To solve the given system of linear equations using the row-echelon form matrix, we perform back substitution.
The augmented matrix is:
[tex]\[
\left[\begin{array}{rrr|c}
1 & -\frac{1}{4} & -\frac{1}{4} & \frac{13}{4} \\
0 & 1 & -\frac{7}{2} & \frac{17}{2} \\
0 & 0 & 1 & -3
\end{array}\right]
\][/tex]
From the third row of the matrix, we have:
[tex]\[ z = -3 \][/tex]
Using this value of \( z \), we substitute it into the second row equation:
[tex]\[ y - \frac{7}{2}z = \frac{17}{2} \][/tex]
[tex]\[ y - \frac{7}{2}(-3) = \frac{17}{2} \][/tex]
[tex]\[ y + \frac{21}{2} = \frac{17}{2} \][/tex]
[tex]\[ y = \frac{17}{2} - \frac{21}{2} \][/tex]
[tex]\[ y = -2 \][/tex]
Now, we substitute the values of \( y \) and \( z \) into the first row equation:
[tex]\[ x - \frac{1}{4}y - \frac{1}{4}z = \frac{13}{4} \][/tex]
[tex]\[ x - \frac{1}{4}(-2) - \frac{1}{4}(-3) = \frac{13}{4} \][/tex]
[tex]\[ x + \frac{1}{2} + \frac{3}{4} = \frac{13}{4} \][/tex]
[tex]\[ x + \frac{5}{4} = \frac{13}{4} \][/tex]
[tex]\[ x = \frac{13}{4} - \frac{5}{4} \][/tex]
[tex]\[ x = 2 \][/tex]
Thus, the solution to the system of linear equations is:
[tex]\[ x = 2, y = -2, z = -3 \][/tex]
This corresponds to choice A:
A. There is one solution. The solution set is [tex]\(\{(2, -2, -3)\}\)[/tex].
