Answer

Assume that $ad -bc \neq 0$. Let us show that $A$ is row equivalent to $I_2$. Assume moreover that $a \neq 0$. Then divide the first row by $a$ to get the new matrix

\begin{displaymath}\left(\begin{array}{ll}
1 &\displaystyle \frac{b}{a} \\
&\\
c & d
\end{array} \right)\end{displaymath}

Now take the second row minus $c$ times the first row to get

\begin{displaymath}\left(\begin{array}{lc}
1 &\displaystyle \frac{b}{a} \\
&\\...
...\\
0 & \displaystyle \frac{ad - bc}{a}
\end{array} \right)\;.\end{displaymath}

Divide the second row by $\displaystyle \frac{ad - bc}{a}$ since it not equal to 0 to get

\begin{displaymath}\left(\begin{array}{lc}
1 &\displaystyle \frac{b}{a} \\
&\\
0 & 1
\end{array} \right)\end{displaymath}

Finally take the first row minus $\displaystyle \frac{b}{a}$ times the first row to get

\begin{displaymath}\left(\begin{array}{lc}
1 & 0 \\
&\\
0 & 1
\end{array} \right) = I_2 \;.\end{displaymath}

Our proof is almost complete, if we show that the conclusion still holds when $a = 0$. In this case, neither $b$ nor $C$ are equal to 0. Switch the first row with the second one to get

\begin{displaymath}\left(\begin{array}{lc}
c & d \\
&\\
0 & b
\end{array} \right)\end{displaymath}

Divide the first row by $c$ and the second row by $b$ to get

\begin{displaymath}\left(\begin{array}{ll}
1 &\displaystyle \frac{d}{c} \\
&\\
0 & 1
\end{array} \right)\end{displaymath}

Take the first row minus $\displaystyle \frac{d}{c}$ times the second row to get

\begin{displaymath}\left(\begin{array}{lc}
1 & 0 \\
&\\
0 & 1
\end{array} \right) = I_2 \;.\end{displaymath}

The proof is now complete.

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