# When using Gaussian elimination to solve a system of linear equations, how can you recognize that the system has no solution? If we are given a system of linear equations, there are a few different situations that can arise. A system can be consistent (there exists some solution) or inconsistent (there are no solutions.) If consistent, the system can be dependent (an infinite number of solutions) or independent (exactly one solution.)

If there are more variables than equations, the system is always consistent and dependent. If the number of equations is the same as the number of variables or greater, then the system can be either consistent or inconsistent.

In order to determine if a system is inconsistent using Gaussian elimination, we create the augmented matrix and perform row operations with the goal of putting the system in row-echelon or reduced row-echelon form.

(a) If the reduced row-echelon form of the augmented matrix consists entirely of rows with leading digit 1 with zeros everywhere else except the last column the system is consistent and independent. This looks like an nxn identity matrix with an nx1 column appended to it.

(b) If the reduced row-echelon matrix has leading digit 1 in every non-zero row with zeros everywhere else besides the last column and has some rows that are all zeros the system is consistent and dependent. The number of rows could exceed the number of variables.

(c) If the reduced row-echelon form of the augmented matrix has a row or rows that consist of all zeros except for the last element, then the system is inconsistent.

For example:

x+y+z=3
x-y-5z=1
2x+3y+5z=-6

The augmented matrix is:

`([1,1,1,|,3],[1,-1,-5,|,1],[2,3,5,|,-6])`

In reduced row-echelon form we get:

`([1,0,-2,|,0],[0,1,3,|,0],[0,0,0,|,1])`

Note the last row in the matrix. This implies that 0x+0y+0z=1, which cannot be true for any real values of x, y, and z. Thus there is no solution.