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I want to make a quick[br]correction or clarification to

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the last video that you may or[br]may not have found confusing.

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You may not have noticed it,[br]but when I did the general

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case for multiplying a row by a[br]scalar, I had this situation

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where I had the matrix A and I[br]defined it as-- it was n by n

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matrix, so it was a11, a12,[br]all the way to a1n.

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Then we went down this way.

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Then we picked a particular row[br]i, so we called that ai1,

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ai2, all the way to ain.

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And then we keep going down ,[br]assuming that this is the last

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row, so an1 all the[br]way to ann.

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When I wanted to find the[br]determinant of A, and this is

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where I made a-- I would call[br]it a notational error.

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When I wanted to find the[br]determinant of A, I wrote that

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it was equal to-- well, we could[br]go down, and in that

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video, I went down this row.

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That's why I kind of highlighted[br]it to begin with,

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and I wrote it down.

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So it's equal to-- do the[br]checkerboard pattern.

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I said negative 1[br]to the i plus j.

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Well, let's do the first term.

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I plus 1 times ai1 times[br]its submatrix.

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That's what I wrote in the last.[br]So if you have ai1, if

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you get rid of that row, that[br]column, you have the submatrix

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right there: ai1.

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That's what I wrote[br]in the last video,

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but that was incorrect.

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And I think when I did the 2 by[br]2 case and the 3 by 3 case,

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that's pretty clear.

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It's not times the matrix, it's[br]times the determinant of

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the submatrix, so this right[br]here is incorrect.

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And, of course, you keep adding[br]that to-- and I wrote

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ai2 times its submatrix[br]like that.

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ai2 all the way to ain[br]times its submatrix.

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That's what I did[br]in the video.

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That's incorrect.

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Let me do the incorrect in a[br]different color to show that

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this is all one thing.

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I should have said the[br]determinant of each of these.

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The determinant of A is equal[br]to minus 1 to the i plus 1

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times ai1 times the determinant[br]of ai1 plus ai2

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times the determinant of ai2,[br]the determinant of the

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submatrix all the way to ain[br]times the determinant of the

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submatrix ain.

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It doesn't change the logic of[br]the proof much, but I just

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want to be very careful that[br]we're not multiplying the

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submatrices because[br]that becomes a

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fairly complicated operation.

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Well, it's not that bad.

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It's a scalar.

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But when we find a determinant,[br]we're multiplying

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times the determinant[br]of the submatrix.

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We saw that when we first[br]defined it using the recursive

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definition for the n by n[br]determinant, but I just wanted

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to make that very clear.

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