 In this video, we're gonna talk about the solution to question number six from exam three for math 2270, for which we're asked to compute the determinant of the matrix one, two, three, four. It's a two by two matrix right here. For which if we wanna be able to compute a determinant, some things to notice here is that you can cofactor or expand across any row or column. Rows or columns with lots of zeros are very helpful. This doesn't have such a thing right here. But after all, it's just a two by two determinant here. So in general, if you have a two by two matrix, A, B, C, D, we know very well that the determinant is just basically this cross product of the diagonals there. You're gonna end up with A, D minus B, C. So we're just gonna do that calculation right here, for which case we get one times four minus two times three. Let me make that look like multiplication there. So we end up with four minus six, we get negative two and do pay attention to the sign. The sign does matter. We see that the correct response will be A, the determinant is negative two.