Let's explore what happens to determinants when you multiply
them by a scalar.
So let's say we wanted to find the determinant of this
matrix, of a, b, c, d.
By definition the determinant here is going to be equal to a
times d minus b times c, or c times b, either way.
ad minus bc.
That's the determinant right there.
Now what if we were to multiply one of
these rows by a scalar.
Let's say we multiply it by k.
So we have the situation a, b.
Let's multiply the second row times k.
So we kc and kd.
Now what's the determinant going to look like?
We're going to have a times kd, or we could just write
that as kad, minus kc times b, or we could right that as kbc.
If we factor out the k, we get that equals k
times ad minus bc.
Now you immediately see that this thing is the same thing
as this thing.
So this is equal to k times the determinant of a, b, c, d.
So when you just multiply a row by a scalar-- just one
row, not the entire matrix-- when you just multiply a row
by some scalar, the resulting determinant will be the
original determinant times that scalar.
Now you might say well what happens if I multiply the
whole matrix times that scalar.
Well that's equivalent to multiplying by a scalar twice.
Let's say I have the matrix A, and the matrix A is equal to
a, b, c, d.
If I were to think about the matrix kA, now I'm not just
multiplying one row.
I'm multiplying the whole matrix by a scalar.
This is going to be equal to ka, kb, kc, and kd.
When you figure out its determinant, the determinant
of k times A is going to be equal to the determinant of
ka, kb, kc, and kd.
Here you're immediately going to end up with k squared
terms. You're going to have k squared times ad.
It's going to be equal to k squared ad minus k squared bc,
or k squared times ad minus bc, or k squared times the
determinant of just A.
So you have to be very careful.
This is only for a two by two case.
You'll find out if this was an n by n matrix that this would
have been k to the n.
So the take away is the only way you can say that this is
going to be some scalar multiple times your original
determinant, is only if you multiply one row times that
scalar multiple, not the whole matrix.
Let's see how this extends to maybe a 3 by 3 case.
You might say, hey Sal, you just picked the second row.
Does it work with the first row?
I'll leave that for you to determine, but it does.
It does work.
It doesn't matter which row I multiplied it by.
Let's take the three by three case.
Let's say we have some matrix.
Let's call this A again.
I'm redefining A.
It's going to be a, b, c, d, e, f, g, h, i.
Then if you take its determinant, the determinant
of A is going to be equal to-- we can do it a couple of
different ways.
But I'll just pick some arbitrary row.
Because that's the row that we're going to multiply by
some scalar.
So let's just take that row right there.
Remember the plus minus pattern.
Remember plus, minus, plus, minus, plus, minus, plus,
minus, plus, that little checkerboard pattern.
So d is a minus right there.
So it's going to be equal to minus d times the determinant
of its submatrix.
So you cross out that column in that row.
It's b, c, h, i.
It's going to be plus e times its submatrix a, c, g, i.
It's going to be minus f times-- you get rid of that
row in that column-- d, e, g, h, the
determinant of d, e, g, h.
That's the determinant of this matrix A.
Now what if we define some new matrix here?
Let's call it A prime.
Let me make this scroll down a little bit.
Let me define A prime right here.
A prime, I'm just going to multiply this row by a scalar.
So it's going to be equal to a, b, c, kd, ke, and kf.
I'm not multiplying the whole matrix times the scalar.
I can't say this is kA.
I'm just multiplying one of its rows.
Then I have g, h, and i.
So what's the determinant of A prime going to be?
I put that prime there.
So it's different than A, but it's derived from A.
I just multiplied one row of A times the scalar.
Well I can go along that same row that I did up here.
I can go along that same row.
The only difference is is that instead of having a d,
I now have a kd.
Instead of an e, I now have a ke.
So instead of a d, I'm going to have a kd there.
Instead of an e, I'm going to have an ke there.
So it's going to be this exact same thing, but I can replace
this guy, this guy, and this guy with them multiplied by k.
So it's going to be equal to minus kd times the determinant
of the submatrix b, c, h, i-- I'm not even look over here
because it's going to be the same thing as that one up
there-- plus ke times the determinant of a, c, g, i,
minus kf times the determinant of d, e, g, h.
What is this equal to?
This is equal to, if you just factor out the k, it's equal
to k times this.
So it's equal to k times the determinant of A.
So our result also worked for three by three case.
I just happened to pick the middle row.
But I encourage you to pick other rows
and to see what happens.
So let's actually do it for the general case.
Because I've just been giving you particular examples, and I
like to show you the general proof when the general proof
isn't too harry.
