Question

Find the inverse of $\left[\begin{array}{ccc}1&2&3\\1&1&5\\2&4&7\end{array}\right]$ by using elementary transformations.

Answer 1

$\left[\begin{array}{ccc}1&2&3\\1&1&5\\2&4&7\end{array}\right]$

Find the inverse using elementary transformations.:

$\left[\begin{array}{ccc}1&2&3\\1&1&5\\2&4&7\end{array}\right| \left \begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right]$

R2 = R2 - R1

$\left[\begin{array}{ccc}1&2&3\\0&-1&2\\2&4&7\end{array}\right| \left \begin{array}{ccc}1&0&0\\-1&1&0\\0&0&1\end{array}\right]$

R3 = R3 - 2xR1

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

R2 = R2 ÷ (-1)

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

R2 = R2 - 2xR3

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

R1 = R1 - 3xR3

$\left[\begin{array}{ccc}1&2&0\\0&1&0\\0&0&1\end{array}\right| \left \begin{array}{ccc}7&0&-3\\-3&-1&2\\-2&0&1\end{array}\right]$

R1 = R1 - 2xR2

$\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right| \left \begin{array}{ccc}13&2&-7\\-3&-1&2\\-2&0&1\end{array}\right]$

The inverse is:

$\left[\left \begin{array}{ccc}13&2&-7\\-3&-1&2\\-2&0&1\end{array}\right]$