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# Obtain the inverse of the following matrix using elementary operation $A = \begin{bmatrix} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{bmatrix}$

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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}0 & 1&2\\1 & 2 & 3\\3 & 1 & 1\end{bmatrix}\$
In order to find the inverse by using row elementary transformation, we write as A=IA.
Step 1: $\begin{bmatrix}0 & 1&2\\1 & 2 & 3\\3 & 1 & 1\end{bmatrix}=\begin{bmatrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0& 1\end{bmatrix}A$
Step 2: Apply $R_1\leftrightarrow R_2$
$\begin{bmatrix}1 & 2&3\\0 & 1 & 2\\3 & 1 & 1\end{bmatrix}=\begin{bmatrix}0 & 1 & 0\\1 & 0 & 0\\0 & 0& 1\end{bmatrix}A$
Step 3: Apply $R_3\rightarrow R_3-3R_1$
$\begin{bmatrix}1 & 2&3\\0 & 1 & 2\\0 & -5 & -8\end{bmatrix}=\begin{bmatrix}0 & 1 & 0\\1 & 0 & 0\\0 & -3& 1\end{bmatrix}A$
Step 4: Apply $R_1\rightarrow R_1-2R_2$
$\begin{bmatrix}1 & 0&-1\\0 & 1 & 2\\0 & -5 & -8\end{bmatrix}=\begin{bmatrix}-2 & 1 & 0\\1 & 0 & 0\\0 & -3& 1\end{bmatrix}A$
Step 5: Apply $R_3\rightarrow R_3+5R_2$
$\begin{bmatrix}1 & 0&-1\\0 & 1 & 2\\0 & 0 & 2\end{bmatrix}=\begin{bmatrix}-2 & 1 & 0\\1 & 0 & 0\\5& -3& 1\end{bmatrix}A$
Step 6: Apply $R_3\rightarrow \frac{1}{2} R_3$
$\begin{bmatrix}1 & 0&-1\\0 & 1 & 2\\0 & 0 & 1\end{bmatrix}=\begin{bmatrix}-2 & 1 & 0\\1 & 0 & 0\\\frac{1}{2}&\frac{ -3}{2}& \frac{1}{2}\end{bmatrix}A$
Step 7: Apply $R_1\rightarrow R_1+R_3$
$\begin{bmatrix}1 & 0&0\\0 & 1 & 2\\0 & 0 & 1\end{bmatrix}=\begin{bmatrix}\frac{1}{2} & \frac{-1}{2}& \frac{1}{2}\\1 & 0 & 0\\\frac{5}{2}&\frac{ -3}{2}& \frac{1}{2}\end{bmatrix}A$
Step 8: Apply $R_2\rightarrow R_2-2R_3$
$\begin{bmatrix}1 & 0&0\\0 & 1 & 0\\0 & 0 & 1\end{bmatrix}=\begin{bmatrix}\frac{1}{2} & \frac{-1}{2}& \frac{1}{2}\\-4 & 3 & -1\\\frac{5}{2}&\frac{ -3}{2}& \frac{1}{2}\end{bmatrix}A$
Step 9: $A^{-1}=\begin{bmatrix}\frac{1}{2} & \frac{-1}{2}& \frac{1}{2}\\-4 & 3 & -1\\\frac{5}{2}&\frac{ -3}{2}& \frac{1}{2}\end{bmatrix}$
answered Apr 8, 2013

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