Question

Find the A inverse using column transformation where A is above matrix?
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Answer 1

Taking  $\text{A \Rightarrow ( A | I )}$,  where I is the identity matrix,

$\left[\begin{array}{ccc}1&2&3\\ 2&4&5\\ 3&5&6\end{array}\right]\ \Rightarrow\ \left[\begin{array}{ccc|ccc}1&2&3&1&0&0\\ 2&4&5&0&1&0\\ 3&5&6&0&0&1\end{array}\right]$

After column transformation operations, we get  $\text{( I | A ^{-1} )}.$

Let's begin.

$\left[\begin{array}{ccc|ccc}1&2&3&1&0&0\\ 2&4&5&0&1&0\\ 3&5&6&0&0&1\end{array}\right]\ \xrightarrow{R_2-2R_1\to R_2}\ \left[\begin{array}{ccc|ccc}1&2&3&1&0&0\\ 0&0&-1&-2&1&0\\ 3&5&6&0&0&1\end{array}\right]\\ \\ \\ \left[\begin{array}{ccc|ccc}1&2&3&1&0&0\\ 0&0&-1&-2&1&0\\ 3&5&6&0&0&1\end{array}\right]\ \xrightarrow{3R_1-R_3\to R_3}\ \left[\begin{array}{ccc|ccc}1&2&3&1&0&0\\ 0&0&-1&-2&1&0\\ 0&1&3&3&0&-1\end{array}\right]$

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

$\left[\begin{array}{ccc|ccc}1&2&3&1&0&0\\ 0&1&3&3&0&-1\\ 0&0&1&2&-1&0\end{array}\right]\ \xrightarrow{R_1-2R_2\to R_1}\ \left[\begin{array}{ccc|ccc}1&0&-3&-5&0&2\\ 0&1&3&3&0&-1\\ 0&0&1&2&-1&0\end{array}\right]\\ \\ \\ \left[\begin{array}{ccc|ccc}1&0&-3&-5&0&2\\ 0&1&3&3&0&-1\\ 0&0&1&2&-1&0\end{array}\right]\ \xrightarrow{R_1+3R_3\to R_1}\ \left[\begin{array}{ccc|ccc}1&0&0&1&-3&2\\ 0&1&3&3&0&-1\\ 0&0&1&2&-1&0\end{array}\right]$

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

So we got  $\text{( I | A ^{-1} )}.$  Hence,

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