To determine the nature of the solutions for a system of linear equations given in row echelon form, we analyze the augmented matrix provided:
[tex]\[
\left[\begin{array}{lll|l}1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 3 \\ 0 & 0 & 0 & 7\end{array}\right]
\][/tex]
Each row in this matrix corresponds to a linear equation. Let's translate each row into its corresponding equation:
1. The first row is [tex]\(1x + 0y + 0z = 1\)[/tex], meaning:
[tex]\[
x = 1
\][/tex]
2. The second row is [tex]\(0x + 1y + 0z = 3\)[/tex], meaning:
[tex]\[
y = 3
\][/tex]
3. The third row is [tex]\(0x + 0y + 0z = 7\)[/tex], which translates to:
[tex]\[
0 = 7
\][/tex]
The third equation, [tex]\(0 = 7\)[/tex], is a contradiction because it is not possible for zero to equal seven. This indicates that there is no set of values for [tex]\(x\)[/tex], [tex]\(y\)[/tex], and [tex]\(z\)[/tex] that will satisfy this system of equations.
Therefore, the system of equations has no solutions.
The correct answer is:
C. no solutions
