Using elementary transformations,find the inv | vedclass

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(A) Let $A = \left[\begin{array}{cc}1 & 2 \ 2 & -1\end{array}ight]$.We write $A = IA$,where $I = \left[\begin{array}{cc}1 & 0 \ 0 & 1\end{array}

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$\left[\begin{array}{cc}\frac{1}{5} & \frac{2}{5} \ \frac{2}{5} & -\frac{1}{5}\end{array}ight]$

Vedclass · Answer

(A) Let $A = \left[\begin{array}{cc}1 & 2 \ 2 & -1\end{array}ight]$.We write $A = IA$,where $I = \left[\begin{array}{cc}1 & 0 \ 0 & 1\end{array}ight]$.$\left[\begin{array}{cc}1 & 2 \ 2 & -1\end{array}ight] = \left[\begin{array}{cc}1 & 0 \ 0 & 1\end{array}ight] A$.Apply $R_2 	o R_2 - 2R_1$:$\left[\begin{array}{cc}1 & 2 \ 0 & -5\end{array}ight] = \left[\begin{array}{cc}1 & 0 \ -2 & 1\end{array}ight] A$.Apply $R_2 	o -\frac{1}{5} R_2$:$\left[\begin{array}{cc}1 & 2 \ 0 & 1\end{array}ight] = \left[\begin{array}{cc}1 & 0 \ \frac{2}{5} & -\frac{1}{5}\end{array}ight] A$.Apply $R_1 	o R_1 - 2R_2$:$\left[\begin{array}{cc}1 & 0 \ 0 & 1\end{array}ight] = \left[\begin{array}{cc}1 - 2(\frac{2}{5}) & 0 - 2(-\frac{1}{5}) \ \frac{2}{5} & -\frac{1}{5}\end{array}ight] A$.$\left[\begin{array}{cc}1 & 0 \ 0 & 1\end{array}ight] = \left[\begin{array}{cc}\frac{1}{5} & \frac{2}{5} \ \frac{2}{5} & -\frac{1}{5}\end{array}ight] A$.Thus,$A^{-1} = \left[\begin{array}{cc}\frac{1}{5} & \frac{2}{5} \ \frac{2}{5} & -\frac{1}{5}\end{array}ight]$.

Using elementary transformations,find the inverse of the following matrix,if it exists: $\left[\begin{array}{cc}1 & 2 \\ 2 & -1\end{array}\right]$

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