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GILBERT STRANG: OK.

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I want to continue
the last video, which

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was about incidence matrices,
and graphs, and networks,

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and flows in the network.

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So that was 5.6.

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This is 5.6b.

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And I'll remember
the same graph.

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You remember a graph is
some nodes, four nodes here,

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and some edges, and in
this case five edges.

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So I have a 5 by 4 matrix,
and that's what it was.

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And I'll remember
how it was created.

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Every row corresponds
to an edge.

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So the first edge there
goes from node 1 to node 2,

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so I put a minus 1 and a
1 in columns one and two.

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That tells me what that
first edge is doing

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and it gives me one row
of the incidence matrix.

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Five edges give me five rows.

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There is the matrix.

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And here I'll multiply by
v, thinking of a vector v

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as voltages at the four
nodes, and I get that answer.

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The 1 and minus 1 produce
this kind of answer.

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OK.

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Now I'm ready for questions
about the matrix A, the 5

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by 4 matrix.

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These matrices, these
incidence matrices,

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are beautiful examples
of rectangular matrices

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where we can ask all the
key questions about a matrix

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and get a nice answer.

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And the key questions
that I have in mind are

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what are their solutions
to Av equals 0?

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Are there--

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That says, are there are
combinations of the columns

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that give the zero column?

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So it's asking, are
the columns dependent?

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If the columns were dependent,
then I'll find some solutions,

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and here I will.

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If the columns are independent,
the only solution I will find

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will be v equals 0.

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But those columns are dependent.

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Now, how can we see that?

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Well, in this case,
we can find a solution

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to Av equals 0,
because I can see

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how do I get all those
differences to be 0?

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Well, not hard.

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v could be the
vector of all 1's.

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Then the differences would all
be 1 minus 1, would all be 0.

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I would be solving Av equals 0.

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And of course, I can
multiply by any constant.

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The voltage is--
So all I'm saying

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is if all the voltages are
equal, there won't be any flow.

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If all the voltages
are equal and I

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don't have any batteries or
other sources in the network,

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there will be no flow.

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And those are all the solutions.

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But the only way I
could make all those 0

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would be for all the
v's to be the same.

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So all the v's have
to be the same.

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v is C, C, C, C.

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And I learn something important.

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Av equals 0 has some solutions.

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And I'll just jump ahead
one electrical moment.

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That's not good if we
want an invertible matrix.

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In the end we would
have A transpose A

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and it won't be invertible
unless we do something.

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And what do we do?

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We want to get rid
of that last column.

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We can have three columns.

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Those will be independent,
but that fourth column

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is a combination of the others.

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And what we do,
in reality, is we

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ground a node, which means we
set one of the v's, maybe v4,

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if we set that to 0, it's like
we're fixing the temperature,

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we're fixing the
voltage, we often

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have to do this on
a sliding scale.

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If we only know
differences in temperature,

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we have to say, where is 0?

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And if we make that
point 0, then we

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have only three unknown
voltages and a 5

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by 3 matrix and all well.

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OK.

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So that's the discussion
of Av equals 0.

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Now, what about A
transpose w equals 0?

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So now I'm asking about the
transpose of that matrix.

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Now, this is a 4 by 5 matrix.

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Again, a beautiful
example, 4 by 5 matrix.

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w, of course.

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It's a 4 by 5 matrix
multiplying w, which is 5 by 1.

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So 4 by 5 times 5 by 1.

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And I want to get all
zeros, four zeros this time.

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Right.

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So first of all, if I
have a 4 by 5 matrix,

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so when I transpose this is
you could say short and wide,

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I think there are
automatically solutions.

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There will be solutions
in a 4 by 5 matrix.

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With five unknowns and
only four equations,

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I'm going to have some
solutions to that system.

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So there will be some solutions.

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Well, the question is how
many different w's could I

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find, how many
different solutions,

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and what do they mean.

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And that's the beauty
of this example,

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that it's not just a bunch
of 20 numbers in the matrix.

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The matrix has a meaning.

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The incidence matrix
takes differences A to Av

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is differences in v, but what's
the meaning of A transpose?

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That's the key question here.

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Why is this equation
very important?

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OK.

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So I have to tell you the
meaning of A transpose.

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And maybe I have to copy
down what A transpose is.

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So let me go to the next board
and copy down A transpose.

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So I'm looking now
at A transpose w.

