$$A:=\begin{pmatrix} 1 & 2 & 3 \\\\ 1 & 3 & 5 \\\\ 1 & 5 & 12 \end{pmatrix}$$
Using $A = A I$ where left hand side is transformed into an identity matrix using elementary row and column operations and the same operations are made on $I$ on the right hand side which will result in $I = A A'$ where $A'$ is inverse of the matrix $A$ . Is
$$A'=\begin{pmatrix} 1 & -2 & 1 \\\\ -1 & 3 & -3 \\\\ \frac{2}{3} & \frac{7}{3} & 3 \end{pmatrix}\ ?$$