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# Using elementary transformations, find the inverse of the following matrix : $\begin{bmatrix} 1 & 2 & 3 \\ 2 & 5 & 7 \\ -2 & -4 & -5 \end{bmatrix}$

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.
Given:
$A=\begin{bmatrix}1 & 2 & 3\\2 & 5 & 7\\-2 & -4 & -5\end{bmatrix}$
In order to find inverse by using row elementary transformation we write as A=IA.
$\begin{bmatrix}1 & 2 & 3\\2 & 5 & 7\\-2 & -4 & -5\end{bmatrix}=\begin{bmatrix}1 &0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{bmatrix}A$
Step 1: Apply $R_2=R_2+R_3$
$\begin{bmatrix}1 & 2 & 3\\0 & 1 & 2\\-2 & -4 & -5\end{bmatrix}=\begin{bmatrix}1 &0 & 0\\0 & 1 & 1\\0 & 0 & 1\end{bmatrix}A$
Step 2: Apply $R_2=2R_1+R_3$
$\begin{bmatrix}1 & 2 & 3\\0 & 1 & 2\\0 & 0 & 1\end{bmatrix}=\begin{bmatrix}1 &0 & 0\\0 & 1 & 1\\2 & 0 & 1\end{bmatrix}A$
Step 3: Apply $R_1=R_1-3R_3$
$\begin{bmatrix}1 & 2 & 0\\0 & 1 & 2\\0 & 0 & 1\end{bmatrix}=\begin{bmatrix}-5 &0 & -3\\0 & 1 & 1\\2 & 0 & 1\end{bmatrix}A$
Step 4: Apply $R_2=R_2-2R_3$
$\begin{bmatrix}1 & 2 & 0\\0 & 1 & 0\\0 & 0 & 1\end{bmatrix}=\begin{bmatrix}-5 &0 & -3\\-4 & 1 & -1\\2 & 0 & 1\end{bmatrix}A$
Step 5: Apply $R_1=R_1-2R_2$
$\begin{bmatrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{bmatrix}=\begin{bmatrix}3 &-2 & -1\\-4 & 1 & -1\\2 & 0 & 1\end{bmatrix}A$
Step 6: $A^{-1}=\begin{bmatrix}3 &-2 & -1\\-4 & 1 & -1\\2 & 0 & 1\end{bmatrix}$
Wrong in step no. 2
right is R3=2R1+R3