Find the inverse,by elementary row operations | vedclass

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(C) Let $A = \left[\begin{array}{cc}1 & -3 \ -2 & 6\end{array}ight]$.To find the inverse using elementary row operations,we write $A = IA$,where $I

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Does not exist

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(C) Let $A = \left[\begin{array}{cc}1 & -3 \ -2 & 6\end{array}ight]$.To find the inverse using elementary row operations,we write $A = IA$,where $I$ is the identity matrix.$\left[\begin{array}{cc}1 & -3 \ -2 & 6\end{array}ight] = \left[\begin{array}{cc}1 & 0 \ 0 & 1\end{array}ight] A$Applying the row operation $R_2 ightarrow R_2 + 2R_1$:$\left[\begin{array}{cc}1 & -3 \ -2 + 2(1) & 6 + 2(-3)\end{array}ight] = \left[\begin{array}{cc}1 & 0 \ 0 + 2(1) & 1 + 2(0)\end{array}ight] A$$\left[\begin{array}{cc}1 & -3 \ 0 & 0\end{array}ight] = \left[\begin{array}{cc}1 & 0 \ 2 & 1\end{array}ight] A$Since all elements in the second row of the matrix on the left-hand side are zero,the matrix $A$ is singular (non-invertible).Therefore,$A^{-1}$ does not exist.

Find the inverse,by elementary row operations (if possible),of the following matrix: $\left[\begin{array}{cc}1 & -3 \\ -2 & 6\end{array}\right]$

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