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Current time:0:00Total duration:2:52

I want to make a quick correction or clarification to the last video that you may or may not have find 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 an N by n matrix so it was a 1 1 a 1 2 all the way to a 1 n then we don't went down this way and then we picked a particular row I so we call that a I 1a i 2 all the way to a I N and then we keep going down assuming that this is the last row so a + 1 all the way to a and N and when I wanted to find the determinant of a this is where I made a I would call it a notational error when I wanted to find the determinant the determinant of a I wrote that it was equal to it was equal to well we could go down in that video I went down this row that's why I kind of highlighted to begin with and I wrote it down so it's equal to 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 a I a I 1 times its sub-matrix side that's what I'd wrote in the last so if you have if you have a y1 you get rid of that row that column you have the sub matrix right there a I 1 that's what I wrote in the last video but that was incorrect and I think you know when I did the 2 by 2 case in the 3 by 3 case that's pretty clear it's not times a matrix it's time's the determinant of the sub-matrix so this right here is incorrect and of course you keep adding that too and I wrote ai ai 2 times its sub-matrix like that a I - all the way to a I n times its sub-matrix that's what I did in the video that's incorrect we do the incorrect in a 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 this guy is equal to the determinant of a is equal to minus 1 to the I plus 1 times AI 1 times the determinant the determinant of a I 1 plus AI two times the determinant of a I to the determinant of the sub-matrix all the way to a I N times the determinant of the sub-matrix AI and it doesn't change the logic of the proof much but I just want to be very careful that we're not multiplying the sub matrices because that becomes a fairly complicated operation what's not that Benja scalar but we find the determinant we're multiplying times the determinant of the sub-matrix we saw that when we first defined it using the recursive definition for the N by n determinate but I just wanted to make that very clear