Q.By using elementary transformation, find the inverse of the following matrix: $A = \begin{bmatrix} 1 & 3 & -2 \\ -3 & 0 & -5 \\ 2 & 5 & 0 \end{bmatrix}$.

For a non-singular square matrix $A$, the inverse can also be built purely from elementary row (or purely from elementary column) transformations, without ever computing a cofactor. The method rests on one rule: whenever a row transformation is applied to the prefactor of a matrix product $AB=C$, applying the same transformation to $C$ keeps the equation valid while $B$ is left untouched (the mirror rule holds for column transformations applied to a postfactor). Starting from the defining equation $AA^{-1}=I$, a sequence of row transformations is applied simultaneously to the left-hand $A$ and to the right-hand $I$; once the left side has been driven all the way to the identity matrix $I$, the right-hand side — which started as $I$ and absorbed every transformation — has automatically become $A^{-1}$. Symbolically: $A\,A^{-1}=I \xrightarrow{\text{row ops}} I\,A^{-1}=B \Rightarrow A^{-1}=B$. If column transformations are preferred instead, the defining equation $A^{-1}A=I$ is used, and the transformations are applied to the columns of $A$ and of the right-hand $I$, again finishing when the left side reaches $I$. A standard order for a $3\times3$ matrix is to reduce $a_{11}$ to $1$, then use it to clear …