- Hi, I'm Jen from
Calcworkshop, and in this video, we're gonna learn about the
most important technique in all of linear algebra, row reduction. So, either the form of
REF, or Row Echelon Form, or Reduced Row Echelon Form or RREF. Both of these techniques
are pivotal to your success in linear algebra. We're gonna walk through
a bunch of examples to make sure that you are successful in understanding this important concept. So, check it out. Alright, so let's look at the
two forms of row reduction. The first one is called Row Echelon Form. Some people refer to it
as just Echelon Form, and others refer to it
as Gauss Elimination, since Gauss was the first
to have actually come up with this idea. We abbreviate and say
Row Echelon Form is REF, which is referred to as
Row Echelon Form, REFing, Gauss Elimination. Here's the idea: We're gonna take a system, and we again, we know we like the identity matrix. That's our main guy that we like, 'cause ones and zeros are the easiest. What we wanna do is to, using techniques that we're gonna talk about in a second, is to reduce our matrix down so that we get ones
along the main diagonal, if you can see that, ones going along the main
diagonal, and if you see, in the lower triangle, we're creating this lower
triangle of all zeros. So, again, the objective is
to create an upper triangle with just numbers, and the
lower triangle has all zeros, with ones along the main diagonal. That is our objective. So, what we're doing is
we're taking a system, we're reducing it down,
so that we get ones down the main diagonal,
zeros in the lower triangle. Now, this is a square matrix,
meaning we have three rows, three columns, so we know we're going to get a main diagonal. If you look here, it's
not a square matrix. We have four rows and only three columns. This is a four by three matrix, so there really isn't a
main diagonal to speak of. We can kinda tell it's
got that diagonal feel, but it's not a nice, cut symmetry that we can see in the
first one, and that's okay. We're still going to try our
best to get this main diagonal if we can kinda fudge it and say, okay, I know it's not square, I know I don't have a main
diagonal going all the way down, but again, what I'm looking for is to get ones coming down diagonally, zeros in the lower triangle. Here's the big key: So, we know that we want to get this lower triangle of zeros. For REF, here's what's gotta happen: you get your ones, and everything below the one must be a zero. Everything below the one in each column must be a zero. So, that is the main objective for REF. Now, what are these little asterisks? These asterisks represent any
number we want them to be; we don't really care,
because what we'll be able to do is do back-substitution
to find our answer, and we're seen how to do substitution; we've been doing it since Algebra 1. So, the asterisks represent any number. What we wanna do is get
ones along a diagonal, zeros below any of the ones. That's for REF, Row Echelon
Form or Gauss Elimination. Now, why don't we like this
one as much as the second one? Because, at the end of
completing this task, we're gonna have to use
the substitution method, and how many of us jumped at the chance to use the substitution
method in Algebra 1? Not many of us. So, we would have to
substitute back in our numbers to simplify an equation,
and substitute again to simplify another equation. Not really helpful, and
we don't like doing it. So, what would be better? Well, not just getting
zeros below the ones, let's get zeros above the ones as well. That way, we don't have
to back-substitute. So, that's the idea behind the second form of row operations, which is
Reduced Row Echelon Form, which is RREF. We refer to this also as
Gauss-Jordan Elimination, 'cause Mr. Jordan came in
after Mr. Gauss and said, hey! Let's clean this up even better. So, it depends on also your
textbook and also your professor who's gonna wanna say which form, but we have REF and RREF, so we're gonna REF or we're gonna RREF. So, we're gonna RREF this guy, but we're gonna row-reduce
it even greater. So again, if you look
at the first example, what you're seeing is we
still wanna get the ones down the main diagonal
just like we saw here, but the difference with this
is that we just love zeros, because zeros are easy. So, we're gonna get the
ones down the main diagonal, but notice, we have this
lower triangle of zeros and we have the upper triangle of zeros. That's beautiful! That means we never
have to back-substitute. We're done, we're simplified, we're happy. So, again, what we see here is that we do have a main diagonal here because we have a square
matrix, which is nice. We get ones in a column;
anything above and below has to be a zero, which we see here. Well, look at this matrix. This is definitely not a square matrix. It is out of control, a lot of variables, a lot of equations. So, what do we see here? Well, we don't have a main diagonal. There's no way for us to kinda slice that, but we can kinda tell that
it is kinda making a diagonal kinda shape to it, which
is what we're trying to do. We're trying to eliminate to the point where we're gonna get that
diagonal feel going down, but what's important is this. Everywhere we see a one, so if
you see this one right here, everything above and
below it must be a zero. Well, there's nothing above it, but everything below has to be a zero. Look at the next column. Do you see a one? I don't see a one. So, I couldn't care less
what's in this column, but here, I see a one in the third column. If I see a one in the third column, that means everything else in
