Linear Algebra: (correction) scalar muliplication of row

Correction of last video showing that the determinant when one row is multiplied by a scalar is equal to the scalar times the determinant

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.