Question

find the inverse using elementary Row operation of A=[011 123 311]​

Answer 1

Given matrix is,

$A=\left[\begin{array}{ccc}0&1&2\\ 1&2&3\\ 3&1&1\end{array}\right]$

We have to find  $A^{-1}$  using elementary row operation.

So first we have to combine this matrix with identity matrix to form an augmented matrix like,

$\left[\begin{array}{ccc|ccc}0&1&2&1&0&0\\ 1&2&3&0&1&0\\ 3&1&1&0&0&1\end{array}\right]$

Now let's begin!

$\left[\begin{array}{ccc|ccc}0&1&2&1&0&0\\ 1&2&3&0&1&0\\ 3&1&1&0&0&1\end{array}\right]\ \xrightarrow{R_1\leftrightarrow R_2}\ \left[\begin{array}{ccc|ccc}1&2&3&0&1&0\\ 0&1&2&1&0&0\\ 3&1&1&0&0&1\end{array}\right]$

$\left[\begin{array}{ccc|ccc}1&2&3&0&1&0\\ 0&1&2&1&0&0\\ 3&1&1&0&0&1\end{array}\right]\ \xrightarrow{3R_1-R_3\to R_3}\ \left[\begin{array}{ccc|ccc}1&2&3&0&1&0\\ 0&1&2&1&0&0\\ 0&5&8&0&3&-1\end{array}\right]$

$\left[\begin{array}{ccc|ccc}1&2&3&0&1&0\\ 0&1&2&1&0&0\\ 0&5&8&0&3&-1\end{array}\right]\ \xrightarrow{R_1-2R_2\to R_1}\ \left[\begin{array}{ccc|ccc}1&0&-1&-2&1&0\\ 0&1&2&1&0&0\\ 0&5&8&0&3&-1\end{array}\right]$

$\left[\begin{array}{ccc|ccc}1&0&-1&-2&1&0\\ 0&1&2&1&0&0\\ 0&5&8&0&3&-1\end{array}\right]\ \xrightarrow{5R_2-R_3\to R_3}\ \left[\begin{array}{ccc|ccc}1&0&-1&-2&1&0\\ 0&1&2&1&0&0\\ 0&0&2&5&-3&1\end{array}\right]$

$\left[\begin{array}{ccc|ccc}1&0&-1&-2&1&0\\ 0&1&2&1&0&0\\ 0&0&2&5&-3&1\end{array}\right]\ \xrightarrow{R_2-R_3\to R_2}\ \left[\begin{array}{ccc|ccc}1&0&-1&-2&1&0\\ 0&1&0&-4&3&-1\\ 0&0&2&5&-3&1\end{array}\right]$

$\left[\begin{array}{ccc|ccc}1&0&-1&-2&1&0\\ 0&1&0&-4&3&-1\\ 0&0&2&5&-3&1\end{array}\right]\ \xrightarrow{\frac{1}{2}R_3\to R_3}\ \left[\begin{array}{ccc|ccc}1&0&-1&-2&1&0\\ 0&1&0&-4&3&-1\\ 0&0&1&2.5&-1.5&0.5\end{array}\right]$

$\left[\begin{array}{ccc|ccc}1&0&-1&-2&1&0\\ 0&1&0&-4&3&-1\\ 0&0&1&2.5&-1.5&0.5\end{array}\right]\ \xrightarrow{R_1+R_3\to R_1}\ \left[\begin{array}{ccc|ccc}1&0&0&0.5&-0.5&0.5\\ 0&1&0&-4&3&-1\\ 0&0&1&2.5&-1.5&0.5\end{array}\right]$

Now we get another augmented matrix consists of identity matrix at left. So,

$A^{-1}=\left[\begin{array}{ccc}0.5&-0.5&0.5\\ -4&3&-1\\ 2.5&-1.5&0.5\end{array}\right]$