that column must be a zero. Well, everything above
and below are zeros; we did a good job. Look at the next column. In the fourth column, I see a one. Everything above and below
must be a zero, and it is. Look at this fifth column. I don't see a one at all,
so it doesn't really matter that we've got zeros and random asterisks which represent numbers. As long as it doesn't have
a one, I don't really care, and we've done our job
and we've row-reduced. Now, let's look at this next column. In column number six, what
we see is we have a one, which means everything
in that column above and below it must be a zero,
which is exactly what we see. In our seventh column, we see a one, and everything else in
the column is a zero. Our last column doesn't have a
one, so we don't really care. So, that is the idea behind
Reduced Row Echelon Form or RREF. So, over there, everything
below had to be zero for REF. For RREF, everything above
and below must be a zero. So, there's the big key. Everything above and below a one must be a zero. That's critical. So, REF, everything below
the one has to be a zero. In RREF, everything above and below the one has to be a zero. We still wanna see if we can
form that triangle going down, but we're not gonna be
always successful to do that. We know the overall goal
is to get the identity, 'cause he's the easiest,
but we're gonna maybe have to kind of fudge that
diagonal as we go, but again, if we can get ones and zeros everywhere, we are on the right track. Now, what is also really
important is to know what these ones actually mean. They have a name. We don't just go around
saying, hey, get your ones, get your ones. They actually are so important, they are pivotal to our answer,
they actually get a name. So what I'm gonna do next: I'm gonna put a new matrix up, and I'm gonna show you
its equivalent matrix in Row Reduced Echelon Form,
and we're gonna talk about what these numbers, these
ones and zeros, really mean. Alright, so we have this system, and what we've done is we've put it into an augmented matrix, and then, what we're going to see
is, after we've completed all the row reduction techniques
that we're going to learn, that it's similar, it
equivalates equivalent values, turns into this matrix, which
is this simplified system. So, we have this equivalent
system to this one, so these two systems are
equivalent to each other. This one looks a lot nicer than that one. It's because we've performed
all of our row operations to the best of our ability. So, we in essence have simplified
our problem significantly. More importantly, we're gonna
see that we've got these ones and zeros, and everybody
loves ones and zeros, 'cause they're so much
easier for us to figure out. So, what do we know? This matrix is more
simplified than that matrix, and what we're seeing is that we've got these ones and zeros. So, what are these things called? Well, this one, and
this one, and this one, we don't call them ones,
we call them pivots. The reason why is because they are pivotal to us getting the right answer. So, these values right here,
these ones that we see, are called pivots. They are also called
basics or basic variables, but you're gonna see pivot more often than you're gonna see basic. They are basic values,
they are pivot values, they are pivotal to our answer. Now, what does that mean? So, how many do we have? We've got three, but wait a minute. We had more than three variables. We had x-sub-one, x-sub-two,
x-sub-three, x-sub-four, and then we had the equals and then we had the Bs, the solutions. Well, that means, we have a
pivot in the x-one column, we have a pivot in the x-two column, and we have a pivot in the x-four column. Why don't we have a pivot in the x-three? Well, what we see here is
that we don't see a one. If we see a one, everybody else in that column must go to zero. We don't have a one,
so we don't need to see that everything goes away. So that means that this guy right here is what we consider a free variable. He is free to be whatever
we want him to be. So what we're seeing here,
in a more general form, is that this little
guy is a free variable. So, I'm going to try to put
that in a different color so we can see it. So, we have free variables, which in essence say they are
free to be whatever we want. They are free to be whatever we choose. So, the pivots, the basic
variables, they are going to give you an exact value,
based off of the mechanics and the chemical make-up
of these equations. The free variables, well,
if we know he's free, well, I like the value of
one, let's plug him in, but if you like the value of
two, you plug that one in. It changes the question just a bit. It's a scalar multiple. So, what's interesting is that the ones are no
longer referred to as ones, they're referred to as
pivots, and the columns that they appear in are
referred to as pivot columns. Now, the pivot columns always go back to the original matrix,
so the pivots are found in the reduced row-reduction matrix. The pivot columns go back to the original. So we saw the pivot columns
appeared in column one, column two, and column four. These are the pivot columns. So the pivots are found
after row reduction. Once you've done the row
reduction, you've found your ones, those are your pivots. Once you've found your
pivots, then you can say, oh, in the original matrix,
these were the pivot columns. So that's how we kind of relate the two. So what's interesting, though, too, is that we can then talk about, well, which variables are basic, are pivots, and which ones are free. If we can't see it here, it's
