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ZK3O402wf1c | And I want to take those
equations one at a time and ask |
ZK3O402wf1c | -- |
ZK3O402wf1c | and make a picture of all
the points that satisfy -- |
ZK3O402wf1c | let's take equation number two. |
ZK3O402wf1c | If I make a picture of all
the points that satisfy -- |
ZK3O402wf1c | all the x, y, z points
that solve this equation -- |
ZK3O402wf1c | well, first of all, the
origin is not one of them. |
ZK3O402wf1c | x, y, z -- it being 0, 0, 0
would not solve that equation. |
ZK3O402wf1c | So what are some points
that do solve the equation? |
ZK3O402wf1c | Let's see, maybe if x is
one, y and z could be zero. |
ZK3O402wf1c | That would work, right? |
ZK3O402wf1c | So there's one point. |
ZK3O402wf1c | I'm looking at this
second equation, |
ZK3O402wf1c | here, just, to start with. |
ZK3O402wf1c | Let's see. |
ZK3O402wf1c | Also, I guess, if
z could be one, |
ZK3O402wf1c | x and y could be zero,
so that would just |
ZK3O402wf1c | go straight up that axis. |
ZK3O402wf1c | And, probably I'd want
a third point here. |
ZK3O402wf1c | Let me take x to be
zero, z to be zero, |
ZK3O402wf1c | then y would be
minus a half, right? |
ZK3O402wf1c | So there's a third point,
somewhere -- oh my -- okay. |
ZK3O402wf1c | Let's see. |
ZK3O402wf1c | I want to put in all the points
that satisfy that equation. |
ZK3O402wf1c | Do you know what that
bunch of points will be? |
ZK3O402wf1c | It's a plane. |
ZK3O402wf1c | If we have a linear
equation, then, fortunately, |
ZK3O402wf1c | the graph of the thing, the plot
of all the points that solve it |
ZK3O402wf1c | are a plane. |
ZK3O402wf1c | These three points
determine a plane, |
ZK3O402wf1c | but your lecturer
is not Rembrandt |
ZK3O402wf1c | and the art is going to
be the weak point here. |
ZK3O402wf1c | So I'm just going to
draw a plane, right? |
ZK3O402wf1c | There's a plane somewhere. |
ZK3O402wf1c | That's my plane. |
ZK3O402wf1c | That plane is all the
points that solves this guy. |
ZK3O402wf1c | Then, what about this one? |
ZK3O402wf1c | Two x minus y plus zero z. |
ZK3O402wf1c | So z actually can be anything. |
ZK3O402wf1c | Again, it's going
to be another plane. |
ZK3O402wf1c | Each row in a three
by three problem |
ZK3O402wf1c | gives us a plane in
three dimensions. |
ZK3O402wf1c | So this one is going to
be some other plane -- |
ZK3O402wf1c | maybe I'll try to
draw it like this. |
ZK3O402wf1c | And those two planes
meet in a line. |
ZK3O402wf1c | So if I have two equations,
just the first two |
ZK3O402wf1c | equations in three dimensions,
those give me a line. |
ZK3O402wf1c | The line where those
two planes meet. |
ZK3O402wf1c | And now, the third
guy is a third plane. |
ZK3O402wf1c | And it goes somewhere. |
ZK3O402wf1c | Okay, those three
things meet in a point. |
ZK3O402wf1c | Now I don't know where
that point is, frankly. |
ZK3O402wf1c | But -- linear
algebra will find it. |
ZK3O402wf1c | The main point is that the three
planes, because they're not |
ZK3O402wf1c | parallel, they're not special. |
ZK3O402wf1c | They do meet in one point
and that's the solution. |
ZK3O402wf1c | But, maybe you can see
that this row picture is |
ZK3O402wf1c | getting a little hard to see. |
ZK3O402wf1c | The row picture was a cinch
when we looked at two lines |
ZK3O402wf1c | meeting. |
ZK3O402wf1c | When we look at
three planes meeting, |
ZK3O402wf1c | it's not so clear and in
four dimensions probably |
ZK3O402wf1c | a little less clear. |
ZK3O402wf1c | So, can I quit on
the row picture? |
ZK3O402wf1c | Or quit on the row picture
before I've successfully |
ZK3O402wf1c | found the point where
the three planes meet? |
ZK3O402wf1c | All I really want to see is
that the row picture consists |
ZK3O402wf1c | of three planes and, if
everything works right, |
ZK3O402wf1c | three planes meet in one
point and that's a solution. |
ZK3O402wf1c | Now, you can tell I
prefer the column picture. |
ZK3O402wf1c | Okay, so let me take
the column picture. |
ZK3O402wf1c | That's x times -- |
ZK3O402wf1c | so there were two xs in the
first equation minus one x is, |
ZK3O402wf1c | and no xs in the third. |
ZK3O402wf1c | It's just the first
column of that. |
ZK3O402wf1c | And how many ys are there? |
ZK3O402wf1c | There's minus one in the first
equations, two in the second |
ZK3O402wf1c | and maybe minus
three in the third. |
ZK3O402wf1c | Just the second
column of my matrix. |
ZK3O402wf1c | And z times no zs minus
one zs and four zs. |
ZK3O402wf1c | And it's those three
columns, right, |
ZK3O402wf1c | that I have to combine to
produce the right-hand side, |
ZK3O402wf1c | which is zero minus one four. |
ZK3O402wf1c | Okay. |
ZK3O402wf1c | So what have we got on
this left-hand side? |
ZK3O402wf1c | A linear combination. |
ZK3O402wf1c | It's a linear combination
now of three vectors, |
ZK3O402wf1c | and they happen to be -- each
one is a three dimensional |
ZK3O402wf1c | vector, so we want to know
what combination of those three |
ZK3O402wf1c | vectors produces that one. |
ZK3O402wf1c | Shall I try to draw the
column picture, then? |
ZK3O402wf1c | So, since these vectors
have three components -- |
ZK3O402wf1c | so it's some multiple -- let
me draw in the first column |
ZK3O402wf1c | as before -- |
ZK3O402wf1c | x is two and y is minus one. |
ZK3O402wf1c | Maybe there is the first column. |
ZK3O402wf1c | y -- the second column has maybe
a minus one and a two and the y |
ZK3O402wf1c | is a minus three, somewhere,
there possibly, column two. |
ZK3O402wf1c | And the third column has -- |
ZK3O402wf1c | no zero minus one four,
so how shall I draw that? |
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