## Elementary Linear Algebra 7th Edition

$$a\neq 0, \quad bc-ad\neq 0.$$
We have the matrix $$\left[ \begin {array}{ccc} a&b\\ c&d \end {array} \right].$$ Multiply the first row by $c$ and adding it to $-a$ times the second row, we get $$\left[ \begin {array}{ccc} a&b\\ 0&bc-ad\end {array} \right].$$ Dividing the second row on $bc-ad$, we get $$\left[ \begin {array}{ccc} a&b\\ 0&1\end {array} \right].$$ Multiply the second row by $-b$ and adding it to the first row, we get $$\left[ \begin {array}{ccc} a&0\\ 0&1\end {array} \right].$$ Dividing the first row on $a$, we get $$\left[ \begin {array}{ccc} 1&0\\ 0&1 \end {array} \right].$$ Now, the matrix $\left[ \begin {array}{ccc} a&b\\ c&d \end {array} \right]$ is row-equivalent to $\left[ \begin {array}{ccc} 1&0\\ 0&1 \end {array} \right]$ provided that $$a\neq 0, \quad bc-ad\neq 0.$$