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(A) To find the inverse using elementary row operations,we write $A=IA$.$\left[\begin{array}{rr}1 & 2 \ 2 & -1\end{array}ight]=\left[\begin{array}{

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$A^{-1}=\left[\begin{array}{ll}\frac{1}{5} & \frac{2}{5} \ \frac{2}{5} & \frac{-1}{5}\end{array}ight]$

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(A) To find the inverse using elementary row operations,we write $A=IA$.$\left[\begin{array}{rr}1 & 2 \ 2 & -1\end{array}ight]=\left[\begin{array}{ll}1 & 0 \ 0 & 1\end{array}ight] A$Applying $R_{2} ightarrow R_{2}-2 R_{1}$:$\left[\begin{array}{rr}1 & 2 \ 0 & -5\end{array}ight]=\left[\begin{array}{rr}1 & 0 \ -2 & 1\end{array}ight] A$Applying $R_{2} ightarrow -\frac{1}{5} R_{2}$:$\left[\begin{array}{ll}1 & 2 \ 0 & 1\end{array}ight]=\left[\begin{array}{cc}1 & 0 \ \frac{2}{5} & \frac{-1}{5}\end{array}ight] A$Applying $R_{1} ightarrow R_{1}-2 R_{2}$:$\left[\begin{array}{ll}1 & 0 \ 0 & 1\end{array}ight]=\left[\begin{array}{ll}\frac{1}{5} & \frac{2}{5} \ \frac{2}{5} & \frac{-1}{5}\end{array}ight] A$Thus,$A^{-1}=\left[\begin{array}{ll}\frac{1}{5} & \frac{2}{5} \ \frac{2}{5} & \frac{-1}{5}\end{array}ight]$.

By using elementary operations,find the inverse of the matrix $A=\left[\begin{array}{rr}1 & 2 \\ 2 & -1\end{array}\right]$.

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