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##### Solving 3 Equations with 3 Unknowns : Solving 3 equations with 3 unknowns (old video from July 2008)

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- You're probably pretty familiar at this point with
- how to solve two equations and two unknowns.
- Let's say we had the equation 2x, I don't know, minus y is
- equal to 10.
- And let's say you had x plus 3y is equal to 5.
- And there are a lot of different ways to solve it.
- And visually, the representation we use in our
- heads, is that this represents a line, and this
- represents a line.
- And if they're not parallel to each other, they're going to
- intersect at a unique point.
- And that unique point will be the solution.
- That unique x and y will be the point where these two
- lines intersect.
- And so one way that we can solve these is we can multiply
- either of these equations by scalars, by numbers.
- And we have to multiply both sides of the equation, and
- then add them to the other equation.
- So what we could do, and we do it in such a way is that we
- eliminate variables.
- So for example, let's say we wanted to get
- rid of the y term.
- So we could take this equation and then add it to what?
- So we have a 3y here.
- So then if we made this into a minus 3y, when you add the
- two, you would get 0.
- The y's would eliminate.
- Let me do it this way.
- If you multiply this top equation by 3, you get 6x
- minus 3y is equal to 30.
- Multiply both sides by 3.
- And the bottom equation we're keeping the same.
- x plus 3y is equal to 5.
- And then you just add the two.
- So 7x.
- That's a 0, is equal to 35.
- I picked these numbers at random, but it seems that they
- actually work out.
- So we get x is equal to 5.
- And then you could go back, and you take this, and you can
- substitute it back into either one of these equations.
- So you would have-- let's do the first one.
- 2 times 5 minus y is equal to 10.
- 10 minus y is equal to 10.
- y would be equal to 10, right?
- 10 minus 0 is 10.
- And it's true here in this equation as well.
- So that's how you solve two equations of two unknowns.
- So how do you do it with three equations?
- So I have here-- let me write it down.
- I designed this equation, so it's actually reasonably
- straightforward to solve.
- You have minus x plus 2y minus z is equal to 9.
- I have 3x minus 7y minus 2z is equal to minus 20.
- And I have 2x plus 2y plus z is equal to 2.
- Now first of all, what's the visual representation of this?
- And if you watched the intuition when I talked about
- the video on the intuition of a three variable integration,
- you know that each of these represent a plane in three
- dimensions.
- And so, if you're trying to figure out what x, y, and z
- satisfy all three of these equations, you're essentially
- saying, what point in three-dimensional space exists
- on all three of these planes?
- And that point is essentially the intersection of these
- three planes.
- And that's very hard to draw, so I won't do it now.
- But you might want to watch that visual intuition--
- intuition video that I did.
- But anyway, let's go to actually solving it.
- So solving three equations and three unknowns, you actually
- do the same thing as you did with two
- equations and two unknowns.
- It's just a lot messier.
- And it's just a little bit more confusing in terms of
- which variable do you need to eliminate, and when.
- And because of that, it's very, very, very important to
- be systematic about it.
- So the way I like to do it is, I like to-- and this is
- actually called Gaussian elimination, or Gauss-Jordan
- elimination, and it's done a little bit more formally in
- linear algebra.
- But essentially, I like to eliminate the two x's in these
- two equations, and then I like to eliminate
- this y in this equation.
- So all you're left with is a z here.
- And then you'd have z equals something.
- Then you can substitute into this equation, then you can
- substitute for y.
- And then you could substitute the y and the z back to this
- equation, and solve for x.
- So let's do that.
- I know that sounds a little daunting, but I think if we
- actually do the problem, it'll be a little bit more
- straightforward.
- So before we break into the mechanics, I want to show you
- something else.
- And this is-- depending when you're doing this, your
- teacher might be upset or impressed that you do this--
- but this is just an easier way to save time.
- It's to write an augmented matrix that represents these
- three equations of three unknowns.
- And what do I mean by that?
- We're going to be rewriting these over and over again.
- And instead of writing an x, a y, a z, and the equal sign
- every time, let's just write the numbers that matter.
- So we can represent this as-- So minus x, we could say this
- is minus 1, 2, minus 1.
- And this is called an augmented matrix.
- Because we'll put a 9 here.
- So this just says, minus x plus 2y minus z is equal to 9.
- And then we say 3x minus 7y minus 2z is equal to minus 20.
- I'm not doing anything here.
- I'm just rewriting it in a form that has a little bit
- less letters in it.
- And then the last equation is 2x plus 2y plus
- z is equal to 2.
- So the first thing I said that wanted to do, is I wanted to
- eliminate these two x terms. Right?
- Remember, these just correspond to these two terms.
- So how do I do that?
- Well I'm going to keep the top equation the same, minus x,
- plus 2y minus z is equal to positive 9.
- That's just that equation.
- And remember, you could write it out.
- You could rewrite these with x's and y's all the time.
- I'm just doing this for a little bit of simplicity.
- And your teacher actually might do it this way.
- So that's more useful.
- And I like to do it this way.
- So how can we eliminate this 3x here, or this 3 right here?
- Well what if we took this equation, and added it to 3
- times this equation?
- Or another way, we substituted this equation with this
- equation, plus 3 times this equation.
- So what happens?
- So we take each of these terms, and add to it 3 times
- each of these terms. So 3x plus 3 times minus x, that is
- equal to 0.
- Right?
- I'm just multiplying this top equation by 3, and adding it
- to this equation.
- So minus 3x plus 3x-- Let me actually do it one time, just
- so you see what I'm doing.
- I just multiplied this top equation by 3.
- So you get minus 3x plus 6y minus 3z is equal to 27.
- And I'm adding that to this.
- So minus 3x plus 3x is 0.