obvious that x-one, x-two, and x-four are the pivots,
and x-three doesn't have a one in it, so he's gotta be the free variable, we can see it even better up here. What we're seeing, if I
rewrite this, we can then say, if I kinda put this down here, if I move everybody over to
the right-hand side and solve, what we're seeing is that x-sub-one equals five
plus three x-sub-three. I'm just moving the three
x-sub-three to the other side. We see that x-sub-two equals, and I'm gonna bring the
other guy over to solve for x-sub-two, is negative
three minus two x-sub-three. Well, I see that x-sub-four equals zero, and the bottom row wiped out and zero equals zero all the time, but what does x-sub-three equal? We didn't have a pivot. That means, x-sub-three is itself. It can be anything we want it to be. So, once again, our pivots
for this specific question are x-sub-one, x-sub-two, and x-sub-four. Our frees is x-sub-three. So, these are pivot positions. Our pivot columns go back
to the original matrix. So, now that we know what
these things are called, what are we gonna do? How are we get from this guy to that guy? Let's talk about the
row-reduction algorithm. Alright, so here are the
steps that we're going to use to perform row operations. There's only three. So, the row-operation rules, here we go. We can replace. What does it mean to replace? We can add two rows together. Now, what's really important is that you're only allowed to add. You can't subtract. We can add a negative,
but we cannot subtract, because subtraction is not
commutative, whereas addition is. So again, if we need to
put a negative in there, we're going to add a negative value, but again, we can replace
by adding two rows. We can interchange; we
can switch two rows. So we can switch two rows, we
can interchange them, saying, why did that equation come up top? Why don't we switch it? Because we can. We can scale, we can multiply
any row by a specific number. We can multiply by a positive,
a negative, a fraction, whatever we need it to be, to either make it bigger
or smaller to what we want. These are all the same
techniques that we used for linear combinations,
and what's really cool is, you technically know how
to row-reduce already. You've been doing it
every time you perform linear combinations, but
what we're gonna see is that we're just gonna
write it differently. So, everything you know about how to do linear combination
method, the elimination method from Algebra 1 to solve a system, is everything you need to
row-reduce for a matrix. We're just going to write
it a little bit differently and show how to work that out. Here are things that you're
not going to find in a textbook that are really, really important
for you to keep in mind. Once we get better at how to row-reduce, you won't always have to be
as rigid with these rules, but these rules will
work 100% of the time. So, here's how we're going to do it. You must work column-by-column. No matter if you're an expert at this, this rule still holds. You have to work column-by-column, and you must work with the first column, complete that column, go to
the next column, and so forth. You can't just randomly say,
I like column number five, so I'm gonna go to column five
and work on that one first. You're going to be doing a
lot of extra work that way, because you're gonna have to fix it later. So, you work column-by-column. You do not move to another column until you've gotten
all the necessary zeros before you go any further. Once you've gotten your zeros, then you can find your pivots. So, pivots, yes, we know
they're pivotal to our answer, but it's a lot easier to take any number and reduce it down to a one. If I have the number
three, I can multiply three by one-third, I can scale, and that three times
one-third becomes a one. It's a lot easier to get a
one than it is to get a zero. So, the first thing
we're gonna do is get all of our zeros that we need, and then the second thing we'll do is to get our pivots by scaling. So that's really, really important for us. Now, here is the massive hint. The hint is this: if you're
working in column one to get your zeros, you're gonna
use row one to manipulate. If you're working in column
two to get your zeros, you're gonna use row two to manipulate. If you're in column five, you're gonna use row five to manipulate. So whatever column you're in, you're gonna use that equivalent row to make the manipulations. This is really, really
important and very helpful in terms of what in the world we're doing. So, these are the rules, and
we're gonna continue to refer to them over and over and over
again, but we can replace, we can switch, and we can scale. So basically, we can
add, switch, and scale, and we're gonna work column-by-column, we're gonna get our zeros
first, and then get our pivots, and whatever column we're in, we're gonna use that equivalent
row to do the manipulation. Now, before we actually
jump into seeing a matrix and how it works, what we're
gonna do is we're gonna walk through a system. In this system we're gonna use for linear combinations,
the way we did in Algebra 1, and then we're gonna do
the exact same system using matrices, and see
exactly how they are the same, just different ways of writing it. So, if I have this system, and I have eight x plus six y equals two, and then, five x plus four
y equals negative one, how would we do this way back in the day? Well, in Algebra 1, what we would say is we wanna pick a variable
we want to eliminate. What we would see is, the
Xs are lined up nicely, the Ys are lined up, the equals sign, and also the constants. So, we're already ready to go
