If possible, using elementary row transformations, find the inverse of the following matrices.

(i) $\left[ {\begin{array}{cccccccccccccccccccc}2&{ - 1}&3\\{ - 5}&3&1\\{ - 3}&2&3\end{array}} \right]$

(ii) $\left[ {\begin{array}{cccccccccccccccccccc}2&3&{ - 3}\\{ - 1}&{ - 2}&2\\1&1&{ - 1}\end{array}} \right]$

(iii) $\left[ {\begin{array}{cccccccccccccccccccc}2&0&{ - 1}\\5&1&0\\0&1&3\end{array}} \right]$

For getting the inverse of the given matrix $A$ by row elementary operations we may write the given matrix as
$A = IA$

(i) $\left[ {\begin{array}{cccccccccccccccccccc}2&{ - 1}&3\\{ - 5}&3&1\\{ - 3}&2&3\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}1&0&0\\0&1&0\\0&0&1\end{array}} \right]A$
$\Rightarrow$ $\left[ {\begin{array}{cccccccccccccccccccc}2&{ - 1}&3\\{ - 3}&2&4\\{ - 3}&2&3\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}1&0&0\\1&1&0\\0&0&1\end{array}} \right]A$

$\Rightarrow$ $\left[ {\begin{array}{cccccccccccccccccccc}2&{ - 1}&3\\{ - 3}&2&4\\0&0&{ - 1}\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}1&0&0\\1&1&0\\{ - 1}&{ - 1}&1\end{array}} \right]A$

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

$\Rightarrow$ $\left[ {\begin{array}{cccccccccccccccccccc}{ - 1}&1&7\\0&{ - 1}&{ - 17}\\0&0&{ - 1}\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}2&1&0\\{ - 5}&{ - 2}&0\\{ - 1}&{ - 1}&1\end{array}} \right]A$

$\Rightarrow$ $\left[ {\begin{array}{cccccccccccccccccccc}{ - 1}&0&{ - 10}\\0&{ - 1}&{ - 17}\\0&0&1\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}{ - 3}&{ - 1}&0\\{ - 5}&{ - 2}&0\\1&1&{ - 1}\end{array}} \right]A$
$\Rightarrow$ $\left[ {\begin{array}{cccccccccccccccccccc}{ - 1}&0&0\\0&{ - 1}&0\\0&0&1\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}7&9&{ - 10}\\{12}&{15}&{ - 17}\\1&1&{ - 1}\end{array}} \right]A$

$\Rightarrow$ $\left[ {\begin{array}{cccccccccccccccccccc}1&0&0\\0&1&0\\0&0&1\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}{ - 7}&{ - 9}&{10}\\{ - 12}&{ - 15}&{17}\\1&1&{ - 1}\end{array}} \right]A$

Therefore, the inverse of $A$ is $\left[ {\begin{array}{cccccccccccccccccccc}{ - 7}&{ - 9}&{10}\\{ - 12}&{ - 15}&{17}\\1&1&{ - 1}\end{array}} \right]$.

(ii) $\therefore$ $\left[ {\begin{array}{cccccccccccccccccccc}2&3&{ - 3}\\{ - 1}&{ - 2}&2\\1&1&{ - 1}\end{array}} \right] = \left[ {\begin{array}{llllllllllllllllllll}1&0&0\\0&1&0\\0&0&1\end{array}} \right]A$
$\Rightarrow$ $\left[ {\begin{array}{cccccccccccccccccccc}0&1&{ - 1}\\0&{ - 1}&1\\1&1&{ - 1}\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}1&0&{ - 2}\\0&1&1\\0&0&1\end{array}} \right]A$

$\Rightarrow$ $\left[ {\begin{array}{cccccccccccccccccccc}0&1&{ - 1}\\0&0&0\\1&1&1\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}1&0&{ - 2}\\2&1&{ - 2}\\0&0&1\end{array}} \right]A$

Since, second row of the matrix A on LHS is containing all zeroes, so we can say that inverse of matrix A does not exist.

(iii) $\therefore$ $\left[ {\begin{array}{cccccccccccccccccccc}2&0&{ - 1}\\5&1&0\\0&1&3\end{array}} \right] = \left[ {\begin{array}{llllllllllllllllllll}1&0&0\\0&1&0\\0&0&1\end{array}} \right]A$
$\Rightarrow$ $\left[ {\begin{array}{cccccccccccccccccccc}2&0&{ - 1}\\3&1&1\\0&1&3\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}1&0&0\\{ - 1}&1&0\\0&0&1\end{array}} \right]A$

$\Rightarrow$ $\left[ {\begin{array}{cccccccccccccccccccc}2&0&{ - 1}\\1&1&2\\2&1&2\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}1&0&0\\{ - 2}&1&0\\1&0&1\end{array}} \right]A$

$\Rightarrow$ $\left[ {\begin{array}{cccccccccccccccccccc}2&0&{ - 1}\\0&1&{\frac{5}{2}}\\4&1&1\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}1&0&0\\{\frac{{ - 5}}{2}}&1&0\\2&0&1\end{array}} \right]A$

$\Rightarrow$ $\left[ {\begin{array}{cccccccccccccccccccc}2&0&{ - 1}\\0&1&{\frac{5}{2}}\\0&1&3\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}1&0&0\\{\frac{{ - 5}}{2}}&1&0\\0&0&1\end{array}} \right]A$

$\Rightarrow$ $\left[ {\begin{array}{cccccccccccccccccccc}2&0&{ - 1}\\0&1&{\frac{5}{2}}\\0&0&{\frac{1}{2}}\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}1&0&0\\{\frac{{ - 5}}{2}}&1&0\\{\frac{5}{2}}&{ - 1}&1\end{array}} \right]A$

$\Rightarrow$ $\left[ {\begin{array}{cccccccccccccccccccc}1&0&{\frac{{ - 1}}{2}}\\0&1&{\frac{5}{2}}\\0&0&1\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}{\frac{1}{2}}&0&0\\{\frac{{ - 5}}{2}}&1&0\\5&{ - 2}&2\end{array}} \right]A$

$\Rightarrow$ $\left[ {\begin{array}{cccccccccccccccccccc}1&0&0\\0&1&0\\0&0&1\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}3&{ - 1}&1\\{ - 15}&6&{ - 5}\\5&{ - 2}&2\end{array}} \right]A$

Hence, $\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}3&{ - 1}&1\\{ - 15}&6&{ - 5}\\5&{ - 2}&2\end{array}} \right]$ is the inverse of given matrix A.