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- (toʻgʻrilash) satrning skalyar koʻpaytmasi
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(toʻgʻrilash) satrning skalyar koʻpaytmasi
Bitta satr skalyar koʻpaytirilganda skalyar marta determinantga teng ekanidagi determinantni koʻrsatuvchi oxirgi videoni toʻgʻrilash. Salmon tomonidan yaratilgan.
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Hozircha izohlar yoʻq.
I want to make a quick correction or clarification to the last video that you may or may not have found confusing. You may not have noticed it, but when I did the general case for multiplying a row by a scalar, I had this situation where I had the matrix A and I defined it as-- it was n by n matrix, so it was a11, a12, all the way to a1n. Then we went down this way. Then we picked a particular row i, so we called that ai1, ai2, all the way to ain. And then we keep going down , assuming that this is the last row, so an1 all the way to ann. When I wanted to find the determinant of a, and this is where I made a-- I would call it a notational error. When I wanted to find the determinant of a, I wrote that it was equal to-- well, we could go down, and in that video, I went down this row. That's why I kind of highlighted it to begin with, and I wrote it down. So it's equal to-- do the checkerboard pattern. I said negative 1 to the i plus j. Well, let's do the first term. I plus 1 times ai1 times its submatrix. That's what I wrote in the last. So if you have ai1, if you get rid of that row, that column, you have the submatrix right there: ai1. That's what I wrote in the last video, but that was incorrect. And I think when I did the 2 by 2 case and the 3 by 3 case, that's pretty clear. It's not times the matrix, it's times the determinant of the submatrix, so this right here is incorrect. And, of course, you keep adding that to-- and I wrote ai2 times its submatrix like that. ai2 all the way to ain times its submatrix. That's what I did in the video. That's incorrect. Let me do the incorrect in a different color to show that this is all one thing. I should have said the determinant of each of these. The determinant of a is equal to minus 1 to the i plus 1 times ai1 times the determinant of ai1 plus ai2 times the determinant of ai2, the determinant of the submatrix all the way to ain times the determinant of the submatrix ain. It doesn't change the logic of the proof much, but I just want to be very careful that we're not multiplying the submatrices because that becomes a fairly complicated operation. Well, it's not that bad. It's a scalar. But when we find a determinant, we're multiplying times the determinant of the submatrix. We saw that when we first defined it using the recursive definition for the n by n determinant, but I just wanted to make that very clear.