# Using elementary transformation, find the inverse of $\begin{bmatrix} 1 & 3 & -2 \\ -3 & 0 & -5 \\ 2 & 5 & 0 \end{bmatrix}$

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.
In order to find the inverse by row elementary transformation we write as $A=I_3A$
$\begin{bmatrix} 1 & 3 & -2 \\ -3 & 0 & -5 \\ 2 & 5 & 0 \end{bmatrix} =\begin{bmatrix}1 & 0 &0\\0 & 1&0\\0 & 0 &1\end{bmatrix}A$
Step 1: Apply $R_2\rightarrow R_2+3R_!$
$\begin{bmatrix} 1 & 3 & -2 \\ 0 & 9 & -11 \\ 2 & 5 & 0 \end{bmatrix} =\begin{bmatrix}1 & 0 &0\\3 & 1&0\\0 & 0 &1\end{bmatrix}A$
Step 2: Apply $R_3\rightarrow R_3-2R_!$
$\begin{bmatrix} 1 & 3 & -2 \\ 0 & 9 & -11 \\ 0 & -1 & 4 \end{bmatrix} =\begin{bmatrix}1 & 0 &0\\3 & 1&0\\-2 & 0 &1\end{bmatrix}A$
Step 3: Apply $R_2\rightarrow R_2+8R_3$
$\begin{bmatrix} 1 & 3 & -2 \\ 0 & 1 & 21 \\ 0 & -1 & 4 \end{bmatrix} =\begin{bmatrix}1 & 0 &0\\-13 & 1&8\\-2 & 0 &1\end{bmatrix}A$
Step 4: Apply $R_1\rightarrow R_1-3R_2$
$\begin{bmatrix} 1 & 0 & -65 \\ 0 & 1 & 21 \\ 0 & -1 & 4 \end{bmatrix} =\begin{bmatrix}40 & -3 &-24\\-13 & 1&8\\-2 & 0 &1\end{bmatrix}A$
Step 5: Apply $R_3\rightarrow R_3+R_2$
$\begin{bmatrix} 1 & 0 & -65 \\ 0 & 1 & 21 \\ 0 & 0 & 25 \end{bmatrix} =\begin{bmatrix}40 & -3 &-24\\-13 & 1&8\\-15 & 1 &9\end{bmatrix}A$
Step 6: Apply $R_3\rightarrow \frac{1}{25}R_3$
$\begin{bmatrix} 1 & 0 & -65 \\ 0 & 1 & 21 \\ 0 & 0 & 1 \end{bmatrix} =\begin{bmatrix}40 & -3 &-24\\-13 & 1&8\\\frac{-3}{5} & \frac{1}{25} &\frac{9}{25}\end{bmatrix}A$
Step 7: Apply $R_1\rightarrow R_1+65R_3$
$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{bmatrix} =\begin{bmatrix}1 & \frac{-2}{5} &\frac{-3}{5}\\-13 & 1&8\\\frac{-3}{5} & \frac{1}{25} &\frac{9}{25}\end{bmatrix}A$
Step 8: Apply $R_2\rightarrow R_2-21R_3$
$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} =\begin{bmatrix}1 & \frac{-2}{5} &\frac{-3}{5}\\\frac{-2}{5} & \frac{4}{25}&\frac{11}{25}\\\frac{-3}{5} & \frac{1}{25} &\frac{9}{25}\end{bmatrix}A$
Step 9: $A^{-1}=\begin{bmatrix}1 & \frac{-2}{5} &\frac{-3}{5}\\\frac{-2}{5} & \frac{4}{25}&\frac{11}{25}\\\frac{-3}{5} & \frac{1}{25} &\frac{9}{25}\end{bmatrix}$