Question

$$\begin{gathered} QIfA\left[\begin{array}{cc}1& \tan x\\ -\tan x& 1\end{array}\right]showthatA\text{'}{A}^{-1}=\left[\begin{array}{cc}\cos 2x& -\sin 2x\\ \sin 2x& \cos 2x\end{array}\right] \\ QIfA=\left[\begin{array}{cc}3& 2\\ 7& 5\end{array}\right]andB=\left[\begin{array}{cc}6& 7\\ 8& 9\end{array}\right]verify(AB{)}^{-1}=B^{-1}{A}^{-1} \end{gathered}$$

Answer 1

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Que 4
$$\begin{gathered} \mathrm{A}=\left[\begin{array}{cc}1& \mathrm{tanx}\\ -\mathrm{tanx}& 1\end{array}\right] \\ \left|\mathrm{A}\right|=\left|\begin{array}{cc}1& \mathrm{tanx}\\ -\mathrm{tanx}& 1\end{array}\right|=1+{\tan }^2\mathrm{x}\ne 0 \\ \mathrm{So},\mathrm{A}\mathrm{is}\mathrm{invertible}. \\ \mathrm{Let}{\mathrm{C}}_{\mathrm{ij}}\mathrm{be}\mathrm{the}\mathrm{cofactor}\mathrm{of}{\mathrm{a}}_{\mathrm{ij}}\mathrm{in}\mathrm{A}=\left[{\mathrm{a}}_{\mathrm{ij}}\right].\mathrm{Then}, \\ {\mathrm{C}}_{11}=1 \\ {\mathrm{C}}_{12}=\mathrm{tanx} \\ {\mathrm{C}}_{21}=-\mathrm{tanx} \\ {\mathrm{C}}_{22}=1 \\ \mathrm{So},\mathrm{AdjA}={\left[\begin{array}{cc}1& \mathrm{tanx}\\ -\mathrm{tanx}& 1\end{array}\right]}^{\mathrm{T}}=\left[\begin{array}{cc}1& -\mathrm{tanx}\\ \mathrm{tanx}& 1\end{array}\right] \\ \mathrm{Now},{\mathrm{A}}^{-1}=\frac{1}{\left|\mathrm{A}\right|}\mathrm{AdjA} \\ =\frac{1}{1+{\tan }^2\mathrm{x}}\left[\begin{array}{cc}1& -\mathrm{tanx}\\ \mathrm{tanx}& 1\end{array}\right] \\ =\left[\begin{array}{cc}\frac{1}{1+{\tan }^2\mathrm{x}}& \frac{-\mathrm{tanx}}{1+{\tan }^2\mathrm{x}}\\ \frac{\mathrm{tanx}}{1+{\tan }^2\mathrm{x}}& \frac{1}{1+{\tan }^2\mathrm{x}}\end{array}\right] \\ =\mathrm{So},{\mathrm{A}}^{\mathrm{T}}{\mathrm{A}}^{-1}=\left[\begin{array}{cc}1& -\mathrm{tanx}\\ \mathrm{tanx}& 1\end{array}\right]\left[\begin{array}{cc}\frac{1}{1+{\tan }^2\mathrm{x}}& \frac{-\mathrm{tanx}}{1+{\tan }^2\mathrm{x}}\\ \frac{\mathrm{tanx}}{1+{\tan }^2\mathrm{x}}& \frac{1}{1+{\tan }^2\mathrm{x}}\end{array}\right] \\ =\left[\begin{array}{cc}\frac{1}{1+{\tan }^2\mathrm{x}}-\frac{\mathrm{tanx}}{1+{\tan }^2\mathrm{x}}& \frac{-\mathrm{tanx}}{1+{\tan }^2\mathrm{x}}-\frac{\mathrm{tanx}}{1+{\tan }^2\mathrm{x}}\\ \frac{\mathrm{tanx}}{1+{\tan }^2\mathrm{x}}+\frac{\mathrm{tanx}}{1+{\tan }^2\mathrm{x}}& -\frac{{\tan }^2\mathrm{x}}{1+{\tan }^2\mathrm{x}}+\frac{1}{1+{\tan }^2\mathrm{x}}\end{array}\right] \\ =\left[\begin{array}{cc}\frac{1-{\tan }^2\mathrm{x}}{1+{\tan }^2\mathrm{x}}& -\frac{2\mathrm{tanx}}{1+{\tan }^2\mathrm{x}}\\ \frac{2\mathrm{tanx}}{1+{\tan }^2\mathrm{x}}& \frac{1-{\tan }^2\mathrm{x}}{1+{\tan }^2\mathrm{x}}\end{array}\right] \\ =\left[\begin{array}{cc}\cos 2\mathrm{x}& -\sin 2\mathrm{x}\\ \sin 2\mathrm{x}& \cos 2\mathrm{x}\end{array}\right] \end{gathered}$$
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