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So now it will be 4 by 5.

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So that first row
becomes a column.

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The second row becomes another
column in the transpose.

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The third row, another column,
the fourth row, is that column.

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And the fifth row is that one.

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And that will multiply w1, 2,
3, 4, and 5 to give 0, 0, 0, 0.

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And that's called the current
law, Kirchhoff's current law.

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And what is that law?

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What does it mean?

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It means that in the network at
a typical node, so at node 1,

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you remember, there
was an edge out.

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Edge 1 went out.

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Three edges went out actually.

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This was to node 2.

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This was to node 3,
and that was to node 4.

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At node 1, three
edges are going out.

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And what does the
current law tell me?

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It tells me that the
total flow out is 0.

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The net flow, any flow in, which
would be negative w's, and any

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flows out, which would
be positive w's-- w,

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that came from the first edge.

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This was maybe the second edge.

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And I think that happened
to be the fourth edge--

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flows out of w.

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And that's what I see here.

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A 1, a 2, and a
4 are multiplying

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w1, w2, and w4 are the-- The
first equation there is minus.

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w1 plus w2 plus w4 equals 0.

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So that came from the first
row of A transpose w equals 0.

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Right?

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I just took those numbers
from the first row.

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I wrote down that
first equation.

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And you see it says exactly
the sum of those three flows

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has to be 0.

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So if there's some
positive flows going out,

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there must be some negative
w's coming in to balance.

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OK.

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And that was at node 1, and
similarly at nodes 2, and 3,

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and 4, currents balance.

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It's the balance equation.

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Kirchhoff's law, it's
the balance equation.

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It's conservation.

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A fundamental
equation in modeling

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applied mathematics is if
a body is sitting there

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in equilibrium, then the
forces on it are in balance.

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If I have steady flow
around the network,

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the currents are in balance.

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Always there's a
balance equation,

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so that things are not
collecting up at a node.

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It's stable.

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OK.

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So that's the meaning of
Kirchhoff's current law.

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That's the meaning of
A transpose w equals 0.

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And what about solutions?

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Solutions w.

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Now, so now we're getting
down into the details.

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Can we actually find the w's?

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Well, there will be some.

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There will be some.

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As I said, we've got five
unknowns here and only

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four equations.

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So we're certainly going
to find a solution.

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And let me suggest one
good way to look for it.

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Suppose the flow-- Let me
put in the other two edges--

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suppose the flow
goes around a loop.

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Loops are the key here.

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The key to the
solution is a loop.

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So that's a flow that sends
a flow of 1 along that edge,

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a flow of 1 going that way
along that edge, which I think

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was w5, and a flow
of 1 going that way.

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Pay attention.

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It's going to send 1
Amp around the loop.

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I go with the arrow, with the
flow, this way and this way,

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but this one is
against the arrow.

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So I'm thinking that a
solution is w1 equals 1.

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You see I'm writing
down a solution

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without doing any elimination
or other linear algebra.

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I'm just understanding
the picture.

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w1 is 1.

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w5 is 1.

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w5 is 1.

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And what is w4?

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Negative 1, because it's
going against the arrow.

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And the other two
w's are 0, w2 and w3.

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This was w3 here.

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Those are not
involved in this loop.

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So there is a solution
with w2 and w3 equal to 0.

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And I think that
how could it fail

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on Kirchhoff's current law?

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Nothing is piling up at a node.

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We're just sending
it around a loop.

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And of course, I
put in that's a 1.

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w2 is a 0.

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w4 is a minus 1.

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I have a 1 and a minus 1.

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I get 0, just right.

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And all the equations
would be solved.

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In other words,
conclusion, the solutions w

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00:13:02,590 --> 00:13:10,960
come from loops in the network.

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Every loop in the
network gives me a w.

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Here's another loop.

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I could send flow down there.

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Now that would be a w4 plus 1.

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This way.

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This way.

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Do you see that second loop?

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Let me draw my
little loopy symbol.

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Flow going around that loop.

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That loop happens to
have four edges on it.

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So I'd have four w's.

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1 minus 1, 1, and minus 1,
and no flow on edge 1, and I

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would have another solution.

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And it would be a
different solution.

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So I'm going from-- Can
I insert here two loops?

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In that graph I see two
loops, two small loops.

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And each of those small
loops gives me a flow, a w,

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that solves the current law,
because it's just continuously

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running around and around.

