## Linear Algebra and Its Applications, 4th Edition

$(\begin{matrix}0 & \frac{1}{3}\\\frac{1}{2} & 0\end{matrix})$, $(\begin{matrix}\frac{1}{2} & 0\\-1 & \frac{1}{2}\end{matrix})$, $(\begin{matrix}\cos{\theta} & \sin{\theta}\\-\sin{\theta} & \cos{\theta}\end{matrix})$
Given a matrix $(\begin{matrix}a & b\\c & d\end{matrix})$, its inverse is defined as $\frac{1}{ad-bc}(\begin{matrix}d & -b\\-c & a\end{matrix})$. We will use this definition to evaluate the following matrices. $$A_{1}=(\begin{matrix}0 & 2\\3 & 0\end{matrix})$$ In this matrix, $a=0$, $b=2$, $c=3$ and $d=0$. Substituting these values into the inverse definition, $$(A_{1})^{-1}=\frac{1}{0-2(3)}(\begin{matrix}0 & -2\\-3 & 0\end{matrix})=-\frac{1}{6}(\begin{matrix}0 & -2\\-3 & 0\end{matrix})=(\begin{matrix}0 & \frac{1}{3}\\\frac{1}{2} & 0\end{matrix})$$ $$A_{2}=(\begin{matrix}2 & 0\\4 & 2\end{matrix})$$ In this matrix, $a=2$, $b=0$, $c=4$ and $d=2$. Substituting these values into the inverse definition, $$(A_{2})^{-1}=\frac{1}{2(2)-0}(\begin{matrix}2 & 0\\-4 & 2\end{matrix})=\frac{1}{4}(\begin{matrix}2 & 0\\-4 & 2\end{matrix})=(\begin{matrix}\frac{1}{2} & 0\\-1 & \frac{1}{2}\end{matrix})$$ $$A_{3}=(\begin{matrix}\cos{\theta} & -\sin{\theta}\\\sin{\theta} & \cos{\theta}\end{matrix})$$ In this matrix, $a=\cos{\theta}$, $b=\sin{\theta}$, $c=-\sin{\theta}$ and $d=\cos{\theta}$. Substituting these values into the inverse definition, $$(A_{3})^{-1}=\frac{1}{{\cos^2{\theta}}-(-{\sin^2{\theta}})}(\begin{matrix}\cos{\theta} & \sin{\theta}\\-\sin{\theta} & \cos{\theta}\end{matrix})=(\begin{matrix}\cos{\theta} & \sin{\theta}\\-\sin{\theta} & \cos{\theta}\end{matrix})\quad(\because\cos^2{\theta}+\sin^2{\theta}=1)$$ (Note: Check if the matrix is invertible before evaluating. A matrix is invertible only when $ad-bc\neq0$.)