 In this video we provide the solution to question number seven from the practice exam number two for math 1050. We're given three augmented matrices which correspond to three systems of linear equations. And so we're asking which of these linear systems corresponding to the three augmented matrices correspond to consistent linear systems. That is which ones have solutions. Now we don't need to distinguish between do we have infinitely many solutions or a unique solution. We need to have a solution to be consistent. So looking at the first one you'll notice that the last column is all just zeros. This is what we could refer to as the homogeneous system of equations there because the right hand side are all the same they're all zero. The homogeneous system is always consistent because you can always just put in as your variables x, y, and z you could always put in zero like if you take for example one time zero plus three times zero times two times zero that adds up to be zero. And so the zero all zero zero zero always is a solution for the homogeneous system. Are there other solutions? I don't really care because we just need to note that there's at least one solution that's what consistent here means. So one is consistent because it's homogeneous. Looking at the next one we do recognize that there's this row of zeros on the bottom. That just means we ignore that row. It doesn't tell us that there's a solution but it also doesn't tell us there's no solution. We should check that because if the number over here were not zero it would have an inconsistency. It would be contradiction. That would mean no solution but let's go on here. Notice that this matrix is also an echelon form. We have a pivot here and a pivot here. And so we have a free variable in the third column. We could set those to be zero and we get solutions that x equals four-thirds and y equals three and then x equals zero that would be a solution to the system. So because the matrix is an echelon form we can very quickly see that it is in fact consistent. Looking at the last one here, number three, number three is not an echelon form but just like the last one we do have a row of zeros on the left hand side. When you look at the other side though we do get this non-zero value and so that's going to give us that contradiction I mentioned just a moment ago. So because of the third row of the matrix we actually have an inconsistency. We have zero equals negative one, which is not possible for the real number system. Therefore, we put an x on that. So we see that the correct solution will be one and two, which then leads us to select choice D.