 In this video, we're going to answer question number two from the practice exam number three for math 2270, for which we're given a matrix A. It's two by two. It consists of the numbers four, one, three, and six. And we're asked, which of the following is an eigenvector of A? So there's sort of two ways we could approach this question. One, we could actually go and try to compute eigenvalues from the characteristic polynomial, then calculate eigenspaces to find eigenvectors. We could do that, but that's a sort of a lengthy process here. And this is a multiple choice question. What we can do is try to answer the question by process of elimination. After all, we have candidates for the eigenvectors here. Why don't we just multiply them by A? Because after all, if we have an eigenvector, what's going to happen is you're going to get A times x. This should equal lambda times x. So that is, we can multiply this matrix by each of the vectors individually. And once we discover we have a scalar multiple of the vector we started with, that means we found an eigenvector. So let's try this out. Worst case scenario, we have to multiply these things together six times, for which if we have a calculator that supports matrix operations, then we could use that to speed up this process. But we'll just do this one by hand. So if you take the first row times the vector right here, we're going to get 4 minus 2, and then we're going to get 3 minus 12 for the second one right there. I'm going to scooch it over a little bit, for which that then adds up to be 2 and negative 9, for which that is not a scalar multiple of the first vector right there. So we're going to back up, we remove that one from the possibility, then we're going to try 1 and 2, for which that's going to give us 4 plus 2, and then we're going to get 3 plus 12, that gives us 6 and 15, for which that's not a scalar multiple of choice B either. So we remove that one from consideration. Let's try choice C, we take 1 and 0, this is going to give us 4 plus 0, 3 plus 0, that gives us 4 and 3, that does not give us a scalar multiple of choice C, so we remove that one. Let's try D, would you give 1 and 1, for which then here we're going to get 4 plus 1, we're going to get 3 plus 6, that gives us 5 and 9, that's not a scalar multiple of 1, 1, so we'll remove that one from consideration. Now we have to be cautious on choice E right here because the vector in consideration is 0, 0. If I stuck 0, 0 into the situation here, what you're going to get is 0 plus 0, plus 0 and 0, which would add up to be 0, 0, in which case it's tempted to be like, oh that's just 1 times 0, 0, so that's an eigenvector, but it's also 2 times 0, 0, it's also 3 times 0, 0, it's also the square root of pi times 0, 0, this is the problem with the 0 vector. If you have the 0 vector, then any scalar multiple of the 0 vector will give you back the 0 vector, in which case we don't want every scalar necessarily to be an eigenvalue, so the 0 vector is never included in this list of eigenvectors, so don't be deceived if you see the 0 vector automatically excluded. So it would seem by process of elimination, f is probably the correct answer, but we might be, we should check it anyways just to make sure because who knows, maybe we made a mistake on one of the previous ones, maybe our arithmetic was too rushed or just human error, so if we try f right here, what do we see? With the matrix product you're going to get 4 minus 1 and you're going to get 3 minus 6, for which that then gives you 3 in minus 3, which notice that that is actually a scalar multiple of f, if you factor out the 3 you get back 1 and negative 1, so we see that 1 negative 1 is an eigenvector associated to the eigenvalue 3, and so on this question choice f was the correct response, and we found that just by elimination.