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# Solving a system of 3 equations and 4 variables using matrix row-echelon form

Sal solves a linear system with 3 equations and 4 variables by representing it with an augmented matrix and bringing the matrix to reduced row-echelon form. Created by Sal Khan.

Video transcript

I have here three equations
of four unknowns. You can already guess, or you
already know, that if you have more unknowns than equations,
you are probably not constraining it enough. You actually are going
to have an infinite number of solutions. Those infinite number of
solutions could still be constrained. Let's say we're in four
dimensions, in this case, because we have four
variables. Maybe we were constrained into a
plane in four dimensions, or if we were in three dimensions,
maybe we're constrained to a line. A line is an infinite number of
solutions, but it's a more constrained set. Let's solve this set of
linear equations. We've done this by elimination
in the past. What I want to do is I want to introduce
the idea of matrices. The matrices are really just
arrays of numbers that are shorthand for this system
of equations. Let me create a matrix here. I could just create a
coefficient matrix, where the coefficient matrix would just
be, let me write it neatly, the coefficient matrix would
just be the coefficients on the left hand side of these
linear equations. The coefficient there is 1. The coefficient there is 1. The coefficient there is 2. You have 2, 2, 4. 2, 2, 4. 1, 2, 0. 1, 2, there is no coefficient
the x3 term here, because there is no x3 term there. We'll say the coefficient on
the x3 term there is 0. And then we have 1,
minus 1, and 6. Now if I just did this right
there, that would be the coefficient matrix for
this system of equations right there. What I want to do is I want to
augment it, I want to augment it with what these equations
need to be equal to. Let me augment it. What I am going to do is I'm
going to just draw a little line here, and write the
7, the 12, and the 4. I think you can see that
this is just another way of writing this. And just by the position, we
know that these are the coefficients on the x1 terms.
We know that these are the coefficients on the x2 terms.
And what this does, it really just saves us from having to
write x1 and x2 every time. We can essentially do the same
operations on this that we otherwise would have
done on that. What we can do is, we can
replace any equation with that equation times some
scalar multiple, plus another equation. We can divide an equation,
or multiply an equation by a scalar. We can subtract them
from each other. We can swap them. Let's do that in an attempt
to solve this equation. The first thing I want to do,
just like I've done in the past, I want to get this
equation into the form of, where if I can, I have a 1. My leading coefficient in
any of my rows is a 1. And that every other entry
in that column is a 0. In the past, I made sure
that every other entry below it is a 0. That's what I was doing in some
of the previous videos, when we tried to figure out
of things were linearly independent, or not. Now I'm going to make sure that
if there is a 1, if there is a leading 1 in any of my
rows, that everything else in that column is a 0. That form I'm doing is called
reduced row echelon form. Let me write that. Reduced row echelon form. If we call this augmented
matrix, matrix A, then I want to get it into the reduced row
echelon form of matrix A. And matrices, the convention
is, just like vectors, you make them nice and bold, but use
capital letters, instead of lowercase letters. We'll talk more about how
matrices relate to vectors in the future. Let's just solve this
system of equations. The first thing I want to do is,
in an ideal world I would get all of these guys
right here to be 0. Let me replace this guy with
that guy, with the first entry minus the second entry. Let me do that. The first row isn't
going to change. It's going to be 1, 2, 1, 1. And then I get a
7 right there. That's my first row. Now the second row, I'm going
to replace it with the first row minus the second row. So what do I get. 1 minus 1 is 0. 2 minus 2 is 0. 1 minus 2 is minus 1. And then 1 minus minus 1 is 2. That's 1 plus 1. And then 7 minus
12 is minus 5. Now I want to get rid
of this row here. I don't want to get rid of it. I want to get rid of
this 2 right here. I want to turn it into a 0. Let's replace this row