in terms of combining things. Well, let's get rid of the Xs. So, if we get rid of the
Xs, that means they have to have the same number, opposite signs. What do we see? What we see is that this guy has an eight and this x has a five. They're not the same number, nor are they even opposite signs. So what we're going to do is we're basically gonna do a switcheroo. So we're gonna say, I'm
going to take this five and I'm going to bring it up top. I'm gonna move that, and I'm gonna put a five up here. Well, I'm gonna do the exact same thing, I'm gonna take this eight and
I'm going to bring it down. So by doing that, I can now
multiply the entire top row by five and the entire
bottom row by eight. That's still not gonna get
me exactly what I want. I still need them to be opposite signs. So, if you notice, do you agree, it looks like we're in the first column? Well, we are, so which row
do we use to manipulate? We would use the top row, so we're gonna make the top one negative. We're always gonna do that. So now what we're gonna do is we're going to distribute these numbers
into the entire equation, so that means, negative five
is going to be distributed to everybody in the top row, and then eight is gonna be
distributed or multiplied by everybody in the bottom row. This is what we did way back in Algebra 1. So if we distribute those numbers in, negative five times eight
gives us negative 40 x; Negative five times six
gives us a negative 30 y; equals negative five times
two, we have a negative 10. What did we do? We didn't change the chemical
make-up of the equation. We just made it bigger. Could we factor out a negative five and come right back to where we started? Yes, which means we haven't
done anything illegal, we just made them easier for us to use. So, if we multiply the entire
bottom equation by eight, eight times five is a positive 40 x, eight times four, what we see
is we've got a positive 32 y, and eight times negative one,
we see is a negative eight. Once again, we haven't
changed the chemical make-up of that equation; he's
still five x plus four y equals negative one,
all we did was make him a little bit bigger by
multiplying by an eight. The reason we wanted to
do that is, now we notice, they not only have the same number, they have opposite signs. So now, we're going to add them, because that's how we
can replace, we can add. So we're going to add
these two equations up. Negative 40 x and a positive
40 x got eliminated. It made a zero, which is what we want. Again, we're looking for our zeros. And if we add those two numbers up, we've got a negative 30 y plus a 32 y, and we see it's a plus two y, equals, negative 10 and negative
eight make a negative 18. Well, now we've got two
y equals negative 18. If we divide by two, y
equals negative nine. So now that we know that
y equals negative nine, all we gotta do is find what x equals. We have a choice. We can take this negative nine and plug it into either of our new
and improved equations, or into either of the original equations. I probably wouldn't plug it
into the new and improved because it's bigger numbers. So, but we get to decide
how we wanna do that. We can take this negative
nine and we can plug it into either the top equation
or the bottom equation; it doesn't really matter. If we choose the top
equation, that means we have eight x plus six times negative nine, and then, equals two. So now what are we gonna do? We're going to simplify. Well, we've got eight x, and
then minus 54, equals two. Alright, so now, let's add the 54 over, and we get eight x equals 56. Divide by eight; x equals seven. Well, that means we've found the solution. We found the value of
seven, negative nine. Well, this is everything that
we're gonna do for matrices. I'm going to leave this up and
I'm going to erase that side, and then I'm gonna take
this exact question and we're gonna put it into a matrix, and we're gonna perform the
same in essence operations. We're just gonna write it
a little bit differently, and we're gonna see that we're
gonna get the same outcome in the end by just
performing row operations using matrix form, which is exactly like linear combinations. Here we go. Alright, same exact system. We already know what the answer is. We already know what we had to do. Let's now put this into
an augmented matrix and see what we can discover. So if we put this into
an augmented matrix, that means we're going
to put our coefficients, eight and five for the Xs,
six and four for the Ys, we're gonna put our bar
for the equals sign, and then, two, negative one. So, what makes this a
little bit different? Well, we're getting rid of the variables, and we don't have to deal with them. We can bring them back in at the end. We're just dealing with the numbers, which is what we need, anyway. Well, if we're working in matrix form, we have to work column-by-column, so that means we are in column one. Our ultimate goal is to get everybody in the column to be ones and zeros. That's what we wanna do. Well, what's more important? The big hint was, get your
zeros first, then get your ones, so get the zeros, then the pivots. So, if you notice, if we're looking for that diagonal to be a one, that means everything above
and below must be a zero. Well, that means the five
really does need to go away, because the eight will
eventually become a pivot; it'll eventually become a one. Again, remember, our
ultimate goal is to get this. This is what we're looking for. That's what our objective is. We wanna get ones down the main diagonal, everywhere else to be zeros. So, that means we eventually
