So in this video, we're going to find the reduced row echelon form of a matrix. So here's
the matrix that we're going to work with. In this problem, the matrix is A, and it has
three rows and four columns, and you can see it right there. We are going to take this
matrix A. We're going to perform row operations on it to get it into what we call reduced
row echelon form, and you'll see what that form means here as we go along. In general,
what it means is that the first entry that you come across in any row needs to be a 1.
So the way I think about this is always start with the first row, and I make sure that the
first non-zero number that I encounter is a 1. So we're kind of off to a good start
already with the one there. And then, below that 1, there has to be 0's. So there have
to be 0's right here. And then, I [INAUDIBLE] so right now, I like the one here, but I don't
like these that are not 0. So after we make these 0, then in the next row, the first entry
that I need to come to is a 1. So I need to have 0's going this way until I hit a 1, and
then once I hit that one, there need to be 0's below that as well. So the idea is to
get 1's with 0's below it, and you do that by doing row operations.
So the first thing I'm going to do so I'm going to get rid of the 4 and the 7 by letting
equation 2 equal equation 2 minus 4 times equation 1 and I'm going to let equation 3
equal itself minus 7 times equation 1 to cancel out these 4 and 7 right there. So equation
1 stays the same. I end up with a 0 there for equation 2 and negative 3 here because
5 minus 2 times 4 is a negative 3. And if you just continue the simple computations
there, you end up with a negative 6 and a 7. And then I'll do similar type operation
here for the third row. The third row minus 7 times the first row gives me a 0 there because
7 minus 7 times 1 is 0, and then I end up with eight minus 14 is a negative 6. Nine
minus 21 is a negative 12, and 5 minus negative 1 times 7 ends up being 5 plus 7, which is
12. So I'm off to a good start. I have a 1 right
here, and I have 0's below it, but now in the second row, the first non-zero thing that
I come across is negative 3. So I don't like that negative 3 there. Same thing here. I
don't like this negative 6 here either. So we're going to start working on this next
column. So if I let equation 3 equal equation 3 minus 2 times equation 2, I can get a 0
right there. So equation 1 and equation 2 remain unchanged.
I keep a 0 there, and then I'll have a negative 6 minus a negative 3 times 2, which would
be a zero and a negative 12 minus a negative 2 times the negative 6 gives me 0 there as
well. And then I get a negative 2 here when I do 12 minus 7 times 2. It's 12 minus 14.
It's a negative 2. OK, so we're getting there. We got some 0's there.
Let's go ahead and equation let equation 3 equal negative 1/2 times equation 3. We're
going to turn this into a 1 because remember in any row, the first thing that I come across
has to be a 1, so as I'm going across this row, I have 0, 0, 0, then I have negative
2. I can't hit a negative 2. I have to hit a 1. So I can turn a negative 2 into a 1 by
multiplying by negative 1/2, and then we'll do something similar here for equation 2.
If we multiply equation 2 by a negative 1/3, that will turn a negative 3 into a negative
1. We'll have that 1 sitting there like we like. So after we do this multiplication,
we have the 1 there in equation, 2 then we get the 1 in equation 3 we would like. Now
we're getting there. We're almost done. I have 1 with 0's below it. I have a 1 with
0's below it. The next thing is I need to do is get rid
of some of these other entries right here. [INAUDIBLE] equation 1 equal equation 1 minus
2 E2. So only equation 1's going to change. I'll only copy down E2 and E3 by themselves
without changing equation 1. I end up with a 0 and with this. Then, we'll go ahead and
let equation 2 equal equation 2 plus 7/2 equation 3. So we're only going to change equation
2. I still have a 0 there, I still have a 1 there,
but I can cancel out that negative 7/2. So after doing this, I end up with this matrix
right here. You might be wondering, why are we trying to get rid of these entries? Well,
one of the other rules about the reduced row echelon form is that any column that has a
pivot, and these are what we call pivots, these are 1's that can be the only non-zero
term there. So we're glad that we have a one right here,
but the fact that these are non-zero is a problem. So we need to get rid of those. So
I've gotten rid of one of those right there. I can keep going. I'll let E1 equal E1 minus
63. We can get rid of this right here, and then we'll end up with this final matrix after
we do the computations that looks like this. And this is, in fact, the reduced row echelon
form of the matrix. I can check this in the first row, the first non-zero entry that I
come across as a 1, and it is the only non-zero entry in this column. In the second row, the
first non-zero entry that I come across is a 1 and it too is the only non-zero entry
in this column. And then finally, in the third row, the first non-zero entry that I come
across as a 1, and it is the only non-zero entry in this column. So this is the reduced
row echelon form of the matrix.