[BLANK_AUDIO]. 
We would like to learn how to solve 
simultaneous equations. 
Now I'm going to give you an example of a 
second order equation, second order 
matrix. 
Third order is a little bit too hard to 
solve by hand. 
So let's look at a general form first. 
Suppose I've got a 2 by 2 system, two 
equations and two unknowns. 
And it's written in matrix form like 
this. 
This is my A matrix, it's a 2 by 2. 
And then this is my unknown vector. 
Call it v1 and v2. 
And this is a known vector. 
Call it b1 and b2. 
This is capital V, meaning this vector. 
This is capital B, meaning this vector. 
So this equation can be written AV is 
equal to B. 
If I want to solve for V, I can write it 
as A inverse B, and I have to be able to 
find that matrix inverse. 
For a 2 by 2, there's actually a fairly 
simple way of doing matrix inversion, and 
I'll just show you the summary of that. 
A inverse. 
So taking this matrix A, I swap the 
elements on the diagonal, a4 and a1. 
I negate the elements that are off 
diagonal. 
So minus a2, I keep them the same but I 
negate them. 
And then I divide the whole thing by the 
determinant of the original matrix, which 
is a1 times a4, minus a2 times a3. 
A1 times a4 minus a2 times a3. 
Let's look at an example. 
So, here's my matrix. 
Zero minus 1, 3, 4, v1, v2 is equal to 1, 
2. 
Following this formula up here, I swap 
the terms on the diagonal, so 4 and a 
minus 1, I swap. 
I negate the other terms, and then I 
divide by the determinant, which is minus 
1 times 4 minus 0, so I've got a minus 4. 
Sometimes it helps to double check 
yourself on this, double check your 
algebra. 
So, off to the side here, I'm going to 
show how we can double check this, this 
algebra. 
Before I go on, before I want to 
continue, I just, I always want to make 
sure that I inverted the matrix 
correctly, I did my algebra correctly. 
So, off to the side, I want to make sure 
that A inverse is correct. 
Well, the way I'm going to check it is to 
note that if I take a, a matrix and 
times, multiply it by the inverse, I 
should get an identity matrix. 
An identity matrix is just a matrix 
that's diagonal with ones on the 
diagonal. 
So let's go back and double check this. 
My A matrix was minus 1, 3, 0, 4. 
Let me multiply it by the inverse, which 
is, let's go ahead and, and divide 
through. 
Well, let me write it this way. 
It's the way I had it. 
The determinant is just a constant that I 
can multiply through by any of it, by all 
of the terms in one of the matrices. 
So if I multiply this out, let me go, 
this is, I will end up with a 2 by 2 
matrix. 
I start with a two by two times a two by 
two, I will get a two by two. 
And the first, the first element is the 
first row times the first column. 
So it's a minus 1 times 4, plus 0 divided 
by 4, divided by a minus 4, and that ends 
up as 1. 
And the first row, second column is 
gotten by multiplying the first row times 
the second column of these two matrices, 
and that's going to be a minus 1 times 0, 
plus 0. 
Second row, first column, I get by 
multiplying the second row times the 
first column, which is 12 minus 12, which 
is 0. 
And similarly, if I multiply the last row 
times the last column and bring in the 
determinant, the denominator there, I get 
a 1. 
So, this checks. 
I have, I, I, it just tells me that I 
inverted my matrix correctly. 
So, going back and using that with the 
original problem, my goal was to solve 
for V1, V2. 
So V1, V2 is equal to this inverse, which 
is, I'm going to pull in this value here. 
It's a constant. 
I multiply by every term. 
So that's going to be a minus 1, 3 4ths, 
0 and a 1 4th times my b vector, which is 
this, is equal to minus 1 and 5 4ths. 
So, that's my solution right there. 
This is a hand way of solving for a 2 by 
2. 
I would not do a 3 by 3 by hand. 
I would use a matrix solver, which a lot 
of calculators do, and there are also 
solvers available on the internet. 
I want to do one more problem, one more 
example. 
2, 1, minus 1, 1. 
And I want to solve for, say, i sub 1 and 
i sub 2, because I've changed the 
variables here. 
It doesn't really matter what I call 
these variables. 
I'm just solving for this as my unknown. 
Okay. 
So If I call this A, then A inverse is 
going to be equal to, reversing these 
two, interchanging these two diagonal 
terms. 
1, 2. 
Negating the off-diagonal terms. 
And then dividing by the determinant. 
In this case, it's 2 minus a minus 1, so 
2 minus, minus 1. 
So, I'm going to end up with 1 3rd there. 
So, I get 1 divided by 3, 1 divided by 3, 
minus 1 divided by 3 and 2 divided by 3. 
So, then if I solve for i1, i 2, it's 
going to be this inverted matrix times 
this b vector. 
And that gives me 1 3rd times 1, and a 0 
there. 
And similarly here, I'm going to have a 1 
3rd, and that's my solution. 
Just to summarize then, to do this, these 
problems, I have to invert the matrix, 
and we gave you a formula for how to 
invert the matrix. 
And then once I inverted it, I multiplied 
it by this vector, which is always on the 
right side of the equation.