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# Exploring the solution set of Ax = b

Exploring the solution set of Ax=b (non homogeneous equations). Created by Sal Khan.

Video transcript

Let's say I have some linear
transformation T from R2 to R2. So that is R2, and
then this is R2. T just maps from any member of
R2 to another member of R2, just like that. And I'm going to define T. It's a linear transformation. I'm going to define it. When I take the transformation
of some member of R2, it's equivalent to multiplying it by
this matrix, by the matrix 1 minus 3, minus 1, 3. So just to kind of understand
this transformation a little bit more, let's think about all
of the values that it can take on in our codomain. So if we say that 1 minus 3,
minus 1, 3, times any vector in our domain-- so x1, x2-- it's
going to be equal to some other vector in our codomain. Let me call that vector b. B is a member of R2. So it's going to be
equal to b1, b2. All I'm doing here is I'm
essentially-- if I call this right here, I call that matrix
A-- I'm just trying to find out what are all the possible
values for Ax is equal to b? I'm trying to find out all of
the possible b's in this case. So if I were to solve this
equation for any particular b, what I would do is I
would put this in reduced row echelon form. Well, first I would make an
augmented matrix of it. So I would put a 1, a minus
3, a minus 1, and a 3. And I would augment it with
the member of our codomain we're trying to be equal to. So b1, b2. And then, I would put it in
reduced row echelon form. So how can we do that? How can we put it in reduced
row echelon form? I'll keep my first
row the same. 1 minus 3, and then b1. And then my second row, I'll
replace it with my second row plus my first row. So minus 1 plus 1 is 0. 3 plus minus 3 is 0. And then b2 plus b1 is--
well I can just write that as b1 plus b2. So this is only going
to have solutions. I have an interesting
situation here. You've seen it before. I have a row of 0's. And the only way that this
is going to actually have solutions is if this thing right
here is also going to be equal to 0. So the only members b that are
members of Rm that have solutions are the ones where
if you add up their two terms-- so let's say b is equal
to b1 and b2-- are the ones where the two terms added
up have to be to 0. b1 plus b2 have to
be equal to 0. Or another way you could write
it is that b2 has to be equal to minus b1. So if we were to actually draw
our codomain-- let's do it. We always stay in the abstract,
but sometimes it's useful to draw an
actual example. Let's say that our
codomain is R2. Let me draw some axes
right here. Let's say that this
is my b1-axis, and this is my b2-axis. I could have called that
x and y, but I called that b1 and b2. What are all of the members
of my codomain that have a solution, that have a mapping? Well, b2 has to be equal
to minus b1. So it's going to
look like this. It's just going to be a line
with a slope of negative 1. This is all the b's that
have a solution. Because if you're not
on this line. If you're a member of your
codomain-- this is the codomain right here, R2. R2 is also our domain, but let
me make it very clear that this is the codomain
that I've drawn. This is what we're
mapping into. It's very clear that if we're
not on this line, if you pick somebody whose two terms don't
add up to equalling 0, or aren't the negative of each
other, if you pick someone over here in our codomain, and
then you try to solve the equation over here, 0 is going
to equal some non-zero number here and you're not going
to have a solution. And we touched on this
in the last video. And so in this case we could
say that this right here is the image of our
transformation. Or even another way of
thinking about it. This is our codomain. Let me draw a domain. So if you take any member of
R2, and it's always mapping onto something onto that line,
clearly each point on that line is going to be mapped to
by more than one vector. So we're not dealing with
an onto transformation. We saw that in the last video. In order for something to be
onto, when you put it in reduced row echelon form,
you cannot have all 0's in one of the rows. Or another way to say it in
reduced row echelon form, every row has to have
a pivot entry. Well, let's focus
on the b's that actually do have a solution. So let's focus on these b's,
so that when you take a b1 plus b2, it actually is going
to be equal to 0. So we could have the b, I don't
know, that could be the b5 minus 5. Obviously the b0, 0
is going to work. You could have 1, negative 1. Maybe that's right there. Let's focus on those for a
second and see how many members of our domain
map into them. So if we take this right here
and then we apply this equation up here, we only
have one constraint. We're assuming that this is
going to be equal to 0. So let's assume that
we're dealing with the b's in our image. Let's assume that we're dealing
with something where we can get a solution. That b1 plus b2 is equal to 0. Then what is our constraints? What will map to our vector
b that we're dealing with? If we just take this topic
equation right here, we have 1 times x1-- let me switch
colors-- minus 3 times x2 is equal to b1. And then this row will give us
no constraints, because it's just going to be
a bunch of 0's. So this is the only constraint
for a member of our domain that will map to some particular
b now that we are picking, some particular
b that satisfies this constraint. So we could write this solution
set-- let me rewrite it as x1 is equal
to b1, plus 3x2. Or if we wanted to write the
entire solution set, it would look like this. x1, x2 is equal to the vector
b1, 0, plus-- x1 is b1, plus 3, times x2, so plus
x2 times 3. And x2 is just going
to be equal to x2. That's a free variable. So x2 is equal to 0
plus x2, times 1. So what we're saying here is
this transformation just maps to this line here for all of the
vectors in our codomain, where their two entries
add up to each other. Let's assume that we actually
have one of those vectors. And so first of all, this is
definitely not an onto transformation. But let's assume that