with this row minus 2 times that row. What I'm going to do is,
this row minus 2 times the first row. I'm going to replace
this row with that. 2 minus 2 times 1 is 0. That was the whole point. 4 minus 2 times 2 is 0. 0 minus 2 times 1 is minus 2. 6 minus 2 times 1 is 6
minus 2, which is 4. 4 minus 2 times 7, is 4 minus
14, which is minus 10. Now what can I do next. You can kind of see that this
row, well talk more about what this row means. When all of a sudden it's all
been zeroed out, there's nothing here. If I had non-zero term here,
then I'd want to zero this guy out, although it's already
zeroed out. I'm just going to move
over to this row. The first thing I want to do is
I want to make this leading coefficient here a 1. What I want to do is, I'm going
to multiply this entire row by minus 1. If I multiply this entire
row times minus 1. I don't even have to
rewrite the matrix. This becomes plus 1,
minus 2, plus 5. I think you can accept that. Now what can we do? Well, let's turn this
right here into a 0. Let me rewrite my augmented
matrix in the new form that I have. I'm going to keep the
middle row the same this time. My middle row is 0, 0, 1,
minus 2, and then it's augmented, and I
get a 5 there. What I want to do is I want to
eliminate this minus 2 here. Why don't I add this row
to 2 times that row. Then I would have minus 2, plus
2, and that'll work out. What do I get. Well, these are just
leading 0's. Then I have minus 2,
plus 2 times 1. That's just 0. 4 plus 2 times minus
2, that is minus 4. That's 4 plus minus 4,
that's 0 as well. Then you have minus
10 plus 2 times 5. Well, that's just minus 10
plus 10, which is 0. That one just got zeroed out. Normally, when I just did
regular elimination, I was happy just having the situation
where I had these leading 1's. Everything below it were 0's. I wasn't too concerned about
what was above our 1's. What I want to do is,
I want to make those into a 0 as well. I want to make this
guy a 0 as well. What I can do is, I can replace
this first row with that first row minus
this second row. What is 1 minus 0? That's just 1. 2 minus 0 is 2. 1 minus 1 is 0. 1 minus minus 2 is 3. 7 minus 5 is 2. There you have it. We have our matrix in reduced
row echelon form. This is the reduced row echelon
form of our matrix, I'll write it in bold, of our
matrix A right there. You know it's in reduced row
echelon form because all of your leading 1's in each
row-- so what are my leading 1's in each row? I have this 1 and
I have that 1. They're the only non-zero
entry in their columns. These are called the
pivot entries. Let me label that for you. That's called a pivot entry. Pivot entry. They're the only non-zero
entry in their respective columns. If I have any zeroed out rows,
and I do have a zeroed out row, it's right there. This is zeroed out row. Just the style, or just the
convention, is that for reduced row echelon form, that
has to be your last row. We have the leading entries are
the only -- they're all 1. That's one case. You can't have this a 5. You'd want to divide that
equation by 5 if this was a 5. So your leading entries
in each row are a 1. That the leading entry in each
successive row is to the right of the leading entry of
the row before it. This guy right here is to
the right of that guy. This is just the style, the
convention, of reduced row echelon form. If you have any zeroed out rows,
it's in the last row. And finally, of course, and I
think I've said this multiple times, this is the only non-zero
entry in the row. What does this do for me? Now I can go back from
this world, back to my linear equations. We remember that these were the
coefficients on x1, these were the coefficients on x2. These were the coefficients on
x3, on x4, and then these were my constants out here. I can rewrite this system of
equations using my reduced row echelon form as x1,
x1 plus 2x2. There's no x3 there. So plus 3x4 is equal to 2. This equation, no x1,
no x2, I have an x3. I have x3 minus 2x4
is equal to 5. I have no other equation here. This one got completely
zeroed out. I was able to reduce this system
of equations to this system of equations. The variables that you associate
with your pivot entries, we call these
pivot variables. x1 and x3 are pivot variables. The variables that aren't
associated with the pivot entry, we call them
free variables. x2 and x4 are free variables. Now let's solve for, essentially
you can only solve for your pivot variables. The free variables we can
set to any variable. I said that in the beginning
