[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.