So let's say I have an n by n matrix.
Let's say that I have matrix A.
Let's say that A is n by n.
You can write it like this.
This is the first row.
This is the first column a11, a12, all the way to a1n.
I'm going to pick some arbitrary row here that I'm
going to end up multiplying by a scalar.
So we can go down here.
Let's say row ai, so this is ai1, ai2, all the way to ain.
This is some row that I'm going to use to determine the
determinant.
Remember we can go to any row to get the determinant.
Then finally you keep going.
You get an1, an2, all the way to ann.
This is as general as you can get for an n by n matrix.
Now let's figure out its determinant, so the
determinant of A.
I'm just going to go down this row right there.
So the determinant of A is equal to what?
Well we have to remember the checkerboard pattern.
We don't know where we are on the checkerboard pattern
because I just picked an arbitrary general row here.
But we can use the general formula that the sign is going
to be determined by negative 1 to the i-- I don't know if i
is even or odd-- so it's going to be i plus
for this term 1 power.
That's it's sign.
This is what gives us the checkerboard pattern.
Let me make that clear.
It looks complicated, but this is just a checkerboard.
Times this term right there, so times ai, so the
coefficient, ai1, and then times this guy's submatrix.
You remember the submatrix.
You get rid of this row, and this column is going to be
everything that's left over.
So times that submatrix of ai1.
Then it's going to be plus-- we just keep doing it-- plus
negative 1 to the i plus 2 times ai2 times its
submatrix-- all the way, you just keep going-- plus minus 1
to the i plus n times ai.
You're in the nth column, and then its submatrix.
This is going to be an n minus 1 by n minus 1 matrix.
All of these are going to be.
Just like that, that's the determinant of A.
We could actually rewrite it in sigma notation.
That'll simplify things a little bit.
So the determinant of A, we can rewrite it as the sum from
j is equal to 1 to j-- I'll write it explicitly here-- j
is equal to n of negative 1 to the i plus j times aij, and
then each of the submatrices, Aij.
This thing right here is just another way of writing this
thing I wrote up there.
I'm just saying the sum.
You just take j equal 1, put them in there.
You get this term right there.
You take j equal 2, you add it.
You get this term right there.
You keep doing it.
You get j equal n.
You get that term right there.
So these two things are equivalent.
So what happens I have some new matrix?
Let me copy and paste my current matrix.
So let me copy and paste it.
Actually let me copy and paste everything.
That'll make things move quickly.
I copied it.
Now let me paste it just like that.
Let me define a new matrix A prime.
It's still an n by n matrix.
But that row that I just happened to use to determine
my determinant, I'm going to multiply it by a scalar k.
So it's k ai1, k ai2 k ain just like that.
So what's the determinate of A prime?
Well we're just going to go down this row again.
But now instead of just an ai1, we have a k ai1 Instead
of an ai2, we have a k ai2 Instead of an
ain we have a k ain.
So it's determinant is just going to be this same thing,
but instead of an aij everywhere, we're going to
have a k aij.
So this is the determinant of A prime.
We could just take out this constant right here.
It has no i's or j's in it.
I has no j's in it in particular, so we can just
take it out.
So it's equal to k times the sum from j is equal to 1 to j
is equal to n of minus 1 to the i plus j times aij.
This is the coefficient.
This is the submatrix for each of those coefficients, aij.
That's a matrix right there, an n minus 1
by n minus 1 matrix.
Then you immediately recognize-- I think you saw
where this was going-- this right here is just the
determinant of A.
So we get the result that the determinant of A prime is
equal to k times the determinant of A.
So we've just shown you in general, if you have any n by
n matrix, if you multiply only one row, not the whole matrix,
only one row by some scalar multiple k, the resulting
determinant will be your original determinant times k.
Now I touched on this in the original video.
What is the determinant of k times A?
So now we're multiplying every row times k.
Or another way to think about is you're multiplying
n rows times k.
So you're doing this n times.
So if you multiply k times itself n
times, what do you get?
You get k to the n.
So this is going to be equal to k to the n times the
determinant of A.
If you just do it once, you get k times the
determinant of A.
Now if you do a second row, you're going to get k times k
times the determinant of A.
If you do a third row, you're going to get k to the third
times the determinant of A.
The fourth row, k to the fourth times the
determinant of A.
If you do them all, all n rows, you're going to have k
to the n times the determinant of A.
Anyway, hopefully you found that interesting.
I encourage you to experiment with these other ways.
Try going down a column and seeing what happens.