Congratulations to the following members for their correct solutions:
1) Sudharaka
Solution:
Here is the way Sudharaka correctly demonstrated that the given matrix inverse for A satisfied the conditions necessary to in fact be its inverse.
[sp]\(\mbox{Let, }A=\begin{bmatrix}a & b \\ c & d\end{bmatrix}\mbox{ and }B=\frac{1}{\det A} \begin{bmatrix}d & -b \\ -c & a\end{bmatrix}\mbox{ where }\det A=ad-bc\neq 0\,.\)
\begin{eqnarray}
AB &=& \frac{1}{\det A}\begin{bmatrix}a & b \\ c & d\end{bmatrix}\begin{bmatrix}d & -b \\ -c & a\end{bmatrix}\\
&=&\frac{1}{\det A}\begin{bmatrix}ad-bc & 0 \\ 0 & ad-bc\end{bmatrix}\\
&=&\frac{1}{ad-bc}\begin{bmatrix}ad-bc & 0 \\ 0 & ad-bc\end{bmatrix}\\
&=&\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}\\
\therefore AB &=&\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}~~~~~~~~~~(1)
\end{eqnarray}
Similarly,
\begin{eqnarray}
BA &=& \frac{1}{\det A}\begin{bmatrix}d & -b \\ -c & a\end{bmatrix}\begin{bmatrix}a & b \\ c & d\end{bmatrix}\\
&=&\frac{1}{\det A}\begin{bmatrix}ad-bc & 0 \\ 0 & ad-bc\end{bmatrix}\\
&=&\frac{1}{ad-bc}\begin{bmatrix}ad-bc & 0 \\ 0 & ad-bc\end{bmatrix}\\
&=&\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}\\
\therefore BA &=&\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}~~~~~~~~~~(1)
\end{eqnarray}
By (1) and (2);
\[AB=BA=I\mbox{ where }I\mbox{ is the identity matrix of order 2}\]
\[\therefore B=A^{-1}=\frac{1}{\det A} \begin{bmatrix}d & -b \\ -c & a\end{bmatrix}\]
Q.E.D[/sp]