 In this video, we present the solution to question number 10 from the practice exam number three for math 2270. We are given a four by four matrix A that you can see right here. We are also given the following useful information that A augment the identity will row reduce to the following four by eight matrix right there. It's a four by four matrix augment with a four by four matrix. And then we're asked to find a basis for the left null space of A. So to find a basis for the left null space of A, you want to find the RREF of A, which you can see that is right here. And then when A is augmented with the identity, the appropriately sized identity, which in this case would be four by four, we want to see what trans the transformations that changed A into its row reduced echelon form. What does that do to the identity? We see that right here. And so then looking at the RREF of A, let's look for any rows of zeros. If A doesn't have any rows of zeros, which is a possibility, it would mean that the left null space of A is trivial, in which case its basis would be the empty set. We do see there in fact two rows of zeros in the row reduced echelon form of A. Therefore, if we look at the corresponding rows in the, on the identity side of things, the right hand side, these vectors right here on the right hand side of these rows of zeros, these vectors correspond to a basis for the left null space, in which case then we record these things down here, you're going to get two negative three, negative one and zero as the first vector. You can run as rows if you prefer. I'm writing this columns right now. You get negative 15, excuse me, positive 15, negative 12, zero, negative three. And this will then give us a basis for the left null space.