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Solving 3 Equations with 3 Unknowns

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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.

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Solving 3 Equations with 3 Unknowns

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