# 1 row of the reduced row-echelon form is all 0s in a matrix to solve equations. How do you interpret this row and what does it imply about the solution of the system?

*print*Print*list*Cite

Suppose you had the system of equations:

`a+b=3`

`2a+2b=6`

These two equations actually are "the same" (if one is true the other must be true... once you know one of these, the second doesn't give you any new information). So, this system has infinitely many solutions. I can pick anything for `a` , and then find a `b` that makes the equations true. For example:

`a=1`, `b=2`

`a=0`, `b=3`

`a=12`, `b=-9`

Since I can pick `a` to be anything, there are infinitely many solutions.

In matrix form, we have:

`[[1,1,|,3],[2,2,|,6]]`

which, in reduced row echelon form becomes:

`[[1,1,|,3],[0,0,|,0]]`

An entire row of zeros means there are infinitely many solutions. For each row of zeros, there is a variable that you can pick to be anything, and then solve the other variables around your pick.

Now:

if instead you had the system:

`a+b=2`

`a+b=3`

This system has no solution. There is nothing you can pick for `a` and `b` and still manage to make both of these equations true. So there are no solutions. In matrix form, this looks like:

`[[1,1,|,2],[1,1,|,3]]`

which, in reduced row echelon form becomes:

`[[1,1,|,0],[0,0,|,1]]`

Now we have a row with all zeros, but it has a 1 in the "extra column"

When this happens in reduced row echelon form, it means there are no solutions. Literally it means: `0a+0b+0c+...+0z=1` which can't happen, no matter how you pick `a`,`b`, etc