A basic principle states that row operations do not affect the solution set of a linear system.
For finding the solution of the system of equations, eliminate one of the variables, either \({x_1}\) or \({x_2}\).
For obtaining \(1\) as the coefficient of \({x_1}\), perform an elementaryrow operationon theaugmented matrix \(\left[ {\begin{array}{*{20}{c}}a&b&f\\c&d&g\end{array}} \right]\).
Multiply row one by \(\frac{1}{a}\), i.e., \({R_1} \to \frac{1}{a}{R_1}\).
\(\left[ {\begin{array}{*{20}{c}}{\frac{a}{a}}&{\frac{b}{a}}&{\frac{f}{a}}\\c&d&g\end{array}} \right]\)
After performing the row operation, the matrix becomes:
\(\left[ {\begin{array}{*{20}{c}}1&{\frac{b}{a}}&{\frac{f}{a}}\\c&d&g\end{array}} \right]\)
Use the \({x_1}\) term in the first equation to eliminate the \(c{x_1}\) term from the second equation.
Perform an elementary row operationon the matrix \(\left[ {\begin{array}{*{20}{c}}1&{\frac{b}{a}}&{\frac{f}{a}}\\c&d&g\end{array}} \right]\), as shown below.
Add \( - c\)times row one to row two, i.e., \({R_2} \to {R_2} - c{R_1}\).
\(\left[ {\begin{array}{*{20}{c}}1&{\frac{b}{a}}&{\frac{f}{a}}\\{c - c\left( 1 \right)}&{d - c\left( {\frac{b}{a}} \right)}&{g - c\left( {\frac{f}{a}} \right)}\end{array}} \right]\)
After the row operation, the matrix becomes:
\(\left[ {\begin{array}{*{20}{c}}1&{\frac{b}{a}}&{\frac{f}{a}}\\0&{\frac{{ad - bc}}{a}}&{\frac{{ag - fc}}{a}}\end{array}} \right]\)
This shows that \(ad - bc\)is non-zero since \(f\) and \(g\) are arbitrary. Otherwise, for some choices of f and g, the second row would correspond to an equation of the form 0 = b, where b is non-zero.
