Question

Find the inverse of A = $\left[\begin{array}{ccc}1&0&1\\0&2&3\\1&2&1\end{array}\right]$ elementary column transformations.

Answer 1

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

Find the inverse:

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

R3 = R3 - R1

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

R2 = R2 - R3

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

R2 = R2 ÷ 3

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

R3 = R3 ÷ 2

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

Swap R2 and R3

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

R1 = R1 - R3

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

The inverse is:

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