Solving the matrix vector equation

Voiceover:In the last video we saw

that we could take a
system of two equations

with two unknowns and represent it

as a matrix equation where the matrix A's

are the coefficients here
on the left-hand side.

The column vector X has our two

unknown variables, S and T.

Then the column vector B is essentially

representing the
right-hand side over here.

What was interesting about it,

then that would be the equation A,

the matrix A times the column vector X

being equal to the column vector B.

What was interesting about that is we saw

well, look, if A is invertible,

we can multiply both the
left and the right-hand sides

of the equation,

and we have to multiply
them on the left-hand sides

of their respective sides by A inverse

because remember matrix,

when matrix multiplication order matters,

we're multiplying the left-hand side

of both sides of the equation.

If we do that then we
can get to essentially

solving for the unknown column vector.

If we know what column vector X is,

then we know what S and T are.

Then we've essentially solved

this system of equations.

Now let's actually do that.

Let's actually figure
out what A inverse is

and multiply that times
the column vector B

to figure out what the column vector X is,

and what S and T are.

A inverse, A inverse is equal to

one over the determinant of A,

the determinant of A for a two-by-two here

is going to be two times
four minus negative

two times negative five.

It's going to be eight minus positive 10,

eight minus positive 10,

which would be negative two.

This would become negative
two right over here.

Once again, two times four is eight minus

negative two times negative five

so minus positive 10 which
gets us negative two.

You multiply one over the determinant

times what is sometimes
called the adjoint of A

which is essentially swapping the top left

and bottom right or at least
for a two-by-two matrix.

This would be a four.

This would be a two.

Notice I just swapped these,

and making these two negative,

the negative of what they already are.

This is from a negative two this is

going to become a positive two,

and this right over
here is going to become

a positive five.

If all of this looks
completely unfamiliar to you,

you might want to review the tutorial

on inverting matrices

because that's all I'm doing here.

So A inverse is going to be equal to,

A inverse is going to be equal to,

let's see, this is negative 1/2 times four

is negative two.

Negative 1/2, negative 1/2 times five

is negative 2.5, negative 2.5.

And negative 1/2 times
two is negative one.

Negative 1/2 times two is negative one.

So that's A inverse right over here.

Now let's multiply A inverse times

our column vector, seven, negative six.

Let's do that.

This is A inverse. I'll rewrite it.

Negative two, negative 2.5, negative one,

negative one times seven and negative six.

Times, I'll just write
them all in white here now.

Seven, negative six.

We've had a lot of practice
multiplying matrices.

So what is this going to be equal to?

The first entry is
going to be negative two

times seven which is negative 14 plus

negative 2.5 times negative six.

Let's see. That's going to be positive.

That's going to be 12 plus another 3.

That's going to be plus 15.

Plus 15.

Negative 2.5 times negative six

is positive 15.

Then we're going to have negative one

times seven which is negative seven plus

negative one times negative six.

Well, that is positive six.

So the product A inverse B

which is the same things
as a column vector X

is equal to,

we deserve a little
bit of a drum roll now,

the column vector one, negative one.

We have just shown that this is equal to

one, negative one or that X is equal to

one, negative one,

or we could even say
that the column vector,

the column vector ST,

column vector with the
entries S and T is equal to,

is equal to one, negative one,

is equal to one, negative one

which is another way of saying

that S is equal to one

and T is equal to negative one.

I know what you're saying.

I said this in the last video

and I'll say it again in this video.

You're like, "Well, you
know, it was so much easier

"to just solve this system directly

"just with using elimination
or using substitution."

I agree with you, but
this is a useful technique

because when you are doing problems

in computation there may be situations

where you have the left-hand side

of this system stays the same,

but there's many, many,
many different values

for the right-hand side of the system.

So it might be easier to just compute

the inverse once and
just keep multiplying,

keep multiplying this inverse times

the different what we have
on the right-hand side.

You probably are familiar with some types,

you have graphics processors,

and graphics cards on computers

and they talk about
special graphic processors.

What these are really all about

are the hardware that is special-purposed

for really fast matrix multiplication

because when you're
doing graphics processing

when you're thinking about modeling things

in three dimensions,

and you're doing all
these transformations,

you're really just doing a lot of

matrix multiplications
really, really, really fast

in real time so that to
the user playing the game

or whatever they're doing,

it feels like they're in some type

of a 3D, real-time reality.

Anyway, I just want to point that out.

This wouldn't be, if I
saw this just randomly

my instincts would be to
solve this with elimination,

but this ability to think of this

as a matrix equation is a
very, very useful concept,

one actually not just in computation,

but also as you go into
higher level sciences

especially physics, you will see a lot of

matrix vector equations like this

that kind of speak in generalities.

It's really important to think about

what these actually represent

and how they can actually be solved.