Suppose \(a,b,c,\) and \(d\) are constants such that \(a\) is not zero and the system below is consistent for all possible values of \(f\) and \(g\). What can you say about the numbers \(a,b,c,\) and \(d\)? Justify your answer.

28. \(\begin{array}{l}a{x_1} + b{x_2} = f\\c{x_1} + d{x_2} = g\end{array}\)

To express a system in theaugmented matrix form, extract the coefficients of the variables and the constants, and place these entries in the column of the matrix.

Thus, the augmented matrix for the system of equations \(a{x_1} + b{x_2} = f\) and \(c{x_1} + d{x_2} = g\) is:

\(\left[ {\begin{array}{*{20}{c}}a&b&f\\c&d&g\end{array}} \right]\)

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.