Welcome to a lesson
on how to write a matrix
as a product of elementary matrices.
If E is an elementary matrix
by applying one elementary row operation
to an identity matrix,
and A is an n by n matrix,
the idea is if we start with matrix A
then determine what elementary matrices
we have to multiply by here on the left,
to obtain
the identity matrix on the right
if we solve this equation for A
we could determine which elementary matrices
would have a product of matrix A.
Remember the inverses of elementary matrices
are also elementary matrices.
So if we wanted to solve this equation for A
the first step would be to multiply
both sides of the equation by the inverse
of matrix Ek.
Which we see here, on the right side of the equation.
And then we just keep multiplying by the inverses
of the elementary matrices on the left
until we finally multiply by the inverse of E2
which we see here in this line
and the last step would be to multiply
by the inverse of E1 which we see here on this line.
Last thing we need to recognize
is that multiplying by the identity matrix here
is really just like multiplying by one
so we can drop this identity matrix.
So it's this product here
that we use to determine which elementary matrices
would have a product of A or a given matrix.
So let's take a look at an example.
We want to find a sequence of elementary matrices
whose product is matrix A.
So to set this up we're going to start with this
2 by 2 identity matrix.
If we multiply this by matrix A
we know we would obtain the given matrix.
And now we're going to perform raw operations
on this matrix here to obtain the 2 by 2 identity
matrix on the right.
We'll perform those same raw operations
on this identity matrix here on the left.
So for example,
if we wanted to start by having a 1 in this position here
we can replace row 1 with -1/2 times row 1.
Let's go ahead and do that.
So we have -1/2 times row 1.
We'll perform this operation on the matrix on the right.
Then we'll perform the same row operation
on the identity matrix here on the left.
Which will give us our first elementary matrix.
So this will be positive 1 and then -2
second row stays the same.
And then on the left side we still have matrix A
and to form our first elementary matrix
we're going to multiply row 1 by -1/2.
So we'd have -1/2, 0, and the second row stays the same.
We'll go ahead and call this matrix E sub 1.
Now on the right side if we want this element here
to be zero we can replace row 2
with -5 times row1 plus row 2.
First row stays the same.
Second row -5 times 1 + 5 that's 0.
5 times -2 that's positive 10
plus -14 that's -4.
And on the left side we already have E sub 1
times matrix A,
but now we'll have another elementary matrix here
that we obtained by performing the same
raw operation here on the original identity matrix.
So we're going to replace row 2 here with -5 times row 1
plus row 2.
So first row stays the same.
Second row will be -5 times 1 plus 0 that's -5.
And then we have -5 times 0 plus 1 is still 1.
This is our E sub 2 or elementary matrix sub 2.
We'll continue this on the next slide
but so far what we have found
is that E sub 2
times E sub 1 times matrix A
would give us the matrix on the right
which is 1, -2, and 0, -4.
We want to keep reforming row operations
until this matrix here is the identity matrix.
Now let's obtain a 1 in this position here.
So we're going to replace row 2 with -1/4 times row 2.
First row stays the same, second row's going to be 0,1.
On the left side we already have E sub 2
times E sub 1 times A.
We're going to multiply this by another elementary matrix.
Which will form by replacing row 2 of the identity matrix
with -1/4th times row 2.
So we'll have 1, 0, 0, -1/4.
This will be elementary matrix step 3.
Looks like we have one more row operation to perform.
We have to throw in a 0 in this position here
we can replace row 1 with 2 times row 2 plus row 1.
So we have 2 times 0 plus 1 that's 1.
Here we have 2 times 1 is 2 plus -2 that's 0.
On the left side
we already have this product.
We need to multiply by one more elementary matrix
that is formed by replacing row 1
with 2 times row 2 plus row 1.
That'd be 2 times 0 plus 1 that's 1,
2 times 1 is 2 plus 0 is 2,
second row stays the same.
So this is E sub 4.
So what we have found is E sub 4
times E sub 3 times E sub 2
times E sub 1 times matrix A is equal to
the 2 by 2 identity matrix.
Which means if we want the elementary matrices
that have a product equal to matrix A
matrix A is going to be equal to
the inverse of E sub 1 times the inverse
of E sub 2 times the inverse of E sub 3
times the inverse of E sub 4.
We can leave off the identity matrix
because remember that's like multiplying by 1.
So this is the product that we're being asked to find.
In order to find these inverse matrices
we first have to go back and find E sub 1 through E sub 4.
So here's E sub 1, here's E sub 2,
here's E sub 3, and here's E sub 4.
So here are the elementary matrices that we just identified.
But the product that we need
is going to involve the inverses
of these matrices which I've already found here.
And since the given matrices were 2 by 2 matrices
remember we can use this shortcut here
to determine the inverses of the 2 by 2 matrices.
So again matrix A is equal to E sub 1 inverse
times E sub 2 inverse times E sub 3 inverse
times E sub 4 inverse.
So just to write this out,
E sub 1 inverse is -2, 0, 0, 1.
E sub 2 inverse is 1, 0, 5, 1.
E sub 3 inverse is 1, 0, 0, -4.
And E sub 4 inverse is 1, -2, 0, 1.
Okay that's all we have time for this video.
I hope you found this helpful.