Welcome to a lesson on performing
row operations using elementary matrices.
The goals of this video are to perform
row operations using elementary matrices,
define row equivalent matrices,
and also to solve a system of equations
using elementary matrices.
If E is an elementary matrix, by applying
one elementary row operation to an identity matrix I,
then if A is an n by m matrix,
E times A will perform
the same row operation to matrix A.
Whatever row operation we use to form the elementary matrix,
E times A will perform that same row operation to matrix A.
It's also true that A and E times A are
row equivalent matrices.
To illustrate this property, we're going to write
this system of equations as an augmented matrix,
as we see here, matrix A, and then solve this
by writing this matrix in echelon form,
using elementary matrices.
To get started, we need to recognize that
the 2 by 2 identity matrix times matrix A
will give us our augmented matrix.
Now, we're going to determine what row operations
we would perform on this matrix to put it in echelon form,
and then we'll do the same to the identity matrix
to find our elementary matrices.
If we wanted to have a zero in this position here,
we could replace row 2 with negative 2
times row 2 plus row 1.
We're going to go ahead an perform this row operation
on this augmented matrix and then we'll perform
the same row operation on this identity matrix
to find our first elementary matrix.
The first row stays the same.
Negative 2 times 1 plus 2, that's zero.
Negative 2 times 4 plus 1, that's negative 7.
We have negative 2 times 6, that's negative 12,
plus five, which should be negative 7.
On the left side of this equation,
we're going to perform the same row operation
on this 2 by 2 identity matrix.
The first row stays the same.
For the second row, we have negative 2
times 0 plus 1, that's 1.
Negative 2 times 1 plus 0 is negative 2.
This matrix here is our first elementary matrix.
If we multiply this by the original matrix A,
we would perform the same row operation
as we did here on the right.
This product would be this matrix here.
If we wanted to have a positive 1 in this position here,
we could replace row 2 with negative 1/7 times row 2.
Again, we're going to perform that operation on this matrix,
and then again on this identity matrix here
to find our second elementary matrix.
The first row stays the same.
The second row would be 0, this should be positive 1,
and so would this element here.
Again, on the left side, so far we have
elementary matrix 1 times matrix A,
but now we're going to have another elementary matrix here
that would perform the same row operation
that we performed here, so now we're going to multiply
row 2 of the original identity matrix
by negative 1/7, so we'd have 1, 0,
this would be 0, and negative 1/7.
This product here would perform the same two
row operations that we performed on the right side.
Let's go ahead and call this elementary matrix 2.
Looking back on the right side,
if we wanted this element here to be positive one,
we'd have to replace row 1 with one-half times row 1.
Again, we'll go ahead and do that here on the right,
and then perform the same row operation
on the original identity matrix here,
which will give us elementary matrix sub 3.
We'd have 1, 1/2, 5/2,
and the second row stays the same.
On the left side, so far we have
E sub 2,
times E sub 1 times A,
but now we're going to have another product here,
where this elementary matrix here will be
the identity matrix, where the first row
is multiplied by 1/2.
We'll have one-half, 0, 0, 1.
Again, this will be E sub 3.
Now we have enough information to solve the system,
but before we do that, let's focus
on the left side of this equation.
What we should recognize is that elementary matrix sub 3
times elementary matrix sub 2 times elementary matrix sub 1
times our original matrix A would perform the same
row operations that we performed here on the right,
so let's call this matrix B.
The product of these matrices here
would perform the same row operations that we performed
here on the right, therefore this would equal
matrix B, which is our 2 by 3
augmented matrix in echelon form.
Again, this product is equal to matrix B,
where A was the original augmented matrix,
and B was the augmented matrix in echelon form.
Let's go ahead and solve the system
using this matrix here.
This tells us that x plus 1/2 y equals
5/2 and that y is equal to 1.
If y is equal to 1, this should be x plus 1/2
equals 5/2, subtract 1/2 on both sides
that would give us x equals positive 2.
Our solution is x equals 2 and y equals 1.
Before we go on, I do want to discuss
this equation here one more time.
If we wanted to solve this equation for matrix A,
notice how the first step would be to multiply
both sides of the equation by the inverse of E sub 3.
This would give us E sub 2 times E sub 1 times A equals
the inverse of E sub 3 times matrix B.
The next step would be to multiply
by the inverse of E sub 2.
That would give us E sub 1 times matrix A equals
E sub 2 inverse
times E sub 3 inverse times matrix B.
For the last step, we'd multiply both sides
by the inverse of E sub 1.
Matrix A, our original matrix would be equal to
the inverse of E sub 1, times the inverse of E sub 2
times the inverse of E sub 3 times matrix B.
It is important to make the connection between
this equation here and this and this equation here
for our next lesson.
Let's go ahead and formally summarize this one more time.
Again, here are the two equations that we just found.
This one we used to solve the system of equations.
This is the equation that we found
when we wanted to solve the above equation for A,
where these matrices here are the inverses
our original elementary matrices, which I provided here
but if you want to, you can verify yourself.
Remember, to find an inverse of a 2 by 2 matrix,
we can use the formula given on this slide here.
Just one more thing I want to mention:
since we found matrix B by multiplying matrix A
by elementary matrices, A and B
are considered row-equivalent matrices.
OK, thank you for watching.