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# Find inverse,by elementary row operations(if possible),of the following matrices$(ii)\quad\begin{bmatrix}1 &- 3\\-2 & 6\end{bmatrix}$

Note: This is part 2 of a 2 part question, split as 2 separate questions here.

Toolbox:
• There are six operations (transformations) on a matrix,three of which are due to rows and three due to columns which are known as elementary operations or transformations.
• Row/Column Switching: Interchange of any two rows or two columns, i.e, $R_i\leftrightarrow R_j$ or $\;C_i\leftrightarrow C_j$
• Row/Column Multiplication: The multiplication of the elements of any row or column by a non zero number: i.e, i.e $R_i\rightarrow kR_i$ where $k\neq 0$ or $\;C_j\rightarrow kC_j$ where $k\neq 0$
• Row/Column Addition:The addition to the element of any row or column ,the corresponding elements of any other row or column multiplied by any non zero number: i.e $R_i\rightarrow R_i+kR_j$ or $\;C_i\rightarrow C_i+kC_j$, where $i \neq j$.
• If A is a matrix such that A$^{-1}$ exists, then to find A$^{-1}$ using elementary row operations, write A = IA and apply a sequence of row operation on A = IA till we get, I = BA. The matrix B will be the inverse of A. Similarly, if we wish to find A$^{-1}$ using column operations, then, write A = AI and apply a sequence of column operations on A = AI till we get, I = AB.
Step1:
Given
$A=\begin{bmatrix}1 & -3\\-2 & 6\end{bmatrix}$
In order to use elementary row operation we may write A=IA.
$\Rightarrow \begin{bmatrix}1 & -3\\-2 & 6\end{bmatrix}=\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}A$
Step2:
Apply $R_1=2R_1+R_2$
$\Rightarrow \begin{bmatrix}0 & 0\\-2 & 6\end{bmatrix}=\begin{bmatrix}2 & 1\\0 & 1\end{bmatrix}A$
In first row of the LHS each element is zero hence $A^{-1}$ does not exist.