Question

Find the inverse of each of the following matrices:

(i) $\left[\begin{array}{cc}\cos \mathrm{\theta }& \sin \mathrm{\theta }\\ -\sin \mathrm{\theta }& \cos \mathrm{\theta }\end{array}\right]$

(ii) $\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]$

(iii) $\left[\begin{array}{cc}a& b\\ c& \frac{1+bc}{a}\end{array}\right]$

(iv) $\left[\begin{array}{cc}2& 5\\ -3& 1\end{array}\right]$

Answer 1

$$\begin{gathered} \left(i\right)A=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right] \\ \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{Now}, \\ {C}_{11}=\cos \theta \\ {C}_{12}=\sin \theta \\ {C}_{21}=-\sin \theta \\ {C}_{22}=\cos \theta \\ \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] \\ \left(ii\right)B=\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right] \\ \left|B\right|=0-1=-1\ne 0 \\ B\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}{b}_{ij}\mathrm{in}B. \\ \mathrm{Now}, \\ {C}_{11}=0 \\ {C}_{12}=-1 \\ {C}_{21}=-1 \\ {C}_{22}=0 \\ \mathrm{adj}B={\left[\begin{array}{cc}0& -1\\ -1& 0\end{array}\right]}^T=\left[\begin{array}{cc}0& -1\\ -1& 0\end{array}\right] \\ \therefore B^{-1}=\frac{1}{\left|B\right|}\mathrm{adj}B=\frac{1}{-1}\left[\begin{array}{cc}0& -1\\ -1& 0\end{array}\right]=\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right] \\ \left(iii\right)C=\left[\begin{array}{cc}a& b\\ c& \frac{1+bc}{a}\end{array}\right] \\ \left|C\right|=1+bc-bc=1\ne 0 \\ C\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}{c}_{ij}\mathrm{in}C. \\ \mathrm{Now}, \\ {C}_{11}=\frac{1+bc}{a} \\ {C}_{12}=-c \\ {C}_{21}=-b \\ {C}_{22}=a \\ \mathrm{adj}C={\left[\begin{array}{cc}\frac{1+bc}{a}& -c\\ -b& a\end{array}\right]}^T=\left[\begin{array}{cc}\frac{1+bc}{a}& -b\\ -c& a\end{array}\right] \\ \therefore C^{-1}=\frac{1}{\left|C\right|}\mathrm{adj}C=\left[\begin{array}{cc}\frac{1+bc}{a}& -b\\ -c& a\end{array}\right] \\ \left(iv\right)D=\left[\begin{array}{cc}2& 5\\ -3& 1\end{array}\right] \\ \left|D\right|=2+15=17\ne 0 \\ D\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}{d}_{ij}\mathrm{in}D. \\ \mathrm{Now}, \\ {C}_{11}=1 \\ {C}_{12}=3 \\ {C}_{21}=-5 \\ {C}_{22}=2 \\ \mathrm{adj}D={\left[\begin{array}{cc}1& 3\\ -5& 2\end{array}\right]}^T=\left[\begin{array}{cc}1& -5\\ 3& 2\end{array}\right] \\ \therefore D^{-1}=\frac{1}{\left|D\right|}\mathrm{adj}D=\frac{1}{17}\left[\begin{array}{cc}1& -5\\ 3& 2\end{array}\right] \end{gathered}$$