Question

Find the inverse of the matrix $\left[\begin{array}{cc}co\mathrm{\theta }& \sin \mathrm{\theta }\\ -\sin \mathrm{\theta }& \cos \mathrm{\theta }\end{array}\right]$.

Answer 1

$$\begin{gathered} A=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right] \\ \therefore \left|A\right|={\cos }^2\theta +{\sin }^2\theta =1\ne 0 \\ A\mathrm{is}\mathrm{a}\mathrm{singular}\mathrm{matrix}.\mathrm{Therefore},\mathrm{it}\mathrm{is}\mathrm{invertible}. \\ \mathrm{Let}{C}_{ij}\mathrm{be}\mathrm{a}\mathrm{cofactor}\mathrm{of}{a}_{ij}\mathrm{in}A. \\ \mathrm{Th}e\mathrm{cofactors}\mathrm{of}\mathrm{element}A\mathrm{are}\mathrm{given}\mathrm{by} \\ {C}_{11}=\cos \theta \\ {C}_{12}=\sin \theta \\ {C}_{21}=-\sin \theta \\ {C}_{22}=\cos \theta \\ \mathrm{Now}, \\ \mathrm{adj}A={\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]}^T=\left[\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right] \\ \therefore A^{-1}=\frac{1}{\left|A\right|}\mathrm{adj}A=\left[\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right] \end{gathered}$$