\;\;\;\;\;\qquad\;\;=\begin{bmatrix}cos^2\alpha+sin^2\alpha & -cos\alpha sin\alpha+cos\alpha sin\alpha\\-sin\alpha cos\alpha+cos\alpha sin\alpha & sin^2\alpha+cos^2\alpha\end{bmatrix}$
$\Rightarrow \begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}$
Hence we get
$\Rightarrow A^{-1}$ exists.
Hence it is true for all real value of $\alpha$
answered Jul 17, 2013 by sharmaaparna1

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