Question

For $\left(i\right)A=\left[\begin{array}{cc}cos\alpha & sin\alpha \\ -sin\alpha & cos\alpha \end{array}\right]$, verify that A'A=I.

For (ii)$A=\left[\begin{array}{cc}sin\alpha & cos\alpha \\ -cos\alpha & sin\alpha \end{array}\right]$, verify that A'A=I.

Answer 1

Here, $A=\left[\begin{array}{cc}cos\alpha & sin\alpha \\ -sin\alpha & cos\alpha \end{array}\right]\Rightarrow A^{\prime }={\left[\begin{array}{cc}cos\alpha & sin\alpha \\ -sin\alpha & cos\alpha \end{array}\right]}^{\prime }=\left[\begin{array}{cc}cos\alpha & -sin\alpha \\ sin\alpha & cos\alpha \end{array}\right]\therefore A^{\prime }A=\left[\begin{array}{cc}cos\alpha & -sin\alpha \\ sin\alpha & cos\alpha \end{array}\right]\left[\begin{array}{cc}cos\alpha & sin\alpha \\ -sin\alpha & cos\alpha \end{array}\right]$

$=\left[\begin{array}{cc}\left(cos\alpha \right)\left(cos\alpha \right)+\left(sin\alpha \right)\left(sin\alpha \right)& \left(cos\alpha \right)\left(sin\alpha \right)+(-sin\alpha )\left(cos\alpha \right)\\ \left(sin\alpha \right)\left(cos\alpha \right)+\left(cos\alpha \right)(-sin\alpha )& si{n}^2\alpha +co{s}^2\alpha \end{array}\right]=\left[\begin{array}{cc}co{s}^2\alpha +si{n}^2\alpha & cos\alpha sin\alpha -sin\alpha cos\alpha \\ sin\alpha cos\alpha -cos\alpha sin\alpha & si{n}^2\alpha +co{s}^2\alpha \end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=I.[\because si{n}^2\alpha +co{s}^2\alpha =1]$

Here, $A=\left[\begin{array}{cc}sin\alpha & cos\alpha \\ -cos\alpha & sin\alpha \end{array}\right]\Rightarrow A^{\prime }=\left[\begin{array}{cc}sin\alpha & cos\alpha \\ -cos\alpha & sin\alpha \end{array}\right]=\left[\begin{array}{cc}sin\alpha & -cos\alpha \\ cos\alpha & sin\alpha \end{array}\right]$
$\therefore A^{\prime }A=\left[\begin{array}{cc}sin\alpha & -cos\alpha \\ cos\alpha & sin\alpha \end{array}\right]\left[\begin{array}{cc}sin\alpha & cos\alpha \\ -cos\alpha & sin\alpha \end{array}\right]=\left[\begin{array}{cc}\left(sin\alpha \right)\left(sin\alpha \right)+(-cos\alpha )(-cos\alpha )& \left(sin\alpha \right)\left(cos\alpha \right)+(-cos\alpha )\left(sin\alpha \right)\\ \left(sin\alpha \right)\left(cos\alpha \right)+\left(sin\alpha \right)(-cos\alpha )& \left(cos\alpha \right)\left(cos\alpha \right)+\left(sin\alpha \right)\left(sin\alpha \right)\end{array}\right]=\left[\begin{array}{cc}si{n}^2\alpha +co{s}^2\alpha & sin\alpha cos\alpha -cos\alpha sin\alpha \\ sin\alpha cos\alpha -sin\alpha cos\alpha & co{s}^2\alpha +si{n}^2\alpha \end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=1.[\because si{n}^2\alpha +co{s}^2\alpha =1]$
Hence, we have verified that A'A=I.