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### Course: Linear algebra > Unit 2

Lesson 6: More determinant depth- 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

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# Determinant when row is added

The determinant when one matrix has a row that is the sum of the rows of other matrices (and every other term is identical in the 3 matrices). Created by Sal Khan.

## Want to join the conversation?

- Does the determinant have to be taken along the row being added? In other words if in the example given the determinant was calculated for the first row, although the second rows were the ones added, would you still end up with the sum of the determinants?

I haven't finished checking for a 3 X 3 but I suspect the answer is yes.(2 votes)- should be the same regardless of row chosen, just find a row that quickest to work with i.e. most zeroes so theres less calculations and less room for error(2 votes)

- Why is this useful? Is it just a lemma for a more substantial proof?(2 votes)
- When you are working with a big matrix and you have already calculated the determinant once, it's much easier to use this lemmas if something modifies your matrix than to have to calculate the determinant all over again.(2 votes)

- (at4:50) I know that this point came up many times in the previous videos, but I am having hard time understanding how he got the checker board patterns of pluses and minuses. Can you prove it? Please help!(1 vote)
- You could understand this way:

If A =

|a11 a12 a13|

|a21 a22 a23|

|a31 a32 a33|

then, the determinant is calculated by:-

Expanding along Column 1

(-1)^1+1 * a11(a22 * a33 - a32 * a23) +

(-1) ^1+2 * a12(a21 * a33 - a31 * a23) +

(-1)^1+3 * a13(a21 * a32 - a31 * a22)

Here in (-1)^i+j ,

i + j represents the sum of suffix of the element i.e. a11

So here (-1)^1+1

and similarly for all the columns.(3 votes)

- At15:20when he says you can immediately see that det(Z) = det(X) + det(Y), is there math you can perform on the sigma notation to express that? Can x_j and y_j factor out somehow?(1 vote)
- Since each x_j and y_j is different, they can't be factored outside the summation symbol. but for each j you can check that:

(x_j+y_j)|A_ij| = x_j|A_ij + y_j|A_ij|, You then have (leaving out some detail) "sum(x_j|A_ij|) + sum(y_j|A_ij|" and "sum(x_j|A_ij + y_j|A_ij|)", which you can verify are equal for each j (for every term in the summation), which makes the summations equal.(1 vote)

- What is |Aij| in the determinant formula? I can follow the formula, but that term looks strange in general form.(1 vote)
- The determinant of the submatrix when you leave out the ith row and jth column.(1 vote)

- At12:52Sal says x_j did he mean x_ij? shouldn't the i still be part of the summation even though he described i as a row separately? Also should the |A_ij| be |(X,Y,Z)_ij| respectively?(1 vote)
- I agree. "i" can still be a subscript even though it doesn't vary, as he did with "A_ij".

I used "X_ij" for all three, instead of "A_ij". But I think it's good to make them all the same to make it clear that all three are the same.(1 vote)

