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

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

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

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

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

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

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

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

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

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

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

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

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

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

The proof is now complete.

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