Question

Suppose the system below is consistent for all possible values of \(f\) and \(g\). What can you say about the coefficients \(c\) and \(d\)? Justify your answer.

27. \(\begin{array}{l}{x_1} + 3{x_2} = f\\c{x_1} + d{x_2} = g\end{array}\)

Short answer

The system is consistent if \(d \ne 3c\).

Step-by-step solution

Convert the system into an augmented matrix

To express a system in the augmented 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 \({x_1} + 3{x_2} = f\) and \(c{x_1} + d{x_2} = g\) is:

\(\left[ {\begin{array}{*{20}{c}}1&3&f\\c&d&g\end{array}} \right]\)

Apply the elementary row operation

A basic principle of this section is that row operations do not affect the solution set of a linear system.

Use the \({x_2}\) term in the first equation to eliminate the \(c{x_1}\) term from the second equation. Perform an elementary row operationon the matrix \(\left[ {\begin{array}{*{20}{c}}1&3&f\\c&d&g\end{array}} \right]\), as shown below.

Add \( - c\) times row one to row two, i.e., \({R_2} \to {R_2} - c{R_1}\).

\(\left[ {\begin{array}{*{20}{c}}1&3&f\\{c - c\left( 1 \right)}&{d - c\left( 3 \right)}&{g - c\left( f \right)}\end{array}} \right]\)

After performing the row operation, the matrix becomes:

\(\left[ {\begin{array}{*{20}{c}}1&3&f\\0&{d - 3c}&{g - cf}\end{array}} \right]\)

Solution of the system of linear equations

For the system of equations to be consistent, the solution must satisfy the system of equations.

The obtained matrix shows that \(d - 3c\) 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.

Thus, the system is consistent if \(d \ne 3c\).