MARTINA BALAGOVIC: Hi.
Welcome back.
Today's problem is about solving homogeneous linear
systems, ax equals 0, but it's also an introduction to the
next lecture and next recitation section, which are
going to be about solving non-homogeneous linear
systems, ax equals b.
The problem is fill the blanks type.
And it says the set S of all points with coordinates xy and
z, such that x minus 5y plus 2z equals 9 is a blank in R3.
It is in a certain relation to the other blank S0 of all the
points with coordinates x, y, and z that satisfy the
following linear equation, x minus 5y plus 2z equals 0.
After we solve this, we have the second part of the
problem, which says all points of x have a specific form, x y
z equals blank 0 0 plus some parameter times blank 1 0 plus
some other parameter times blank 0 1.
And we need to fill out all six blanks.
Now you should pause the video, fill in the blanks, and
then come back and see some pretty pictures that I
prepared for you.
And we're back.
So you probably picked this up in lectures already.
If you have a three dimensional space with three
degrees of freedom, and put in one constraint, so put in one
equation, you get something that has two degrees of
freedom, something that's two dimensional.
If this equation is linear, rather than quadratic or cubic
or exponential, this something is something two
dimensional and flat.
Something two dimensional and flat in R3 is also called a
plane, or a two plane.
Similarly, S0 is also a plane.
Now, what's the relation between S and S0 if they're
given by these two equations?
Well first let's look at the general positions in which two
planes in R3 can be.
First one is that they're intersecting along a line.
What's going to happen here is that all points on this plane
are points whose coordinates satisfy the
equation of this plane.
The points in this plane are points whose coordinates
satisfy the equation of this plane.
And the points on the line are points whose coordinates
satisfy the system of this equation and this equation.
The other position in which two planes can be is that
they're not intersecting at all, that they're parallel.
So let's start by trying to find this line here.
The equation of one plane is x minus 5y plus 2z equal 9.
The equation of the other one is x minus 5y
plus 2z equals 0.
Now you can just look at it and see how many solutions
it's supposed to have, or you can try doing elimination, and
after one step of elimination get 0 equals
9, which never happens.
There cannot exist numbers x, y, and z such that this
combination of them produces 0, and the same combination of
them produces 9 at the same time.
So this red line here doesn't exist, and the situation of
these two planes S and S0 is this one, they're parallel.
So let's add the word parallel in here.
And let's move on to the other half of the problem.
The other half said all points of S have this specific form.
Now let me call this point here P0.
If all points of S have this form, we can plug in any
parameter c1 and c2 here and we're going to get a
point of the plane.
So in particular, we can plug in c1 and c2 equal to 0.
What we get then is that the point x y z equals P0 is a
point of the plane S. So P0 is in S.
What do we know about the point P0?
Well the fact that it's in S means the its coordinates, x
minus 5y plus 2z equal 9.
That's the equation of S. But we also know that y and z are
equal to 0 and 0.
Solving this system we get that the x-coordinate of this
point P0 is 9, and we can just add 9 here.
So we just have two blanks left to fill.
Before we'll fill them, let me show you a
picture that I drew here.
So we have these two planes, S0 and S, which are parallel.
They're given by these equations.
And the plane S0 has a point 0 in it, because the equation is
x minus 5y plus 2z equals 0, so it satisfied by 0 0 0.
The plane S has this point P0 in it, which is 9 0 0-- we
just figured this out.
And there's this vector connecting one
plane to the other.
Now, since those two planes are parallel and there's this
vector going between them, what we can see is that a good
way to get any point in S is to go to any point in S0 and
go up by this vector.
Now let me write this down.
What I just said is that any point in S is of the form--
use this vector to go up--
plus any point in S0.
And if we compare this to this expression here, we also get
P0 plus this linear combination.
So this here has to be a point in S0.
Now we're left with a question of how to parameterize all
points in S0.
What are all the points in S0, and what does this problem
have to do with solving homogeneous linear equations?
Well let me write this equation of S0 in a slightly
different way.
Let me write it as 1, minus 5, 2, x y z, equals 0.
And let me think of this as a matrix of the system.
It's a very tiny matrix, but it's a matrix.
And think of it as a matrix dot a vector equals 0, and
trying to find all solutions of the system.
Well let's do row reductions here.
It's already as upper triangular as these tiny
matrices get.
This is a pivot.
So we have a pivot variable x.
These are free variables, y and z.
And if you remember how to solve these systems, for each
free variable we get one particular solution.
So we get one particular solution when we plug in y is
1 and all the other free variables are 0.
Plugging it in here, we just get that in that case, x--
so we get x minus 5 times 1 plus 2 times 0 equals 0.
So x is equal to 5.
And the other solution is for setting all free variables
equal to 0, except z which we set equal to 1.
And then we get x minus 5 times 0 plus 2
times 1 equals 0.
So we get that in this case, x equals minus 2.
And any solution of this system is going to be of the
form some constant times this plus some other
constant times this.
And if we walk back to our original problem here, we see
that these parameters, these numbers here, have been set up
exactly so that we can just take these numbers and just
copy them over, 5 and minus 2.
And this is the general form of any point of the plane S.
It's go up this vector, and then add a point in S0, in the
parallel plane that passes through the origin.
This finishes our problem.
But what I would encourage you to do now is to go on to the
next lecture, watch the next recitation video, and then
come back here and think about what is it that we really did
here on this half of the board.
Thank you.