Use elementary row transformation, find the inverse of the matrix $A = \left[ {\begin{array}{*{20}{c}}
  1&2&3 \\
  2&5&7 \\
  { - 2}&{ - 4}&{ - 5}
\end{array}} \right]$

Hint: If we have to find ${A^{ - 1}}$ using row operations, write $A = IA$ and apply a sequence of row operations on $A = IA$ till we get, $I = BA$.Then, the matrix $B$ will be the inverse of $A$. Here $I = \left[ {\begin{array}{*{20}{c}}
  1&0&0 \\
  0&1&0 \\
  0&0&1
\end{array}} \right]$ is the identity matrix.

Complete step-by-step answer:
 Write $A = IA$, i.e.,
$\left[ {\begin{array}{*{20}{c}}
  1&2&3 \\
  2&5&7 \\
  { - 2}&{ - 4}&{ - 5}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  1&0&0 \\
  0&1&0 \\
  0&0&1
\end{array}} \right]A$
Applying \[{R_2} \to {R_2} - 2{R_1}\],
$ \Rightarrow \left[ {\begin{array}{*{20}{c}}
  1&2&3 \\
  0&1&1 \\
  { - 2}&{ - 4}&{ - 5}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  1&0&0 \\
  { - 2}&1&0 \\
  0&0&1
\end{array}} \right]A$
Applying \[{R_3} \to {R_3} + 2{R_1}\],
$ \Rightarrow \left[ {\begin{array}{*{20}{c}}
  1&2&3 \\
  0&1&1 \\
  0&0&1
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  1&0&0 \\
  { - 2}&1&0 \\
  2&0&1
\end{array}} \right]A$
Applying \[{R_1} \to {R_1} - 2{R_2}\],
$ \Rightarrow \left[ {\begin{array}{*{20}{c}}
  1&0&1 \\
  0&1&1 \\
  0&0&1
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  5&{ - 2}&0 \\
  { - 2}&1&0 \\
  2&0&1
\end{array}} \right]A$
Applying \[{R_1} \to {R_1} - {R_3}\],
$ \Rightarrow \left[ {\begin{array}{*{20}{c}}
  1&0&0 \\
  0&1&1 \\
  0&0&1
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  3&{ - 2}&{ - 1} \\
  { - 2}&1&0 \\
  2&0&1
\end{array}} \right]A$
Applying \[{R_2} \to {R_2} - {R_3}\],
$ \Rightarrow \left[ {\begin{array}{*{20}{c}}
  1&0&0 \\
  0&1&0 \\
  0&0&1
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  3&{ - 2}&{ - 1} \\
  { - 4}&1&{ - 1} \\
  2&0&1
\end{array}} \right]A$
Since it is of the form $I = BA$, where matrix $B$ will be the inverse of $A$.
$\therefore {A^{ - 1}} = \left[ {\begin{array}{*{20}{c}}
  3&{ - 2}&{ - 1} \\
  { - 4}&1&{ - 1} \\
  2&0&1
\end{array}} \right]$

Note: A rectangular matrix does not possess an inverse matrix. It means the inverse of a matrix is defined only for a square matrix. If $B$ is the inverse of matrix $A$, then $A$ is also the inverse of matrix $B$.