Okay, this interesting question came up
in class and so let's take a look at it.

We want to be able to find all
of the currents in the circuit,

I1, 2, 3, 4, 5, and 6.

So our total number of unknowns, N = 6.

The first thing I'm going to do is
define the polarity of all of these,

plus on the tail minus on the tip.

Plus to minus, plus to minus, and plus
to minus, plus to minus, plus to minus.

Then the next thing I'll do
is color all of my nodes so

that we can kinda more
clearly see the circuit.

There's one node right there,
here's another node here.

Another node right there.

And there are two ordinary nodes here
we're not going to end up using them.

And we could use this node right here,
except the other nodes are already going

to have taken care of it by
the time we get to that point.

So let's start by drawing
our loop equations,

Kirchhoff's loops, loop equations.

This is loop number one,
let's start right here.

We're going to the minus part of I2,

so it's -I2(9 ohms) + I4(6

ohms) + I6(6 ohms) + 6 volts = 0.

Now let's do this loop,
let's call that loop number 2.

As we're moving through here,

we can see we're going to -I6(6

ohms) + I5(6 ohms)- I1(9
ohms) + 6 volts = 0.

Now we need to do one more loop,
we can either do a loop all the way around

the outside, or
we can do a small loop right here.

Let me just do that small loop just for
fun, so

we'll start there, this is loop number 3,

-I4(6 ohms)- I3(9 ohms),
and -I5(6 ohms) back to 0.

Okay that's three loop equations and
they have covered

all of the elements in my circuit, every
single element has been touched by a loop.

So if I do any more loop equations,
I will have redundant loops.

Now, it's time for
me to do node equations.

So let's do this node equation right here,

that would be equation number 4,
that is the V2 node.

What we'll do is say all
the currents coming in,

I4 is equal to all the currents going out,
I5 + I6.

Now let's do the V3 node that
would be equation number 5,

what's coming in there, looks like I3 is
coming in and going out is I2 and I4.

Our last equation is going to
be this green node, the V4 node.

And what's coming in, I5 and
I1 are coming in and I3 is going out.

So those were our Kirchhoff's equations,
let's convert them now to matrix form.

I'm going to have a 6 by 6 matrix
because I'm going to have unknowns,

I1, I2, I3, I4, I5, and I6.

I'm just writing them for convenience
across the top, and equations 1,

2, 3, 4, 5 and 6, that's just written for
convenience as well.

And then I write my unknowns here
in this vector, 3, 4, 5 and 6 and

this will be my vector of constants.

Okay, so let's look at our first equation,
I1 times 0, I2 times -9,

I3 times 0, I4 times 6,
I5 times 0, I6 times 6 ohms.

And then I need to bring this 6 over
here in order to be in the constants.

My second equation has 9 ohms- 9 ohms,

times I1, nothing times I2 or I3 or I4.

It has 6 ohms times I5, and
it has -6 ohms times I6, and

this 6 volt also has to come
over to the other side.

Right here we have 0, 0, I3 has -9,

I4 has -6, I5 has -6, and I6 has 0,

and there are no constants on this side.

Now let us do equation number 4,
nothing times I1, I2, or I3.

We have to get these on the same
side of the equation, so

I'm gonna bring this over here, so
that whole equation will be equal to 0.

So what I will be writing is
0 is equal to -I4 + I5 + I6,

so there's a -1 here,
a 1 there and a 1 there.

Similarly, I'm going to bring this
over here, 0, 1, -1, 1, 0, 0.

And in this last equation,
let's bring that over here, so

we'll have 1 times I1, 0,
-1, 0, 1 and 0, equal to 0.

So there's my matrix equation, just
a minute, I'll put that into MATLAB and

give you a final answer but there's
our solution to our matrix problem.