Using elementary transformations,find the inv | vedclass

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(A) Let $A = \left[\begin{array}{cc}7 & 4 \ 1 & -2\end{array}ight]$.To find the inverse using elementary row transformations,we write $A = IA$,wher

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$\frac{1}{18}\left[\begin{array}{cc}2 & 4 \ 1 & -7\end{array}ight]$

Vedclass · Answer

(A) Let $A = \left[\begin{array}{cc}7 & 4 \ 1 & -2\end{array}ight]$.To find the inverse using elementary row transformations,we write $A = IA$,where $I = \left[\begin{array}{cc}1 & 0 \ 0 & 1\end{array}ight]$.$\left[\begin{array}{cc}7 & 4 \ 1 & -2\end{array}ight] = \left[\begin{array}{cc}1 & 0 \ 0 & 1\end{array}ight] A$.Apply $R_1 \leftrightarrow R_2$:$\left[\begin{array}{cc}1 & -2 \ 7 & 4\end{array}ight] = \left[\begin{array}{cc}0 & 1 \ 1 & 0\end{array}ight] A$.Apply $R_2 	o R_2 - 7R_1$:$\left[\begin{array}{cc}1 & -2 \ 0 & 18\end{array}ight] = \left[\begin{array}{cc}0 & 1 \ 1 & -7\end{array}ight] A$.Apply $R_2 	o \frac{1}{18}R_2$:$\left[\begin{array}{cc}1 & -2 \ 0 & 1\end{array}ight] = \left[\begin{array}{cc}0 & 1 \ \frac{1}{18} & -\frac{7}{18}\end{array}ight] A$.Apply $R_1 	o R_1 + 2R_2$:$\left[\begin{array}{cc}1 & 0 \ 0 & 1\end{array}ight] = \left[\begin{array}{cc}\frac{2}{18} & \frac{4}{18} \ \frac{1}{18} & -\frac{7}{18}\end{array}ight] A$.Thus,$A^{-1} = \frac{1}{18}\left[\begin{array}{cc}2 & 4 \ 1 & -7\end{array}ight]$.

Using elementary transformations,find the inverse of the following matrix,if it exists: $\left[\begin{array}{cc}7 & 4 \\ 1 & -2\end{array}\right]$

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