Hello. I'm Professor Von Schmohawk and welcome to Why U.
In the last several lectures, we saw how to represent a system of linear equations
as a mathematical object called an "augmented matrix".
When working with an augmented matrix it is important to keep in mind
that the entries which lie to the left of the vertical line
are the equation's variable coefficients
and the entries to the right of the vertical line
are the equation's constants.
We will therefore refer to the entries on the left as the "coefficient entries"
and the entries on the right as the "constant entries".
In the previous lecture, we saw how through the process of "Gaussian elimination"
an augmented matrix can be reduced to "row echelon form".
At that point, to solve the system of equations we have two choices.
As we saw, one choice is to convert the matrix back into equations
which are now simpler and easier to solve using "back substitution".
Another choice, as we will see in this lecture
is to continue to reduce the augmented matrix
through a process called "Gauss-Jordan elimination".
Gauss-Jordan elimination uses the same elementary row operations as Gaussian elimination
but with Gauss-Jordan elimination, the matrix is simplified to the point
that the solutions can be determined directly from the matrix
without having to convert it back into equations.
Once a matrix has been reduced using Gauss-Jordan elimination
it is said to be in "reduced row-echelon form".
So what exactly is reduced row echelon form?
In the previous lecture, we saw that matrices in row echelon form
must satisfy three requirements.
The first requirement is that the left-most nonzero entry in each row
called the row's "leading entry"
must be "one".
The entries to the right of a leading entry can have any value, zero or otherwise.
The second requirement is that each row's leading entry
must lie to the right of the leading entry in the row above it.
And the third requirement is that
if there are any rows which contain all zeros
those rows must be positioned last at the bottom of the matrix.
Reduced row echelon form includes these same three requirements
but it also includes one additional requirement.
In reduced row echelon form
each leading entry must be the only nonzero entry in its column.
Therefore, if a column includes a leading entry
all the other entries in that column must be zero.
In the matrix example shown here
since every column to the left of the vertical line contains a leading entry
all the coefficient entries must be either zeros
or ones which fall on the main diagonal of the matrix.
An augmented matrix in this form
always represents a system of equations with a single unique solution.
For example, this matrix
corresponds to a system of equations with the solution "nine, negative seven, two".
This form of augmented matrix always results in a single unique solution
since each equation in the system equates one variable
to a constant.
However, some augmented matrices in reduced row echelon form
may have columns to the left of the vertical line which do not include leading entries.
A matrix like this is still in reduced row echelon form
since all the leading entries are ones
each leading entry lies to the right of the leading entry above it
and each leading entry is the only non-zero entry in its column.
In this case, some rows will have more than one non-zero coefficient entry
and these non-zero entries may have values other than one.
If all the coefficient entries are not ones which fall on the main diagonal
then the system does not have a single unique solution.
However, in this lecture we will only present systems which have a single solution.
So starting with an augmented matrix
we use elementary row operations through Gaussian elimination
to reduce the matrix to row echelon form
and then we continue to use elementary row operations
through the process of Gauss-Jordan elimination
to further reduce the matrix to reduced row echelon form.
Since the three elementary row operations "swap", "scale", and "pivot"
change the form of an augmented matrix but not the solutions of the system
both Gaussian elimination and Gauss-Jordan elimination
result in systems which are "equivalent" to the original system of equations.
These systems will typically have different graphs than the original system
but since elementary row operations don't affect the solutions of the system
the intersection points of the graphs will be the same.
So let's see how Gauss-Jordan elimination transforms a matrix in row echelon form
to reduced row echelon form.
As we saw in the previous lecture
in the process of Gaussian elimination we work from left to right, and top to bottom
modifying the entries in each column until the matrix is in row echelon form.
With Gauss-Jordan elimination
we continue to modify matrix entries working from right to left, and bottom to top
modifying the entries in each column until the matrix is in reduced row echelon form.
In the last lecture, we started with a system of linear equations which we wished to solve
and represented this system as an augmented matrix.
Then, we used the elementary row operations, "swap", "scale", and "pivot"
to reduce this matrix to row echelon form
assuring that the leading entry of each row was a "one"
and that the entries below those leading entries were zeros.
To further reduce this matrix to reduced row echelon form
we must now continue to use row operations
to change the coefficient entries above those leading entries to zeros.
This can be accomplished using only pivot operations.
We start with the entry directly above the one.
This entry can be zeroed by using a pivot operation
to add negative two times row three
to row two
replacing row two.
We notate this operation by writing
"R subscript two" minus two times "R subscript three"
followed by an arrow and "R subscript two".
Multiplying row three by negative two
and adding that multiple of row three to row two
we replace row two with the result.
Now that the entry above the one in the bottom row is zero
the entry above that must also be changed to zero.
We can accomplish this by using another pivot operation
to add negative six times row three
to row one
replacing row one.
We notate this by writing "R one" minus six times "R three"
followed by an arrow and "R one".
Multiplying row three by negative six
and adding that multiple of row three to row one
we replace row one with the result.
Now that all the entries above the one in the third column are zero
we move over to the second column
where the entry above the one in that column must be changed to zero.
Once again, we use a pivot operation, adding negative four times row two
to row one
replacing row one.
We notate this by writing "R one" minus four times "R two"
followed by an arrow and "R one".
Multiplying row two by negative four
and adding that multiple of row two to row one
we replace row one with the result.
Now that all the leading entries are one
and the rest of the coefficient entries are zeros
this matrix is in "reduced row echelon form".
As we saw earlier in this lecture
a matrix in this form
where the leading entries all fall on the main diagonal
corresponds to a system of equations with a single unique solution
and the system represented by this matrix has the solution "nine, negative seven, two".
As we mentioned before
one advantage of reduced row echelon form
is that these values can be read directly from the matrix.
The process we used in the previous lecture to reduce a matrix to row echelon form
is called "Gaussian elimination".
And the process we used in this lecture
to reduce a matrix in "row echelon form" to "reduced row echelon form"
is called "Gauss-Jordan elimination".
However, the term "Gauss-Jordan elimination" can also be used to describe the entire process
using elementary row operations to reduce an augmented matrix in any form
to "reduced row echelon form".
The row echelon form of an augmented matrix is generally not unique.
Depending upon the sequence of row operations used
a given matrix can typically be reduced to many different equivalent row echelon forms.
In fact, any row echelon form matrix
can be transformed to an equivalent row echelon form matrix
by adding a multiple of any row to any row above it.
For example, we can modify the row echelon form matrix from the beginning of this lecture
by adding row two to row one.
If we replace row one of our matrix with this sum
we will have a different but equivalent matrix which is also in row echelon form.
This matrix and the matrix we just modified
are both row echelon form matrices and equivalent to the original matrix.
So a row echelon form matrix representation of a given system of linear equations
is not unique.
Using Gaussian elimination
many different row echelon form matrices can usually be produced from the same matrix
depending upon the sequence of row operations used.
On the other hand
a reduced row echelon form matrix representation of a given system is unique.
The same reduced row echelon form matrix will always be produced
regardless of the sequence of row operations used.
In this lecture, we have seen how a system of linear equations
can be represented as an augmented matrix
which can be reduced to reduced row echelon form
through the process of Gauss-Jordan elimination.
If the system of equations has a single unique solution
then that solution can be read directly from the augmented matrix.
In the next lecture
we will see what happens to the planes which graphically represent this system
during the process of Gauss-Jordan elimination.