Answer:
The answer is "[tex]d \neq 3c,[/tex]f and g are arbitrary".
Step-by-step explanation:
The matrix of the device is increased
[tex]\left[\begin{array}{ccc}1&3&f\\ c&d&g\\ \end{array}\right][/tex]
Reduce the echelon row matrix
[tex]\left[\begin{array}{ccc}1&3&f\\ c&d&g\\ \end{array}\right] \\\\R_1 \leftrightarrow R_2 \\\\\frac{R_2 -1}{C R_1 \to R_2} \sim \left[\begin{array}{ccc} c&d&g\\ 0 & \frac{3c-d}{c}& \frac{cf-g}{c}\\ \end{array}\right][/tex]
Therefore, if 3c[tex]\neq[/tex] 0 is d [tex]\neq[/tex] 3c, the device is valid. Therefore d [tex]\neq[/tex] are arbitrary 3c, g and f.
The values of the coefficients c and d are 1 and 3, respectively
The system of equations are given as:
[tex]x_1 + 3x_2 = f[/tex]
[tex]cx_1 + dx_2 = g[/tex]
From the question, we understand that the system of equations is consistent.
Assume that the system of equations is also dependent, then it means that:
[tex]c =1[/tex]
[tex]d =3[/tex]
[tex]g =f[/tex]
Hence, the values of the coefficients c and d are 1 and 3, respectively
Read more about systems of equations at:
https://brainly.com/question/13729904