we are dealing with one of those guys. So if we are picking a
particular one of those guys for a particular b-- let me
write this down-- that has a solution to Ax is equal to b,
the solution set will be equal to this thing right here. It'll be equal to x1, x2
is equal to the first entry of your b. b1, 0 plus x2, times
the vector 3,1. And if you think about this, if
you pick a particular b-- so let's say we pick our-- let
me draw this out, because I think it's nice to
visualize it all. Actually, maybe I'll
draw it like this. I don't want to draw
our blobs anymore. So let me draw my axes. So my axes look like this. We know that the image of our
transformation is the line with a negative 1 slope, because
the two entries have to be equal to the negative
of each other. Let's pick a particular
b that has a solution. So let's say we pick
that b right there. In order for it to have a
solution, its entries have to be the negative of each other. So let's say that's the
entry 5, minus 5. That is our b. So what we just showed is that
the solution set, if we want to say, hey, what in our domain
maps to this guy? So let's think about what
in our domain maps to this right here. Let's think about what in our
domain maps to this point right here, to this
particular b. So that's going to be all of
the x's that satisfy Ax is equal to 5, minus 5. So x1, x2 is going to be equal
to b1, it's going to be equal to 5, 0 plus x2, plus any
scale or multiple of the vector 3, 1. So our solution set is going to
be, you take the vector 5, 0-- so maybe the vector 5, 0,
specifies this position right here-- and then you're going to
add to it multiples of the vector 3, 1. The vector 3, 1 looks like this,
1, 2, 3 and you go up 1. The vector 3, 1 is going
to look like this. So if you add multiples of
this-- multiples of that could stretch out like that or could
go negative like that-- to this vector 5, 0, you're
essentially going to-- let me see if I can draw this neatly--
you're going to end up with a solution set that
looks like that right there. So if you pick a particular
b right there that has a solution, we just said that
everything on this line will map to that point in
our solution set. And in fact, if you picked
another point, let's say you picked the point minus 5, 5. Then the solution set that maps
to that would actually be-- this first term would
be the minus 5. It would be here. And all of these guys
would map to that. Well, this all interesting. I mean, we've been doing a
lot of abstract things. And I think it might be
satisfying that you're actually seeing something more
concrete in this example. But I'm doing all of
this for a reason. Because I want to understand
what the solution set is to a general non-homogeneous
equation like this. And to understand it a little
bit better, let's imagine what is the solution set
if I were to pick this guy, the 0 vector? Then what's going to be
the solution set? So if we say that Ax is equal to
0, then our solution set is going to the vector 0, 0,
plus x2, times 3, 1. So it's going to be what? This is just the 0 vector,
so it's going to be here. It's going to be here, and
it's going to just be multiples of 3, 1. It's going to look something
like that. What's this? What is the solution set to the
equation of Ax equal to 0. This is the null space. This, by definition, is
the null space of A. So notice-- and this is the big
takeaway from this video-- that for any solution-- we're
picking b's that actually do have solutions, because we're
picking them on this line. We're picking them in the
image of our codomain. The solution set to any Ax is
equal to some b where b does have a solution, it's
essentially equal to a shifted version of the null set,
or the null space. This right here is
the null space. That right there is the null
space for any real number x2. Any scale or multiple of
3, 1 is the null space. I just showed it right there. It's going to be that. And so all of these other
solution sets are just some particular vector, some
x particular, plus the null space. Obviously, this vector by
itself would also be a solution to Ax is equal to b,
because you can just set x2 to be equal to 0. So in general-- and I haven't
proven this to you rigorously, but hopefully you kind of get
the intuition behind it. The solution-- and I'll do this
in the next video, just because I realize I'm running
long on time. Assuming Axb has a solution-- in
the example we just did, we can assume it has, if we pick
one of these points here. So assuming it has a solution,
if we pick a point off of our image, then we're not going
to have a solution. But assuming Axb has a solution,
the solution set is going to be equal to some
particular vector-- so you could just think of it as a set
with just one vector right there-- with, or combined with,
or the union of that set, with your null space of
this matrix right here. I haven't proven this to you
yet, but hopefully you get the intuition of why this is true. We just solved it for
particular cases that do have solutions. We say, hey, it's going
to take this form. And I just showed you
that this is the form of the null space. And the reason why we're doing
that is because we've been talking about invertibility. And in order to be invertible
you have to be onto, and 1 to 1. And for something to be 1 to
1, you have to have at most one solution that maps to
a particular vector. You might have none. But you might have,
at most, one. So in order to have, at most,
one solution-- and the solution set is always going to
be equal to this, so you're always going to have this
solution-- so in order to have, at most, one solution,
your null space can't have anything in it. Or it can just have
the zero vector. So that means that your null
space of A has to be trivial, or to be empty, or to just
have the zero vector. I'll do this a little bit more
rigorously in the next video, but I think when you do it
sometimes with the rigor, you don't necessarily get
the intuition. But this is going to be a very
interesting takeway, and I think you already understand
it's leading up to conditions for invertibility.