The first column represents the variable x, the second row represents the variable y, the third column represents the variable z and the fourth column represents the independent value.
Therefore, we have the following equations:
First row: x = -7
Second row: y = -1
Third row: 0 = -1
Since the system doesn't define or use the variable z, it can assume any value. That means the system has infinite solutions.
Therefore the correct option is B:
[tex]\lbrace(x,y,z)|x=-7,y=-1,z\text{ is any real number}\rbrace[/tex]