## Video transcript

Let's keep messing with our
determinants to see if we can get more useful results. And they might not be obviously
useful right now, but maybe we'll use them later
when we are exploring other parts of linear algebra. So let's say I have
some matrix, let's call it matrix X. Matrix X is equal to-- I'll just
start with a 3 by 3 case because I think the 2 by 2
case is a bit trivial. Actually, why don't I just
start with a 2 by 2 case. Let's say matrix X is a, b,
and then it has x1, x2. I could have called these c and
d, but you'll see why I called them x1 and
x2 in a second. And I'll say I have
another matrix. Let's say matrix Y is identical
to matrix X except for this row. So matrix Y is a, b y1 and y2. And let's say we have
a third matrix Z. That's identical to the first
two matrices on the first row. So a, b. but on the second row it's
actually the sum of the two rows of x and y. So it's going to be, this
entry's going to be x1 plus y1, and this entry right
here is x2 plus y2. Just like that. I want to be very clear,
Z is not X plus Y. All of the terms of Z are
not the sum of all the terms of X and Y. I'm only focusing on
one particular row. And this is just a general theme
that you'll see over and over again, and we saw it in
the last video and I guess you'll see it here, is that
determinants or finding the determinants of matrices
aren't linear on matrix operations, but they are linear
on operations that you do just to one row. So in this case, everything else
is equal except for this row, and Z has the same first
row as these guys, but its second row is the sum of the
second row of these guys. So let's explore how
the determinants of these guys relate. So the determinant-- let
me do it in X's color. The determinant of X-- I'll
write it like that-- is equal to a ax2 minus bx1. You've seen that
multiple times. The determinant of Y is equal
to ay2 minus by1. And the determinant of Z is
equal to a times x2 plus y2 minus b times x1 plus y1, which
is equal to ax2 plus ay2-- just distributed the
a-- minus bx1 minus by1. And if we just rearrange things,
this is equal to a-- let me write it this
way-- this is equal to ax2 minus bx1. That's that term and that term,
we switched colors. So that's those two guys. And then plus ay2 minus by1. Now what is this right here? That is the determinant of X. And this right here is
the determinant of Y. So there you have it. If we have matrices that are
completely identical except for one row-- and in this case
it's a 2 by 2 matrix, so it looks like half of the matrix--
and Z's, that row that we're referring to that's
different, Z's is the sum of the other two guys' rows, then
Z's determinant is the sum of the other two determinants. So this is a very
special case. I want to keep reiterating it. It only works in the case where
this row and only this row is the sum of this row and
this row, and the matrices are identical everywhere else. Let me show you the 3 by 3 case,
and I think it'll be a little bit more general. And then we'll go to n by n. The n by n is actually, on some
level, the easiest to do, but it's kind of abstract so I
like to save that for the end. So let's redefine all those
guys into the 3 by 3 case. So let's say that X is equal
to a, b, c-- let's just do, let's make the third row the
row we're going to use to determine our determinant. a,
b, c, d, e, f-- actually let me do the middle row, because I
don't want to make you think it always has to be
the last row. So let's say it's x1, c2, x3,
and you have d, e, f. And what's the determinant
of X going to be? The determinant of X is going
to be equal to-- let's say we're going along this
row right here, that's the row in question. It's going to be equal to--
well you remember your checkerboard pattern-- so it's
going to be-- remember, plus, minus, plus, minus, plus--
you remember all the rest how it goes. So it's going to start with
a minus x1 times the sub matrix-- you get rid of that
column, that row-- b, c, e, f. Then you have plus x2 times the
sub matrix-- get rid of that column, that row--
a, c, d, f. And then finally minus x3--
you get rid of its row and column-- you have a, b, d, e. Now let me define another matrix
Y that is identical to matrix X, except for that row. So it's a, b, c. Then down here d, e, f. That middle row is different. It's y1, y2, and y3. What's the determinant
of Y going to be? Determinant of Y? Well it's going to be identical
to the determinant of X because all the
sub-matrices are going to be the same when you cross
out this row and each of the columns. But the coefficients are
going to be different. Instead of an x1,
you have a y1. So it's going to be equal to
minus y1 times the determinant b, c, e, f plus y 2 times the
determinant of a, c, d f minus y3 times the determinant
of a, b, d, e. I think you see where
this is going. Now I'm going to create
another matrix. I'm going to create another
matrix Z just like that that is equal to-- it's identical to
these two guys on the first and third rows, a,
b, c, d, e, f. Just like that. But this row just happens
to be the sum of this row and this row. And when we figured out this
determinant we went along that row-- you can see that
right there. So this row right here is going
to be x1 plus y1, that's its first term. x2 plus y2, and then you
have x3 plus y3. Now what's the determinant
of Z going to be? Well, we can go down this
row right there. So it's going to be minus x1
plus y1 times its sub-matrix-- get rid of that row, that
column-- you get b, c, e, f. I think you definitely see
where this is going. Plus this coefficient, plus x2
plus y2 times its sub-matrix-- get rid of that row, that
column-- a, c, d, f. And then you have minus this
guy right here, x3 plus y3 times its sub matrix-- get
rid of that column and row-- a, b, d, e. Now what do you have
right here? This is the determinant of Z. This right here is the
determinant of Z. This thing right here. I think you can see immediately
that if you were to add this to this you would
get this right here, right? Because you have this
coefficient and this coefficient on that. If you added them up you would
get minus x1 plus y1. This guy and this guy
add up to this guy. And then if I were to do
this guy and this guy, add up to that guy. Let me do another one. And then finally that
term plus that term add up to that term. So you immediately see that the
determinant, or hopefully you immediately see, that the
determinant of X plus the determinant of Y is equal
to the determinant of Z. So we did it for the 2 by
