Question

For the multiplication of matrices as a binary operation on the set of all matrices of the form $\left[\begin{array}{cc}a& b\\ -b& a\end{array}\right]$, a, b ∈ R the inverse of $\left[\begin{array}{cc}2& 3\\ -3& 2\end{array}\right]$ is

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

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

(c) $\left[\begin{array}{cc}2/13& -3/13\\ 3/13& 2/13\end{array}\right]$

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

Answer 1

(c) $\left[\begin{array}{cc}2/13& -3/13\\ 3/13& 2/13\end{array}\right]$

To find the identity element,

$$\begin{gathered} \text{Let}A=\left[\begin{array}{cc}a& b\\ -b& a\end{array}\right]\text{and}I=\left[\begin{array}{cc}x& y\\ -y& x\end{array}\right]\text{such that} \\ A.I=I.A=A \\ A.I=A \\ \left[\begin{array}{cc}a& b\\ -b& a\end{array}\right]\left[\begin{array}{cc}x& y\\ -y& x\end{array}\right]=\left[\begin{array}{cc}x& y\\ -y& x\end{array}\right] \\ \left[\begin{array}{cc}ax-by& ay+bx\\ -\left(ay+bx\right)& ax-by\end{array}\right]=\left[\begin{array}{cc}x& y\\ -y& x\end{array}\right] \\ \Rightarrow ax-by=x...\left(1\right) \\ \Rightarrow ay+bx=y...\left(2\right) \\ \text{Solving these two equations, we get} \\ x=1\text{and}y=0 \\ \text{Thus,} \\ I=\left[\begin{array}{cc}x& y\\ -y& x\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right](\text{which is usually an identity matrix)} \\ \mathrm{Let}\left[\begin{array}{cc}m& n\\ -n& m\end{array}\right]\mathrm{be}\mathrm{the}\mathrm{inverse}\mathrm{of}\left[\begin{array}{cc}2& 3\\ -3& 2\end{array}\right]. \\ \therefore \left[\begin{array}{cc}2& 3\\ -3& 2\end{array}\right]\left[\begin{array}{cc}m& n\\ -n& m\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right] \\ \Rightarrow \left[\begin{array}{cc}2m-3n& 2n+3m\\ -3m-2n& -3n+2m\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right] \\ \Rightarrow 2m-3n=1...(3) \\ 2n+3m=0...(4) \\ -3m-2n=0...(5) \\ -3n+2m=1...(6) \\ \mathrm{From}\mathrm{eq}.(4) \\ n=\frac{-3m}{2}...(7) \\ \mathrm{Substituting}\mathrm{the}\mathrm{value}\mathrm{of}n\mathrm{in}\mathrm{eq}.(3) \\ 2m-3\left(\frac{-3m}{2}\right)=1 \\ \Rightarrow 2m+\frac{9m}{2}=1 \\ \Rightarrow \frac{13m}{2}=1 \\ \Rightarrow m=\frac{2}{13} \\ \mathrm{Substituting}\mathrm{the}\mathrm{value}\mathrm{of}m\mathrm{in}\mathrm{eq}.(7) \\ \Rightarrow n=\frac{-3}{2}\times \frac{2}{13}=\frac{-3}{13} \\ \mathrm{Hence},\mathrm{the}\mathrm{inverse}\mathrm{of}\left[\begin{array}{cc}2& 3\\ -3& 2\end{array}\right]\mathrm{is}\left[\begin{array}{cc}\frac{2}{13}& \frac{-3}{13}\\ \frac{3}{13}& \frac{2}{13}\end{array}\right]. \end{gathered}$$
So, the answer is (c).