want to turn the eight into a one, but that's not
important to us right now. The most important thing
is to get that zero. So, what are we gonna do? We need to get rid of him;
how do we get rid of him? Well, we can add/replace,
we can switch/interchange, or we can scale. Well, what did we do over here? We added; maybe we
should do the same thing. So, how did we do it over here? Well, we got the numbers to
be the same, opposite signs. Maybe we should do the
exact same technique, and that's exactly what we're gonna do. What we're gonna say is,
well, if I am in column one, I'm gonna use row one to
manipulate, so once again, what I'm going to see is
I wanna keep this guy, but I want this one to go away. I'm going to switch those
two numbers once again. So, I'm gonna say, this is
a five and this is an eight. Now, we did the exact
same thing over there, but we recognized over
there that we needed one to be positive and one to be negative. Well, whatever column
you're in, that's the row that you're going to use
predominantly to manipulate. So, we're in column one, so
we're gonna manipulate the most using row one, and so,
if we need a negative, he gets the negative. If we were in column two
and we needed a negative, that means row two would get it, because that would be the
column-row combination. So, what do we do now? Well, we're about to
manipulate, meaning we're about to distribute into everybody that we see. We're not changing the make-up
of the original matrix, we're just gonna make it
bigger, just like we saw here. Now, every time we perform an operation, we need to put some command lines down, meaning you have to tell the
reader, whoever it is looking at your work, what did you do? I don't understand; how
can I follow you quickly so that I can see every
manipulation you make? Now notice, we just wrote
down that we're gonna take and multiply negative five
by the entire top row, and eight by the entire second row. Well, that's the command
line we're gonna write. We're gonna say, negative five row one, plus eight times row two. Well, we're gonna take negative five and multiply entirely through row one, and eight entirely by row two, and we want to eventually add
them, just like we did here. What do we wanna do? We want this this guy to go away, which is exactly what we saw there. We want him to disappear,
which means we're gonna say, I want us to eventually change, let's change row two. Why row two? 'Cause that's the location
that we need to go away. That's the elimination part
that we're looking for. So we're going to multiply two rows and then add them together,
just like we did there. This is the command line,
which is what your professor or anybody looking at
your work wants to see. Now, every time you manipulate
the matrix, we need to say, I am manipulating, I am
performing a row operation, I'm gonna put the wiggle in there, saying, hey, operation is about to happen. So, that means I'm going to,
in small little blue writing, going to write up high
above those numbers. Again, when you get really good at this, you can do it all mentally in your head. So, if you can multiply and
add numbers quickly and easily and arithmetic is no problem for you, then you don't need to write this down. For the rest of us mere mortals, we might need to write it down, because we don't wanna
make a silly mistake. I make silly mistakes with
arithmetic all the time, so, make sure we get it
right by writing it down, if we have to; you can write it in pencil, you can write it really faintly,
you can put it on the sides and no one will ever know that
you wrote down these numbers, but we just gotta make it right. So, I'm gonna do it in small
little writing in blue, so you can see the manipulation. If I take this negative
five and multiply by eight, I get a negative 40; Negative five times six,
we get a negative 30; and negative five times
two, we get a negative 10. Oh, we saw that over here. Alright, if we take the
eight and we multiply him all the way through, eight
times five is a positive 40; eight times four was a positive 32; and eight times negative one
gave us a negative eight. Well, we did exactly what this
blue command line said to do. We multiplied the first
row by negative five; we multiplied the second
row by a positive eight. What do they want us to do now? Add it. Okay, I can add these little blue numbers. Who do we want to change? We want row two to change. Well, that means row two
is the guy that's changing, which means the original row
one is gonna stick around. Now, that's interesting. How come the original row one is staying and I'm not writing
negative 40, negative 30, and negative 10? Because, what would be
easier to work with later? Big numbers, little numbers. So whichever row we need to replace, that's the only one that gets affected, because I don't wanna
have to eventually deal with negative 40, negative
30, and negative 10. You could if you wanted to,
but it's so much easier to say, you know what, I never
change the chemical make-up over here; I just made it bigger, so I don't have to keep
using these big numbers; let's go back to the original. The only one that's
getting changed is row two, 'cause that's who I'm looking for. So let's add the little
blue numbers together. Negative 40 and positive 40 made the zero, just like we saw here, just
like we're looking for there, in terms of our matrix. Negative 30 and a
positive 32 gave us a two. Well, what do we see? Negative 10 and negative
eight make a negative 18. What you're seeing is exactly
what we see here in our line when we simplified using
