Welcome to a video
that will show how to transform an augmented matrix
into reduced row echelon form
in order to solve a system of equations.
Let's begin by reviewing reduced row echelon form.
A matrix is in reduced row echelon form if these
four conditions have been met.
And here are several examples of matrices
in reduced row echelon form.
Step one, the first non-zero element in each row,
called the leading entry, is one.
Two, each leading entry is in a column to the right
of the leading entry in the previous row.
Three, rows with all zero elements, if any,
are below rows having non-zero elements.
And then finally, four,
in each column that contains a leading entry,
which again is a one,
all other elements of the column are zeros.
So all of these matrices satisfy these four conditions
and are in reduced row echelon form.
Let's also review how we transform an augmented matrix,
often referred to as Gaussian elimination.
Number one, any two rows can be interchanged.
Number two, the elements of any row can be multiplied by
a nonzero real number.
Number three, any row can be changed by adding
or subtracting the corresponding element
with another row.
So let's go ahead and give it a try on first,
a system of two equations with two unknowns.
We'll start by writing this as an augmented matrix.
The first row would be two, one, one.
The second row would be three, negative two, negative 16.
I want to have a zero in this position and this position.
Let's first try to get a zero here.
If this was a positive two,
we could add it to row two to obtain a zero.
So let's go ahead and replace row one
with two times row one plus row two.
The second row stays the same.
So two times row one plus row two.
Two times two plus three,
that would be seven.
Two times one plus negative two,
there's the zero.
And two times one plus negative 16,
that's two plus negative 16,
that would be negative 14.
Now we want a zero in this position,
so since the corresponding element is a seven,
the LCM, or least common multiple,
of three and seven would be 21.
So if we make this a 21 and this a 21,
we can then subtract these two rows.
So we'll replace row two with seven times row two,
minus three times row one.
Row one stays the same.
So we'll have seven times three,
minus three times seven.
That's 21 minus 21 or zero.
Now we'll have seven times negative two, that's negative 14.
Minus three times zero, that's negative 14.
Now this is a little more challenging here.
We have seven times negative 16, that's negative 112.
Minus three times negative 14,
so that'll become plus 42.
So we have negative 112 plus 42,
which will give us a negative 70.
And the last step is
to have the main diagonal equal to ones.
So multiply row one by 1/7 or divide by seven.
And multiply row two by negative 1/14,
or divide by negative 14.
Multiplying by 1/7 we'd have one, zero,
and then 1/7 times negative 14, that's negative two.
This would be a zero.
Negative 1/14 times negative 14 is one.
And then negative 1/14 times negative 70,
or just divide negative 70 by negative 14,
would give us positive five.
We convert this back to equations,
we have one x equals negative two, or x equals negative two.
And we have one y equal to five, or y equals five.
And the system has been solved
using reduced row echelon form.
Let's take look at a system of three equations
and three variables.
We'll start with the augmented matrix.
First row would be four, negative one, two, zero.
Second row would be two, one, negative one, negative 11.
The third row is two, negative two, one, three.
Let's see if we can get zeros here and here.
So if you take a look at row one and row two,
if this was a negative four,
we could add it to row one to get zero.
So we'll replace row two with
negative two times row two plus row one.
And if we took row three and subtracted row two,
this would be a zero,
so we'll replace row three with row three minus row two.
First row stays the same.
So multiply this by negative two
and then add it to four.
So negative two times two, that's negative four plus four.
We have zero.
Negative two times one, plus a negative one,
that would be negative three.
Negative two times negative one,
that's two, plus two would be a four.
Negative two times negative 11,
that's positive 22 plus zero, 22.
Now for row three, we'll have row three minus row two.
So two minus two is zero.
Negative two minus one, negative three.
One minus negative one, that would be one plus one for two.
And then three minus negative 11
becomes three plus 11, or 14.
Now let's see if we can get a zero here
and a zero here.
If we take a look at the first two rows,
if this was a positive three,
we could add it to row two to get a zero.
So we'll multiply this by negative three,
and then add it to row two.
So replace row one with
negative three times row one plus row two.
And if you take a look at row three and row two,
these two are the same.
So if we replace row three with row three minus row two,
that would give us a zero here.
So row three minus row two.
Okay, the second row stays the same.
So now we're going to multiple row one by negative three
and then add it to row two.
So negative three times four,
that's negative twelve plus zero.
Negative three times negative one,
that's positive three plus negative three, that's zero.
Negative three times two,
that's negative six plus four, that's negative two.
And then we have negative three times zero, plus 22.
That's 22.
And now for row three,
we'll replace it with row three minus row two.
Zero minus zero is zero.
Negative three minus negative three
becomes negative three plus three, which is zero.
Here we have two minus four, that's negative two.
And then we have 14 minus 22, that's negative eight.
Let's go ahead and take this matrix
over to the next screen and continue.
Now what we want to do
is have a zero here and a zero here,
and then the only thing left would be
to make the main diagonal equal to one.
Now we have to be a little bit careful here,
because we don't want to lose this zero here
to get a zero in this position,
so we're going to have to work with row one and row three
to make this equal to zero.
Notice the corresponding elements are the same number,
so if we subtract them, that would give us zero.
So we're going to replace row one
with row one minus row three.
And then looking at row two, we want a zero here.
If this was a negative four and we added it this four,
we'd have a zero here.
So we're going to replace row two with
row two plus two times row three.
The third row stays the same.
Now we're going to subtract row one and row three.
negative 12 minus zero, that's negative 12.
Zero minus zero.
Negative two minus negative two
becomes negative two plus two, that's zero.
And then 22 minus negative eight becomes
22 plus eight, that's thirty.
And then we're going to replace row two
with row two plus two times row three.
So zero plus two times zero, that's zero.
Negative three plus two times zero,
that's negative three still.
And then four plus two times negative two,
there's the zero we wanted.
And then 22 plus two times negative eight,
that'll be 22 plus negative 16,
which is equal to six.
Okay, we're almost there.
Now we just need to make the main diagonal consist of one.
So we'll multiply row one by negative 1/12,
multiple row two by negative 1/3,
and multiply row three by negative 1/2.
Remember, multiplying by negative 1/12.
is the same as dividing by negative 12,
so the first row would be one, zero, zero, 30
divided by negative 12 is equal to negative 2.5.
Multiplying by negative 1/3 is the same
as dividing by negative three,
so we'd have zero, one, zero, negative two.
And then multiplying by negative 1/2
is the same as dividing by negative two.
So we'd have zero, zero, one, and then positive four.
Okay, we should be done,
let's go ahead and translate this back to equations.
The first row would be x equals negative 2.5.
The second row would be y equals negative two.
And the third row is z equals positive four.
And now we have obtained the solution using
reduced row echelon form.
The next video will show how you can use
your graphing calculator to check your work.
Thank you for watching.