- 6y minus 7y is equal to minus y.
- And then minus 3z plus minus 2z is equal to minus 5z.
- This is very computationally intensive, you can tell.
- And then 27 plus minus 20, that's just equal to 7.
- I'm going to dwell on this step a little bit, because I
- really want you to understand what I'm doing.
- Because if you understood what I just did in this step,
- you'll understand the rest of this video, frankly.
- So I just replaced this equation-- which is the same
- thing as this equation-- with this equation
- plus 3 times this.
- So I said, 3 plus 3 times negative 3 is 0.
- Minus 7 plus 3 times 2 is minus 1.
- Minus 2 plus 3 times minus 1 is minus 5.
- Minus 20 plus 3 times 9 is 7.
- Hopefully that makes sense.
- So now, let's do the same thing here.
- Let's try to eliminate this 2x.
- So if you want to eliminate this 2x, we could take this
- equation and add to it 2 times this equation.
- So what will happen?
- We'll have 2 plus 2 times minus 1.
- That's 0.
- 2 plus 2 times 2.
- Well that's 6.
- And then we have 1 plus 2 times minus 1.
- Well that's minus 1.
- Right?
- 2 times minus 1 is minus 2.
- 1 minus 2 is minus 1.
- And then 2 plus 2 times 9.
- So that's 2 plus 18.
- So that's 20.
- And once again, we could rewrite this
- equation right here.
- This middle equation right here is the same thing as
- minus y minus 5z is equal to 7.
- Just so you remember, this is just another representation of
- these actual linear equations.
- I just don't feel like writing the x's, y's, and z's and the
- equal signs.
- But anyway, now let's see if we can eliminate
- this term down here.
- So how could we do that?
- Well we could add this equation, we could substitute
- this equation with this equation plus 6 times this
- one, right?
- Because if you have this plus 6 times this, you'd
- have 6 minus 6.
- And you'd get a 0 here.
- So let's do that.
- Let's write the augmented matrix.
- We're not changing the top line.
- So we get minus 1x plus 2y minus z is equal to 9.
- We're actually not even changing this equation.
- So we're getting 0x minus y minus 5z is equal to 7.
- And now we're replacing this with this plus 6 times this.
- So the 0 plus 6 times 0 is 0.
- 6 plus 6 times minus 1.
- That's 6 minus 6.
- That's 0.
- Minus 1 plus 6 times minus 5.
- That's minus 1 minus 30.
- So it's minus 31.
- And then we have 20 plus 6 times 7.
- So that's 6 times 7 is 42.
- So that's 62.
- Now we're actually almost done solving.
- What can we do here?
- So I just want to make sure you understand our position.
- What is this last equation?
- This 0, 0, minus 31, line 62?
- Well, this last equation is just minus 31z is equal to 62.
- And so we can divide both sides of this equation by
- minus 31, and we'd get z is equal to minus 2.
- But if we were to do that using our augmented matrix
- notation, we would just say, well let's just divide both
- sides of this bottom equation by minus 31.
- And we would get minus 1, 2, minus 1, 9, 0,
- minus 1, minus 5, 7.
- So we're dividing this one by minus 31.
- So it's 0, 0, 1, minus 2.
- And there you have it.
- So we have at least solved for z.
- Right?
- This equation is the same thing as this one. z is equal
- to minus 2.
- Now at this point, what we have just done using this
- augmented matrix, this is called Gaussian elimination.
- Now we can continue this.
- And we can eliminate this term, this
- term, and this term.
- And it would be called Gauss-Jordan elimination.
- Or, we could use the fact that z is equal to minus 2.
- And then we could substitute back into this equation to
- solve for y.
- And then we'd have our values for y and z.
- And we could substitute back into this equation.
- Let's do it that way, just for kicks.
- So this is called upper row echelon form.
- I always forget the terminology, but let's keep
- this equation in mind.
- And let me erase all of this, so we have some
- room to work with.
- I can't do anything without some room to work with.
- Let me even erase this up here.
- And as you can tell, this is very
- computationally intensive.
- And it's not my cup of tea.
- It's not what I would want to do if I had
- the weekend to myself.
- But anyway, let me write these equations in kind of this
- format, so we can kind of go back to our traditional world.
- So this equation right here, this is
- actually the same equation.
- We never changed it.
- Minus x plus 2y minus z is equal to 9.
- This equation is minus y minus 5z is equal to 7.
- And then this equation is z is equal to minus 2.
- So we solved for z.
- This equation essentially solves for z.
- And now we can substitute back into this one.
- So we have minus y minus 5 times z, which is minus 2, is
- equal to 7.
- So minus y plus 10 is equal to 7.
- Minus y is equal to minus 3.
- y is equal to 3.
- There we have it.
- We solved for z and y.
- And now we could substitute the y, the z into this top
- equation, and solve for x.
- So what do we have?
- And then we're done.
- Let's see.
- Minus x plus 2 times y, so 2 times 3, minus z-- minus minus
- 2, that's z-- is equal to 9.
- So minus x plus 6 plus 4 is equal to 9.
- Mine x plus 10 is equal to 9.
- Minus x is equal to minus 1, or x is equal to 1.
- So the solution-- oh my God.
- My phone is ringing.
- So many people calling me.
- I think it's my wife.
- She might be downstairs.
- We're supposed to go eat out.
- But anyway, let me just finish this.
- So the solution to this 3 by 3 equation is the point x is
- equal to 1, y is equal to 3, and z is equal to minus 2.
- And that's the unique point in three-dimensional space that
- lies on all three of these planes.
- Or the unique point where these three planes intersect.
- Anyway, I hope you got that.
- And it's time for me to go have dinner with my wife.
- See you soon.