19:


19. If $A=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right] and B= \left[\begin{array}{cc}0& 1\\ -1& 0\end{array}\right]$ when write (aA+bB) (aA-bB) in the form of matrix. 
 

Dear Student,

$$\begin{gathered} We have: A = \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right] and B = \left[\begin{array}{cc}0& 1\\ -1& 0\end{array}\right] \\ \therefore aA + bB = a\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right] + b\left[\begin{array}{cc}0& 1\\ -1& 0\end{array}\right] = \left[\begin{array}{cc}a& 0\\ 0& a\end{array}\right] + \left[\begin{array}{cc}0& b\\ -b& 0\end{array}\right] = \left[\begin{array}{cc}a& b\\ -b& a\end{array}\right] \\ and, aA - bB = a\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right] - b\left[\begin{array}{cc}0& 1\\ -1& 0\end{array}\right] = \left[\begin{array}{cc}a& 0\\ 0& a\end{array}\right] - \left[\begin{array}{cc}0& b\\ -b& 0\end{array}\right] = \left[\begin{array}{cc}a& -b\\ b& a\end{array}\right] \\ \therefore \left(aA + bB\right)\left(aA - bB\right) = \left[\begin{array}{cc}a& b\\ -b& a\end{array}\right]\left[\begin{array}{cc}a& -b\\ b& a\end{array}\right] = \left[\begin{array}{cc}{a}^2 + b^2& \cancel{-ab} + \cancel{ba}\\ \cancel{-ba} + \cancel{ab}& b^2 + a^2\end{array}\right] = \boxed{\left({\mathbf{a}}^{\mathbf{2}}\mathbf{ }\mathbf{+}\mathbf{ }{\mathbf{b}}^{\mathbf{2}}\right)\left[\begin{array}{cc}\mathbf{1}& \mathbf{0}\\ \mathbf{0}& \mathbf{1}\end{array}\right]\mathbf{ }\mathbf{=}\mathbf{ }\left({\mathbf{a}}^{\mathbf{2}}\mathbf{ }\mathbf{+}\mathbf{ }{\mathbf{b}}^{\mathbf{2}}\right)\mathbf{A}\mathbf{ }\mathbf{=}\mathbf{ }\left({\mathbf{a}}^{\mathbf{2}}\mathbf{ }\mathbf{+}\mathbf{ }{\mathbf{b}}^{\mathbf{2}}\right)\mathbf{I}} \end{gathered}$$

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