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Using elementary transformations, find the inverse of the matrix if it exists - $ \begin{bmatrix} 2 & 1 \\ 4 & 2 \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:$A=\begin{bmatrix}2 & 1\\4 &2\end{bmatrix}$
Step 1: In order to use row elementary transformation we write as A=IA.
$\begin{bmatrix}2 & 1\\4& 2\end{bmatrix}=\begin{bmatrix}1 & 0\\0&1\end{bmatrix}A$
step 2:Applying $R_2\rightarrow R_2-2 R_1$
$\begin{bmatrix}2 & 1\\4-2(2)& 2-2(1)\end{bmatrix}=\begin{bmatrix}1 & 0\\0-2(1)&1-2(0)\end{bmatrix}A$
$\begin{bmatrix}2 & 1\\0& 0\end{bmatrix}=\begin{bmatrix}1 & 0\\-2&1\end{bmatrix}A$
In LHS ,the elements of the 2nd row is zero hence the $A^{-1}$ does not exist.
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