- Determinant when row multiplied by scalar
- (correction) scalar multiplication of row
- Determinant when row is added
- Duplicate row determinant
- Determinant after row operations
- Upper triangular determinant
- Simpler 4x4 determinant
- Determinant and area of a parallelogram
- Determinant as scaling factor
Correction of last video showing that the determinant when one row is multiplied by a scalar is equal to the scalar times the determinant. Created by Sal Khan.
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- Good job making a correction. Unfortunatly you made another error this time: You forgot the signs, (-1)^(i+j), on every term after the first. Maybe it's because you confused some definitions, because in my material the determinant is defined as: det(A)=a_11*A_11+...+a_1n*A_1n where A_ij=(-1)^(i+j)*det(M(A)_ij) where M(A)_ij is the ij-minor of A. Is the definition Sal uses the most common or does it differ from country to country? :O(32 votes)
- I don't think an answer after such a long time would do you any good, however, I can assure you that it's the same in all the countries where determinants is taught. :)(6 votes)
- why does it say correction in the title?(1 vote)
- He made a computational error while finding the determinant of a n x n matrix in the previous video.(1 vote)
- At 1: 12, what is submatrix?(1 vote)
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.