 In this video, we are going to provide the solution to question number four from the exam one in math 2270 in this question. We are told that the following augmented matrices correspond to systems of linear equations, which are the following augmented matrices correspond to consistent systems. So because it's consistent, we're looking for those which have a solution. Now we can quickly identify whether a matrix is consistent or inconsistent from its echelon form. If we look at choice I, choice number one there, that matrix is not an echelon form, but I can also notice that this matrix is homogeneous. Every homogeneous system is consistent. That is every homogeneous system has a solution because I could just take the zero vector 0000 as a solution that would solve that one. So choice number one is a consistent system because it is homogeneous. Then I look at choice number two, I can see that this matrix is an echelon form, right? You see the pivots in the first and second column, the third column has no pivots whatsoever. So if it was consistent, it would have multiple solutions to it to determine consistency, though I have to be looking for contradictions or not. There is a row of zeros, which does give me some concern, right? Because if you have a row of zeros, that might be an inconsistency, it might be like a contradiction. But this would just translate to the equation zero equals zero, which that's just an identity that's always true. And so that actually has no bearing on the solution set whatsoever. But what we can see is that this matrix in echelon form has no contradictions. Therefore it will be a consistent system. It'll have multiple solutions, but it will be consistent. And so therefore choice number two is also consistent. Now, just because one of the choices or two of the choices consistent doesn't mean anything about the third one. This is completely independent of each other. If we go on to matrix three, we can also see that this one's an echelon form. So there's a, so there is a pivot in the first and second column. Likewise, there's no pivot in the third column. So if this matrix is consistent, if the system is consistent, then it would have multiple solutions. But unlike choice two, though, we do see a stark difference right here. If you look at the last equations, which there is a row of zeros, this would suggest that zero equals negative one. And that actually is a contradiction telling us that choice three is inconsistent. There is no solution because there is no choice of the variables x one x two x three that can force zero to equal negative one. And so one choice one and choice two are in fact consistent choice three is not. So we look for that option that is that is one and two, and that's going to be choice D right here. So D is the correct answer.