linear combinations. Everything is identical,
just written differently. So now, we have finished,
completed, column one. Notice column one, we
need to keep everybody to be ones and zeros. What's most important? Getting your zeros, and at
the very end, get your ones. So, we don't need to change
the eight into a one just yet. We just need to get our zeros. Now what we're gonna
do is we're gonna move to our second column. If we're in our second column, we want to manipulate
using the row two column. So notice, it's mimicking what we want in terms of the identity. We want this two to stick around. We want the six to go away. Hmm, so if we want the six to go away, we want him to become a
zero, like we see here. Well, if we are in column two, that means we're gonna
use row two to manipulate. So, instead of using
the eight and the zero like we see here, we're gonna
use the six and the two, because when we're now in column two, we're gonna use row two. So, what we're gonna do is, we're gonna make a little
bit of a switcheroo. We're going to switch
those numbers around. Now, I'm gonna do something
that you probably can recognize and say, why is she
multiplying by six and two? Two goes into six three times. Why can't we just use that? We absolutely can, and we're
gonna get better and better at it, but what we're gonna do is we're just gonna switch those numbers. So now, we know that
we're gonna switch the two and we're gonna switch the six. Here's the question:
are they opposite signs? We just switched the numbers,
but they're not opposites, so one of them has to be a negative. Which one is it gonna be? Well, if you're in column two, that means you're gonna
use row two to manipulate, so the row two is going
to get the negative if we have to have a
negative, which we do. So, again, we're in column two, so row two is the manipulation part; row two is gonna get the negative. What are we about to do? Well, we're gonna write our command line. We're gonna say, I want us
to multiply negative six by everybody in row two, and
I wanna multiply everybody in row one by a value of two, and I want us to add it together. So, once again, we want to
do an elimination method. We wanna basically eliminate
the y variables a little bit. So, what's happening is, we're gonna multiply the entire second row by a negative six; we're gonna
multiply the entire top row by a two, because those
are the two manipulations we're gonna see; and we
want to change row one. Why row one this time? Because that's the zero
that we're looking for. That's what we're trying to gain. Now, notice something really cool. You notice, the first guy tells
you what column you're in, the second guy is who's getting changed. The first guy is what column you're in, the second guy goes twice, because he's the one
we're trying to change. That's the way all of our
commands will always be, to keep things organized. So, we're going to
perform a row operation. We're gonna put the wiggle in there. Now, again, if you were really
good at using arithmetic and adding and multiplying numbers, go for it; that's awesome. If not, write it down. So, we're going to do
exactly what it says. We're going to multiply
the entire second row by a negative six. Here we go. Zero times negative six, well, is zero. I'm gonna erase this so
we can write it down. So we have zero. Two times negative six is a negative 12. Negative 18 times a negative six is 108. Now, if we multiply the
entire top row, row one, by a two, let's see what happens. Eight times two, we get 16; six times two, we have 12; and two times two, we get four. Notice that these are the
little orange numbers. That's what we're going to be adding. If we can do this in our head, great; otherwise, just write it down. So, we're going to manipulate. Which row are we looking to change? We wanna change row one, which
means row two isn't changing. We were happy with row two; we'd already affected him
on the previous example, so we're going to keep
zero, two, and negative 18. Why? I don't really wanna keep
108 if I don't have to. I mean, it's up to you, but I'd rather keep the smaller numbers. Again, by just multiplying
an entire row by a number, you're not changing that equation, you're just making it
either bigger or smaller, so use the numbers that are easier, only affect one thing at a time. So now, let's add up the
little orange numbers exactly like we see here. 16 plus zero makes 16. Wait a minute; my eight, he changed! Who cares? If I take an eight and I
multiply it by one-eighth, we get one. If I take 16 and multiply
it by one over 16, we still get a one. It doesn't really matter that
we haven't changed him yet, and that's why we don't change our pivots 'til the last minute. So, 12 and negative 12 make zero, which is exactly what we were looking for; and then, four plus 108 makes 112. Alright, so now, we have
completed the hardest part of row reduction. What we see is that we've got the zeros everywhere we're supposed to have a zero. We've got zero in the lower triangle and we've got zero in the upper triangle, just like the identity
matrix we were looking for. Perfect; what's the only
thing left for us to do? Scale; we wanna get these
numbers along this diagonal to become one, so how do we
get a 16 to become a one? We multiply by its inverse, one over 16. How do we get a two to become a one? Multiply by one-half. So, that's the operation
we're gonna perform, and we're gonna write that command line. We're going to say, I want us to multiply by one over 16 times row one, and it's going to change
row one to make it easier. We're then going to say, one-half row two is
