 In this video, we present the solution to question number 11 from the practice midterm exam number three for Math 2270, in which case we're asked to compute the following determinant. It's four by four determinant we see right here, and we're asked to show all the steps. We can't just throw it in our calculator and just write the right answer and expect this to be correct. Such a solution would give you no credit whatsoever. We need to show all the subsequent steps. Now there's a couple of approaches we could take. We could just cofactor expand across our favorite row or column, in which case if that was the approach I would probably take the first row, because it does have a number one and it has a zero. Zeroes are very nice in that regard. But it turns out using row reduction can help us out a little bit better, right? Because with our row operations, we can simplify the matrix to get into something triangular, probably upper triangular, and then it would be a lot easier to compute the determinant from there. So what we're going to do is taking our first pivot position, we want to get ones below the one in that situation. So I'm going to take row two, and I'm going to add to it two times row one. I'm going to take row three and add to it negative three times row one, and then I'm going to take row four and subtract from it row one. So we're going to get a plus, a plus two, a plus six. You add zero and then you're going to get a plus four right there. For the next one, we're going to get minus three, minus nine, zero, and then minus six. And then for the last one, we're going to go minus one, minus three, minus zero, and minus two. Now, when you do replacement row operations, that is when you just when you replace a row with that with its row plus a multiple of the row, the replacement row operation doesn't change the determinant. So the term is going to equal what it was before, but we have a much simpler matrix this time. So we're going to get zero. You're going to get six minus five, which is one, you're going to get seven and you're going to four plus four, which is eight. For the next row, you're going to get zero. You're going to get five minus nine, which is a negative four. You're going to get a two and you're going to get a five, one minus six. For the third row, you're going to get zero. You're going to get a negative four. You're going to get a two and then you're going to get a negative five like so for which then since we have these all these zeros right here, a couple of things we could do. We could just postpone any calculation of the the determinant at this moment. If we wanted to, we could cofactor expand the first row. We'll get one times the minor, which you see right here and then zeros everywhere else. So that's kind of what you can do. You could you can cofactor expand along the way or you can just wait until the end. So the next thing to do normally would be to just look at the next pivot position and start row reducing right here. But the reason I say normal is because I'm going to take a different approach here, I'm I notice right here that my third and fourth row are identical to each other. Kind of like the Weasley twins there were identical. If you take the fourth row and you subtract from it, the third row, you're going to get a row of zeros in your matrix one, three, zero, two, zero, one, seven and eight, zero, negative four, two and negative five. And then the fourth row becomes zero, zero, zero, zero, which now if we were to cofactor expand across that row, we're going to get zeros times all the minors and we see that the determinant turns out to be zero in this situation. So our original matrix was a singular matrix and we discovered these this row of zeros along the way that gives us a determinant of zero.