In Exercises 7–10, the augmented matrix of a linear system has been reduced by row operations to the form shown. In each case, continue the appropriate row operations and describe the solution set of the original system.

7. \(\left( {\begin{aligned}{*{20}{c}}1&7&3&{ - 4}\\0&1&{ - 1}&3\\0&0&0&1\\0&0&1&{ - 2}\end{aligned}} \right)\)

The augmented matrix of a linear system is given as

\(\left( {\begin{aligned}{*{20}{c}}1&7&3&{ - 4}\\0&1&{ - 1}&3\\0&0&0&1\\0&0&1&{ - 2}\end{aligned}} \right)\)

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

Ordinarily, the next step would be to interchange the third row and the fourth row to have 1 in the third row and the third column.

However, in this case, all three elements of the third row of the augmented matrix are zero.

The third row can be written in the equation form as

\(\begin{aligned}{c}0{x_1} + 0{x_2} + 0{x_3} = 1\\ \Rightarrow 0 = 1.\end{aligned}\)

This is a contradiction and is not a possible condition.

All elements of the third row of the matrix are zero, and it is well known that zero never equals one.

Thus, the solution set is empty, or the given system of linear equations has no solution.