going to change row two. Again, we're just telling the reader what you're about to do,
and then, if we do it, we're going to manipulate
it one more time. We're gonna say, okay, I'm
about to perform an operation. There's my wiggle, and we're going to change both of those rows. We're gonna multiply the
entire top row by one over 16, so if we say, one over 16, and we distribute him in, 16 times one over 16 becomes a one; zero times one over 16 stays zero; and then, 112 divided by 16, well, look at that; you get seven. So, 112 divided by 16, we
get a seven value; perfect! So now, let's take one-half and multiply it by the entire second row. So we have one-half and
we're going to distribute, in essence, scale him in. Zero times a half, zero; two times a half, one. Perfect, I've got my pivots. I've got my ones down the main diagonal, zeros everywhere else, top and bottom, and then, negative 18 divided
by 2 or times one-half, we get a negative nine. Well, would you look at that. This guy looks just like that guy; it's just written a
little bit differently. Notice we have x equals
seven, y equals negative nine. If we remember, this
is the x and y columns, and we know this is the equals,
well, we're ready to go. That means x plus zero y equals seven, and zero x plus y equals negative nine, well, x equals seven,
y equals negative nine, just like we saw over there. So, linear combinations
is the exact same way that we perform row operations. We've now proven it. What's interesting is that we were able to do the same manipulations,
write some command lines, and get our overall objective. We took our time to make
sure that we could guarantee that we're never gonna get this wrong. Now, what did we perform? We performed RREF,
Reduced Row Echelon Form, Gauss-Jordan Elimination,
because we got ones down the main diagonal, our
pivots, zeros everywhere else. Did we have to? No, we could have actually
stopped right here, and we could have actually
done back-substitution like we did on the previous question, where we saw four linear combinations, and we would have gotten the same answer, but we're just walking
ourselves through techniques. What we're gonna do next is
we're gonna look at an example of how to just do REF, Row Echelon Form, and then for the next examples,
which we're gonna see, we're gonna continue to perform RREF. Out of the two operations, RREF is the more-powerful
and the more-useful. REF is every effective if you only need to determine certain things, but again, it would still require back-substitution. So, let's look at how to do REF, and then, for the next bunch of examples, we'll do RREF, which is the more-powerful. Here we go. Alright, so for this example, it says, use REF, Row Echelon Form,
or Gauss Elimination, to solve the system. So what we see is, we're
given this basic system, two equations, two unknowns,
and what we're gonna do is we're gonna put it into
an augmented matrix. In order to perform REF or RREF, you must be using an augmented matrix. So, here we go. We're gonna put it into
an augmented matrix. We're gonna keep our
columns nicely organized and pull out the coefficients. So we say, we've got one and negative one for the x-one column; we
have negative two and three for the x-two column;
we're gonna put our bar in for the equals sign; and
then, negative one and three. So we have now successfully
created that augmented matrix. Now, what is one of the
most common mistakes is we transcribe wrong. I've done it myself, and so,
how important is it that, to make sure we copy
correctly to get our matrix, because if we copy it
wrong, then everything we do after that's gonna be wrong. So, take your time, double-check yourself, to make sure that you've
transcribed it correctly. I've made the same mistake myself. So, what we've done is, we've now written the augmented matrix, and now we gotta think, what is the overall goal
that we're looking for? If we know we're looking for REF, that means we are looking for
ones down the main diagonal, zeros below all of the ones,
but we don't really care what happens in the upper triangle. So that means, we only
really need to get one zero and scale on down. That's not so bad; it just
means extra back-substitution to get our answer. So, what do we wanna do? Well, we wanna get this guy,
negative one, to become a zero. Well, let's go ahead and put
that in there, so we can say, hey, I want him to become a zero, so I'm gonna actually put it in and say, that's what I'm really wanting to get. Now, what column are we in? We are in column one, which means, we're gonna use row one to manipulate, so we're going to switch those numbers. Well, they happen to be
the same, one and one. Do we need a negative? Well, one's positive and
one's negative already, so we don't to put a negative in there. All we gotta do is switch those values. So, what do we wanna do? We want to switch the numbers
and then add them together using our addition property,
our replacing property. We wanna replace this guy with a zero, so we're gonna write our command line. Our command line says, hey, I wanna multiply the entire
top row by one, so we'll say, one row one, or just row
one, plus one row two. So, I want to add these
two things together, just like I would see
in linear combinations. If we just put our big bar
and put a big plus there, and we can add 'em up,
that's what I wanna do. Well, who wants to be replaced? Who are we trying to manipulate? We want the row two to go away. We want this guy to be
replaced, so we say, row two. Again, we're gonna say, whoever comes first is