of this equation. We have fewer equations
than unknowns. This is going to be a not well
constrained solution. You're not going to have just
one point in R4 that solves this equation. You're going to have
multiple points. Let's solve for our pivot
variables, because that's all we can solve for. This equation tells us, right
here, it tells us x3, let me do it in a good color, x3
is equal to 5 plus 2x4. Then we get x1 is equal to
2 minus x2, 2 minus 2x2. 2 minus 2x2 plus, sorry,
minus 3x4. I just subtracted these from
both sides of the equation. This right here is essentially
as far as we can go to the solution of this system
of equations. I can pick, really, any values
for my free variables. I can pick any values for my
x2's and my x4's and I can solve for x3. What I want to do right now is
write this in a slightly different form so we can
visualize a little bit better. Of course, it's always hard to
visualize things in four dimensions. So we can visualize things a
little bit better, as to the set of this solution. Let's write it this way. If I were to write it in vector
form, our solution is the vector x1, x3, x3, x4. What is it equal to? Well it's equal to-- let
me write it like this. It's equal to-- I'm just
rewriting, I'm just essentially rewriting this
solution set in vector form. So x1 is equal to 2-- let
me write a little column there-- plus x2. Let me write it this way. Plus x2 times something plus
x4 times something. x1 is equal to 2 minus 2 times
x2, or plus x2 minus 2. I put a minus 2 there. I can say plus x4
times minus 3. I can put a minus 3 there. This right here, the first
entries of these vectors literally represent that
equation right there. x1 is equal to 2 plus x2 times minus
2 plus x4 times minus 3. What does x3 equal? x3 is equal to 5. Put that 5 right there. Plus x4 times 2. x2 doesn't apply to it. We can just put a 0. 0 times x2 plus 2 times x4. Now what does x2 equal? You could say, x2 is equal
to 0 plus 1 times x2 plus 0 times x4. x2 is just equal to x2. It's a free variable. Similarly, what does
x4 equal to? x4 is equal to 0 plus 0 times
x2 plus 1 times x4. What does this do for us? Well, all of a sudden here,
we've expressed our solution set as essentially the linear
combination of the linear combination of three vectors. This is a vector. You can view it as
a coordinate. Either a position vector. It is a vector in R4. You can view it as a position
vector or a coordinate in R4. You could say, look, our
solution set is essentially-- this is in R4. Each of these have four
components, but you can imagine it in r3. That my solution set
is equal to some vector, some vector there. That's the vector. Think of it is as a
position vector. It would be the coordinate
2, 0, 5, 0. Which obviously, this is four
dimensions right there. It's equal to multiples
of these two vectors. Let's call this vector,
right here, let's call this vector a. Let's call this vector,
right here, vector b. Our solution set is all of this
point, which is right there, or I guess we could call
it that position vector. That position vector will
look like that. Where you're starting at the
origin right there, plus multiples of these two guys. If this is vector a, let's do
vector a in a different color. Vector a looks like that. Let's say vector a looks like
that, and then vector b looks like that. This is vector b, and
this is vector a. I don't know if this is going to
be easier or harder for you to visualize, because obviously
we are dealing in four dimensions right here, and
I'm just drawing on a two dimensional surface. What you can imagine is, is that
the solution set is equal to this fixed point, this
position vector, plus linear combinations of a and b. We're dealing, of
course, in R4. Let me write that down. We're dealing in R4. But linear combinations
of a and b are going to create a plane. You can multiply a times 2,
and b times 3, or a times minus 1, and b times
minus 100. You can keep adding and
subtracting these linear combinations of a and b. They're going to construct
a plane that contains the position vector, or contains
the point 2, 0, 5, 0. The solution for these three
equations with four unknowns, is a plane in R4. I know that's really hard to
visualize, and maybe I'll do another one in three
dimensions. Hopefully this at least gives
you a decent understanding of what an augmented matrix is,
what reduced row echelon form is, and what are the valid
operations I can perform on a matrix without messing
up the system.