2 case, we just did it for the 3 by 3 case. Might as well do it for
the n by n case so we know that it works. But the argument is identical
to this 3 by 3 case. So that's good to keep in your
mind because 3 by 3 is easy to visualize, n by n is sometimes
a little bit abstract. So let me re-define
my matrices again. I'm just going to do the
same thing over again. So I'm going to have
a matrix X. But it's an n by n matrix. So let me write it this way. Let's say it is a 1, 1, a 1,
2, all the way to a 1, n. And there's some row here, let's
say that there's some row here on row i-- let's call
this row i right here-- and here it has the terms x1, x2,
all the way to xn, but everything else is just
the regular a's. So then you have a-- let
me make this as a21, all the way to a2n. And then if you went all the way
down here you would have an1, and you'd go all
the way to ann. So essentially you could imagine
our standard matrix where everything is defined in
a, but I replaced row i with certain numbers that are maybe
a little different. And I think you'll see
where I'm going. Now let me define
my other matrix. Let me define matrix Y. Let me define matrix Y to be
essentially the same thing. This is a11-- it's
the same a11. This is a12, all
the way to a1n. This is a21, we could go
all the way to a a2n. And then on row i, the same row,
this is n by n, this is the same n by n-- if this was
10 by 10, this is 10 by 10. If this is row seven, then
this is row seven. It has different terms. It's
identical to matrix X except for row i. In row i it is y1, y2,
all the way to yn. And if you keep going down, of
course, you have an1, all the way to ann. Fair enough. Now let's say we have
a third matrix. Let's have a third matrix. Let me draw it right here. So you have Z, Z is equal to--
I think you could imagine where this is going. Z is identical to these two
guys except for row i. So let me write that out. So Z looks like this. You have a11, a12, all
the way to a1n. And then you go down and then
row i happens to be the sum of the row i of matrix
X and matrix Y. So it is x1 plus y1, x2 plus y2,
all the way to xn plus yn. And then it you keep going
down, everything else is identical, an1 all
the way to ann. So all of these matrices are
identical except for row X has a different row i than
matrix Y does. And row Z is identical
everywhere except its row i is the sum of this row
i and that row i. So it's a very particular case,
but we can figure out their determinants. So what are the determinants? The determinant of X, the
determinant of matrix X-- and hopefully you're maybe a bit
comfortable with writing sigma notation, we did this
in the last matrix. We can go down this row right
there, and for each of these guys we can say so the
determinant is going to be equal to the sum. Let's say we start from j is
equal to 1-- j's going to be the column, so we're going to
take the sum of each of these terms from j is equal
to 1 to n. And then remember our
checkerboard pattern, so we don't know if this is a
positive or negative. We can figure it out by taking
negative 1 to the i plus j-- remember, this is the ith row
that we're talking about-- times xj-- xj is the
coefficient, xsubj, times the sub matrix for xsubj. So if you get rid of this
guy's row and this guy's column, what is it
going to be? We could say that that's the
same thing as the sub matrix, if we called this guy-- let me
write it this way-- if we got rid of this guy's row and this
guy's column, if we had just our traditional matrix where
this wasn't replaced. If we just had an ai1 here, ai2,
its sub matrix would be the same thing, because we're
crossing out this row and this column. So it would be all of these
guys and all of these guys down here. So it would be the sub matrix--
this is a n minus 1 by n minus 1 matrix-- it would
be the sub matrix for aij. That's for the first term--
sorry, the determinant. Don't want to lose the
determinant there-- times the determinant of the
sub matrix aij. And so that's for the first
term, and then you're going to add it to the second term, and
then you're just going to keep doing that. That's what this sigma
notation is. That's the determinant of X. Now what's the determinant
of Y? The determinant of Y is equal
to the sum-- we could do the same thing-- j is equal to
1 to n of negative 1 to the i plus j. We're going to go along this row
right here, the ith row. So we're going to have ysubj--
right, we're going to start with ysub1, then plus ysub2,
times the determinant of its sub matrix, which is
the same as the determinant of this sub matrix. So you get rid of that row and
that column for each of these guys, that everything else on
the matrix is the same. So aij. The matrix of aij. Now what is the determinant
of Z. I'm pretty sure you know exactly
where this is going. The determinant-- this should
be a capital Y right there-- the determinant of Z is equal to
the sum, from j is equal to 1 to n, of negative
1 to the i plus j. We're going along this row. But now the coefficients are xj,
that's what we're indexing along, xj plus yj. And then times its sub matrix,
which is the same as these sub matrices. So aij, which you might
immediately see is the sum of these two things. If I, for every j, I just summed
these two things up, you're having two coefficients--
you could have this coefficient and that
coefficient on your aij term, and then when you add them up
you can factor this guy out and you will get this
right here. So you get the determinant of X
plus the determinant of Y is equal to the determinant of Z. So hopefully that shows
you the general case. But I want to make it very
clear, this is just for a very particular scenario where three
matrices are identical, except for on one row. And one of the matrices on that
special row just happens to be the sum of the other two
matrices for that special row, and everything else
is identical. That's the only time where the
determinant of-- not the only time, but that's the only time
we can make the general statement where the determinant
of Z is equal to the determinant of X plus
the determinant of Y. It's not the case-- let me write
what is not the case-- so not the case that if Z is
equal to X plus Y, it is not the case that the determinant
of Z is necessarily equal to the determinant of X plus
the determinant of Y. You cannot assume this. Determinant operations are not
linear on matrix addition. They're linear only on
particular rows getting at it. Anyway, hopefully you found
that vaguely useful.