the column you're in. Row one corresponds to the column one. Row two better come twice, because that's who we're
going to be affecting. So if we do that, we're about
to manipulate, so we say, wiggle, so equivalent
matrix is about to come up. We're going to only replace row two. We're going to keep row
one exactly the way it is. So, we're going to rewrite row
one, so we don't mess it up and don't get anything weird, and what we're gonna do is, we're gonna now perform the
operations they want us to. Well, this one's pretty nice, but just because I wanna
make sure that we see that we're doing something,
I'm gonna still write it in the little blue color. So it says, multiply the
entire top row by one, so one times one is one; one times negative two is negative two; negative one times one is a negative one. I've successfully multiplied. What about the second row? We are gonna multiply
the entire second row by a positive one. Okay, so negative one times
one is a negative one; three times one is three; and then, one times three is three. I've done exactly what it says. Now all we gotta do is add them together. Okay, one plus negative one makes zero. Perfect, I got exactly what I wanted. So now, negative two plus
three, what we see is one. Alright, and negative one plus
three, we get a value of two. So, we have now completed column one. We have gotten the zero
that we were looking for, and if you notice, we are actually perfect in terms of REF manipulation. We have exactly what we're looking for. We have our pivots down the main diagonal, and we didn't even have to scale, and we've got our zero
in the lower triangle, and it doesn't really matter
what that number is above. What we now see is that
we can pull this back out into a system. This is called general form. So, general form now says,
rewrite what you've got. Again, this is x-one,
x-two, there's the equals, and again, these are the constants. So we see that we've got x-one minus two x-two equals negative one, but what we see here is that the second equation
has gotten better. We now have just x-two equals two. We have performed our general-form
operation by using REF. Well, what are we gonna do now? Unfortunately, since we didn't do RREF, we still have to find out what x-one is by back-substituting. So now, back-substitute. We're gonna take this value,
and we're going to plug it in. So what we have is, x-sub-one minus two times
two equals negative one; so x-sub-one minus four
equals negative one; x-sub-one equals three
if I add the four over, four plus negative one equals three. Now what I've got is
x-sub-one equals three, x-sub-two equals two, or it's
the ordered pair three, two. So, REF is pretty straightforward. All we're doing is getting our
zeros in the lower triangle, and then we would get
our pivots if we need to, which what we saw was a pretty
straightforward question, but the irritating thing at the end is that you still have to
back-substitute, and no one wants to do that, especially if
you've got a bigger matrix. This one wasn't hard, but
imagine if you had five equations and you had to back-substitute five times. Ooh, that's be really frustrating. So, REF is definitely effective, but it's not as powerful as RREF, and here's something
that's very, very important for you to know. REF, there are an infinite
number of possibilities on how to solve using REF, because
what was stopping me from saying, you know, I don't
like the way it's written. I could have said, well, why did that top equation come first and the second equation come second? What happens if we then
said, okay, if we had this, one, negative one, and
then negative two three, and then negative one, three, well, how come that row's on top and that row's on the bottom? Why? Well, it's because the
textbook or the professor or I chose it that way, but it doesn't have to stay that way. I wanna switch 'em, because
that's what I wanna do. So, we could switch. I wanna take the second row
and move it to the top row. I can say, I want row one
to switch with row two. Notice it's the double arrow
that we're seeing here. I want them to switch; why not? Because we can. There's nothing saying that we could put one before the other. So, we would then say, well, this would be negative
one, three, and three, and then one, negative
two, and negative one. I can switch; I can
interchange any way I want to. So, what's really important is,
there are an infinite number of ways to solve a system using REF, so that means, I can get one value and then you can get
another matrix in the end. So this matrix right here I could get, but I can get something totally
different if I wanted to if I had changed the way this was written. I can totally get different possibilities. Well, why did I multiply
both of them by one? I wanna multiply both top and
bottom, row one and row two, by 17, 'cause I like the number 17. So we can have an infinite
different possibilities for REF. What's so cool, though,
that every single person in the world will get the
same exact RREF in the end, no matter what we do to get there. So, I could take seven steps
to get to the final outcome and you can take two. We will still arrive at the same answer, and all of our answers
will be exactly the same if we're going to use RREF. Now, if we use REF, just Row Echelon Form, we can have an infinite
number of possibilities, but every single person
will get the same exact RREF in the end. I hope you've enjoyed the lesson so far on row-reduction techniques. For an additional
question, where we look at how to row-reduce a four by three matrix, check out calcworkshop.com.