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Now, there's another
question to ask you,

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and that is what about
the big loop, w1, w3--

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I think that is-- and minus w2?

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What if I send flow
around the big loop?

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No problem.

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That gives me
another set of w's .

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Those satisfy
Kirchhoff's current law.

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They satisfy these equations.

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They satisfy A
transpose w equals 0.

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But I don't want that big loop.

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I don't want to include that in
my list of w's, because I was

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only looking for two w's.

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I was only looking for two w's.

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And linear algebra told me that
was the number to look for.

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And here you're suggesting--
I'll blame you--

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00:15:07,770 --> 00:15:10,370
a third around the big loop.

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So what's up?

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00:15:12,410 --> 00:15:14,350
Well, do you see it?

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The flow around that big
loop does solve A transpose w

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equals 0, but it's not new.

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It's the sum of a
flow around that loop

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plus a flow around that.

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Do you see?

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If I add together the flow
vector, the loop vector for w

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for that loop and
for that loop, they

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will cancel on the edges
that are in both loops,

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and I'll just be left with the
flow there, the flow there,

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and the flow there, and
that's the big loop.

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In other words, that
big loop doesn't give me

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a new vec-- It doesn't
give me-- It gives me

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a vector w that's a combination
of what I already have.

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And in linear algebra,
that's always the question.

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You want the number
of independent w's,

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and this big loop
is a dependent w,

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because it's a combination
of the other two.

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OK.

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So that's the picture for
one particular example.

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I'll just end with linear
algebra facts, linear algebra

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facts.

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OK.

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So how many-- So if I
have an m by n matrix,

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and suppose Av equals 0 has
how many independent solutions

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shall I say?

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k independent solutions.

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And in my example,
the incidence matrix,

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the answer was, for A equal
incidence matrix, k was 1.

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So if I know the number of
solutions to that equation,

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then how many
solutions do I expect

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to-- This has-- So how many
solutions do I expect there?

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The difference between m and n
comes in it, and then plus k.

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So independent solutions.

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That's a basic fact
of linear algebra

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00:17:50,190 --> 00:17:52,120
that I never wrote down before.

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00:17:52,120 --> 00:17:55,400
I never wrote it
in this notation.

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00:17:55,400 --> 00:18:00,480
I'll make that a question on
a future linear algebra exam.

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00:18:00,480 --> 00:18:10,370
What I'm saying is that if I
know how many solutions Av has,

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00:18:10,370 --> 00:18:12,640
how many combinations,
these are combinations

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00:18:12,640 --> 00:18:15,560
of the columns of
A that give 0, then

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00:18:15,560 --> 00:18:19,330
I know how many combinations
of the rows of A.

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00:18:19,330 --> 00:18:26,450
Let's just check that this
counting theorem was correct.

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00:18:26,450 --> 00:18:28,970
This was k equals 1, right?

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00:18:28,970 --> 00:18:34,610
The only solution to Av equals
0 was the constants, 1, 1, 1, 1.

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Then m was 5.

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00:18:37,040 --> 00:18:38,400
n was 4.

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00:18:38,400 --> 00:18:39,720
k was 1.

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00:18:39,720 --> 00:18:42,680
5 minus 4 plus 1 is 2.

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00:18:42,680 --> 00:18:52,020
And that's the number
of loop solutions

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to Kirchhoff's current law.

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OK.

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00:18:56,610 --> 00:18:58,750
We have voltages.

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00:18:58,750 --> 00:18:59,740
We have currents.

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00:18:59,740 --> 00:19:07,510
And there's a lot of beautiful
linear algebra involved

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00:19:07,510 --> 00:19:09,950
with these matrices.

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00:19:09,950 --> 00:19:16,760
I'll also include a video about
RLC circuits, which are totally

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00:19:16,760 --> 00:19:18,490
an application of this.

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00:19:18,490 --> 00:19:24,560
And there I'll begin with
just one loop, one RLC loop.

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00:19:24,560 --> 00:19:31,420
But the reality of
modern electronics

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00:19:31,420 --> 00:19:37,560
is thousands of nodes,
thousands of edges,

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00:19:37,560 --> 00:19:43,490
maybe tens of thousands of
edges, and many, many loops.

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00:19:43,490 --> 00:19:44,390
Good.

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00:19:44,390 --> 00:19:45,940
Thank you.