Question

if A=[123115247] then find A-1 by elementary column transformation.​

Answer 1

We need to recall the following definition of an inverse matrix.

• A $n\times n$ matrix $A$ is said to be invertible if there is a $n\times n$ matrix $A^{-1}$ such that $AA^{-1}=A^{-1}A=I$

Given:

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

The determinant of the matrix $A$ is,

$|A|=1(1\cdot7-4\cdot5)-2(1\cdot7-2\cdot5)+3(1\cdot4-2\cdot1)$

$|A|=1(7-20)-2(7-10)+3(4-2)$

$|A|=-13-2(-3)+3(2)$

$|A|=-13+6+6$

$|A|=-1$

As $|A|\neq 0$, $A$ is a non-singular matrix.

$\therefore\ A^{-1}$ exist.

consider, $A^{-1}A=I$

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

Apply the column operations. we get

$C_2\rightarrow C_2-2C_1$ , $C_3\rightarrow C_3-3C_1$

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

$C_2\rightarrow -C_2$

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

$C_1\rightarrow C_1-C_2$ , $C_3\rightarrow C_3-2C_2$

$A^{-1}\left[\begin{array}{ccc}1&0&0\\0&1&0\\2&0&1\end{array}\right] =\left[\begin{array}{ccc}-1&2&-7\\1&-1&2\\0&0&1\end{array}\right]$

$C_1\rightarrow C_1-2C_3$

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

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