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$ (i) A = \begin{bmatrix} cos\alpha & sin\alpha \\ -sin\alpha & cos\alpha \end{bmatrix}$ then verify that $A'A = I$

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  • 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: $\begin{bmatrix}AB\end{bmatrix}_{i,j} = A_{i,1}B_{1,j} + A_{i,2}B_{2,j} + A_{i,3}B_{3,j} ... A_{i,n}B_{n,j}$
  • An identity matrix or unit matrix of size n is then n*n square matrix with ones on the main diagonal and zeros elsewhere. An identity matrix of order 2, $I_{2}= \begin{bmatrix} 1 &0 \\ 0&1 \end{bmatrix}$
  • $sin^2\alpha+cos^2\alpha$=1
$(i)A= \begin{bmatrix} cos\alpha & sin\alpha \\ -sin\alpha & cos\alpha \end{bmatrix}$
$A' = \begin{bmatrix} cos\alpha & -sin\alpha \\ sin\alpha & cos\alpha \end{bmatrix}$
$AA' = \begin{bmatrix} cos\alpha & sin\alpha \\ -sin\alpha & cos\alpha \end{bmatrix}\begin{bmatrix} cos\alpha & -sin\alpha \\ sin\alpha & cos\alpha \end{bmatrix}$
$\;\;\;= \begin{bmatrix} sin^2\alpha+cos^2\alpha & cos\alpha sin\alpha-sin\alpha cos\alpha \\ sin\alpha cos\alpha-cos\alpha sin\alpha & cos^2\alpha+sin^2\alpha \end{bmatrix}$
We know that $cos^2\alpha+sin^2\alpha=1$ and $sin^2\alpha+cos^2\alpha$=1
$\;\;\;=\begin{bmatrix}1 &0\\0& 1\end{bmatrix}=I$
Hence we get the value as $2\times 2$ matrix which is equal to identity matrix.
Hence LHS=RHS.


answered Mar 5, 2013 by sharmaaparna1
edited Mar 13, 2013 by balaji.thirumalai

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