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# (toʻgʻrilash) satrning skalyar koʻpaytmasi

## Video sharhi

i want to make a quick correction or clarification to the last video that you may or may not have find compound 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 1 a i 2 all the way to a i n and then we keep going down assuming that this is the last row so a and 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 that's what i wrote in the last so if you have if you have a y 1 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 3x3 case that's pretty clear it's not times the matrix it's times the determinant of the sub matrix so this right here is incorrect and of course you keep adding that too and i wrote a i a i2 times its sub matrix like that a i2 all the way to a i n times its sub matrix that's what i did in the video that's incorrect let me 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 the determinant of this guy is equal to the determinant of a is equal to minus 1 to the i plus 1 times a i 1 times the determinant the determinant of a i 1 plus a i 2 times the determinant of a i2 the determinant of the sub matrix all the way to a i n times the determinant of the sub matrix a i n 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 well it's not that bad it's a scalar but we're what 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 determinant but i just wanted to make that very clear