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Voiceover:In the last video we saw

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that we could take a[br]system of two equations

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with two unknowns and represent it

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as a matrix equation where the matrix A's

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are the coefficients here[br]on the left-hand side.

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The column vector X has our two

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unknown variables, S and T.

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Then the column vector B is essentially

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representing the[br]right-hand side over here.

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What was interesting about it,

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then that would be the equation A,

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the matrix A times the column vector X

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being equal to the column vector B.

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What was interesting about that is we saw

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well, look, if A is invertible,

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we can multiply both the[br]left and the right-hand sides

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of the equation,

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

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of their respective sides by A inverse

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because remember matrix,

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when matrix multiplication order matters,

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we're multiplying the left-hand side

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of both sides of the equation.

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If we do that then we[br]can get to essentially

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solving for the unknown column vector.

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If we know what column vector X is,

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then we know what S and T are.

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Then we've essentially solved

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this system of equations.

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Now let's actually do that.

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Let's actually figure[br]out what A inverse is

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and multiply that times[br]the column vector B

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to figure out what the column vector X is,

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and what S and T are.

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A inverse, A inverse is equal to

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one over the determinant of A,

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

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is going to be two times[br]four minus negative

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two times negative five.

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It's going to be eight minus positive 10,

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eight minus positive 10,

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which would be negative two.

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This would become negative[br]two right over here.

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Once again, two times four is eight minus

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negative two times negative five

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so minus positive 10 which[br]gets us negative two.

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You multiply one over the determinant

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times what is sometimes[br]called the adjoint of A

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which is essentially swapping the top left

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

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This would be a four.

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This would be a two.

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Notice I just swapped these,

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and making these two negative,

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the negative of what they already are.

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This is from a negative two this is

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going to become a positive two,

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and this right over[br]here is going to become

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a positive five.

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If all of this looks[br]completely unfamiliar to you,

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you might want to review the tutorial

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on inverting matrices

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because that's all I'm doing here.

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So A inverse is going to be equal to,

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A inverse is going to be equal to,

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

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is negative two.

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Negative 1/2, negative 1/2 times five

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is negative 2.5, negative 2.5.

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And negative 1/2 times[br]two is negative one.

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Negative 1/2 times two is negative one.

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So that's A inverse right over here.

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Now let's multiply A inverse times

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our column vector, seven, negative six.

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Let's do that.

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This is A inverse. I'll rewrite it.

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Negative two, negative 2.5, negative one,

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negative one times seven and negative six.

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

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Seven, negative six.

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We've had a lot of practice[br]multiplying matrices.

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So what is this going to be equal to?

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The first entry is[br]going to be negative two

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times seven which is negative 14 plus

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negative 2.5 times negative six.

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

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That's going to be 12 plus another 3.

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That's going to be plus 15.

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Plus 15.

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Negative 2.5 times negative six

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is positive 15.

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Then we're going to have negative one

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times seven which is negative seven plus

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negative one times negative six.

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Well, that is positive six.

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So the product A inverse B

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which is the same things[br]as a column vector X

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is equal to,

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we deserve a little[br]bit of a drum roll now,

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the column vector one, negative one.

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We have just shown that this is equal to

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one, negative one or that X is equal to

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one, negative one,

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or we could even say[br]that the column vector,

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the column vector ST,

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

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is equal to one, negative one,

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is equal to one, negative one

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which is another way of saying

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that S is equal to one

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and T is equal to negative one.

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I know what you're saying.

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I said this in the last video

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and I'll say it again in this video.

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

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"to just solve this system directly

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"just with using elimination[br]or using substitution."

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I agree with you, but[br]this is a useful technique

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because when you are doing problems

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in computation there may be situations

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where you have the left-hand side

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of this system stays the same,

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but there's many, many,[br]many different values

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for the right-hand side of the system.

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So it might be easier to just compute

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the inverse once and[br]just keep multiplying,

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keep multiplying this inverse times

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the different what we have[br]on the right-hand side.

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You probably are familiar with some types,

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you have graphics processors,

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and graphics cards on computers

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and they talk about[br]special graphic processors.

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What these are really all about

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are the hardware that is special-purposed

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for really fast matrix multiplication

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because when you're[br]doing graphics processing

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when you're thinking about modeling things

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in three dimensions,

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and you're doing all[br]these transformations,

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you're really just doing a lot of

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matrix multiplications[br]really, really, really fast

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in real time so that to[br]the user playing the game

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or whatever they're doing,

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it feels like they're in some type

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of a 3D, real-time reality.

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Anyway, I just want to point that out.

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

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my instincts would be to[br]solve this with elimination,

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but this ability to think of this

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as a matrix equation is a[br]very, very useful concept,

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one actually not just in computation,

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but also as you go into[br]higher level sciences

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especially physics, you will see a lot of

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matrix vector equations like this

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that kind of speak in generalities.

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It's really important to think about

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what these actually represent

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and how they can actually be solved.