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Using elementary transformations, find the inverse of the matrix if it exists - $ \begin{bmatrix} 6 & -3 \\ -2 & 1 \end{bmatrix} $

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  • 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.
Given:$A=\begin{bmatrix}6 & -3\\-2 &1\end{bmatrix}$
Step 1: In order to use row elementary transformation we write as A=IA.
$\begin{bmatrix}6 & -3\\-2& 1\end{bmatrix}=\begin{bmatrix}1 & 0\\0&1\end{bmatrix}A$
Step 2: Applying $R_1\leftrightarrow R_2$
$\begin{bmatrix}-2 & 1\\6& -3\end{bmatrix}=\begin{bmatrix}1 & 0\\0&1\end{bmatrix}A$
Step 3: Applying $R_2\Rightarrow R_2+3R_1$
$\begin{bmatrix}-2 & 1\\6+3(-2)& -3+3(1)\end{bmatrix}=\begin{bmatrix}0 & 1\\1+3(0)&0+3(1)\end{bmatrix}A$
$\begin{bmatrix}-2 & 1\\0& 0\end{bmatrix}=\begin{bmatrix}0 & 1\\1&3\end{bmatrix}A$
In the second row of LHS elements are zero .
Hence $A^{-1}$ does not exist.
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