Question

[tex] If A=\left[\begin{array}{ccc}Sinθ&Cosθ\\-Cosθ&Sinθ\end{array}\right], Prove that A⁻¹=Aᵀ. [\tex]

Answer 1

Answer:

$\bf\:A^{-1}=A^T$

Step-by-step explanation:

$A=\left[\begin{array}{cc}sin\theta&cos\theta\\-cos\theta&sin\theta\end{array}\right]$

$A^T=\left[\begin{array}{cc}sin\theta&-cos\theta\\cos\theta&sin\theta\end{array}\right]$ ...................(1)

$|A|=\left|\begin{array}{cc}sin\theta&cos\theta\\-cos\theta&sin\theta\end{array}\right|$

$|A|=sin^2\theta+cos^2\theta=1\neq\:0$

$\therefore\:A^{-1}\:exists$

$adjA=\left[\begin{array}{cc}sin\theta&-cos\theta\\cos\theta&sin\theta\end{array}\right]$

Then,

$A^{-1}=\frac{1}{|A|}adjA$

$A^{-1}=\frac{1}{1}adjA$

$A^{-1}=adjA$

$A^{-1}=\left[\begin{array}{cc}sin\theta&-cos\theta\\cos\theta&sin\theta\end{array}\right]$ ..............(2)

From (1) and (2), we get

$\bf\:A^{-1}=A^T$

Hence proved