Main content

## Algebra (all content)

### Course: Algebra (all content) > Unit 20

Lesson 16: Solving equations with inverse matrices# Matrix word problem: vector combination

Sal finds the appropriate combination of two given vectors in order to obtain a third given vector. This is done by representing the problem with a single matrix equation and solving that equation. Created by Sal Khan.

## Want to join the conversation?

- Can anyone point me to a video of vectors introduction? I would like to review basic operation and concepts. Something like what Khan starts doing at3:20(7 votes)
- Hi joaocunhajeronimo.

You can watch the introduction to vectors and scalars in the Physics list: http://www.khanacademy.org/video/introduction-to-vectors-and-scalars?playlist=Physics

I can’t recommend you other videos about vectors because I didn’t see the physics list yet, but I hope you can find them. You can also read a little bit on Wikipedia: http://en.wikipedia.org/wiki/Euclidean_vector (I recommend you the “Representations” section)

Also you should remember that the matrices are arrangements of numbers that represent something else; in this case it represents a vector. When you have a matrix with just one column it is also called a column vector; analogously happens the same with matrices whit just one row (row vector). In this case, the first number represents the “x” value and the second represent the “y” value. If you have a 3x1 matrix (or vector) or a 1x3 matrix (or vector), the third value represents the “z” value. When you have more than 3 values the thing gets funnier because it can represent time or temperature or anything else but I don’t want to confuse you.

Sorry for my poor english, not my native language. Hope this help you, see you.(16 votes)

- Since order matters with matrix multiplication, why is Sal multiplying the adjoint of vector A with vector [7, 6] before multiplying the reciprocal of A's determinant with it's adjoint? Is this allowed by some property of matrix inversions? I tried multiplying through in both orders, and got the same result; why does this work in this case?(5 votes)
- That's because A's determinant (and it's reciprocal) is a
**scalar**while the adjugate is a**matrix**. you can play a bit with scalars to simplify calculations.(6 votes)

- what is the difference between linear algebra and original algebra(5 votes)
- I've never heard of "original algebra," but "linear algebra" is a sub-topic of "algebra" that deals with linear equations (ones that make straight lines when graphed) and systems of these equations.(4 votes)

- I am dealing with something called a Tensor, but for me it is just like a matrix, in fact is is a matrix as far as operations and math goes, and for what we use them in class. But my teacher and books of engineering keep calling them tensors, and the definitions I found in internet are quite useless, they keep using the word tensor and tensor field to define tensor... So can anyone give me an easy to go and understand explanation of what a tensor is, and why a tensor of second order can be viewed as a matrix, and what would be a tensor of third order and the difference with a matrix... Is it really helpful the definition of tensor or it is just another jargon or nomenclature more in the world of mathematics...(5 votes)
- TO MY UNDERSTANDING, tensors represent vectors, scalars, and other tensors. Watch http://www.youtube.com/watch?v=f5liqUk0ZTw for a fairly simple explanation.(3 votes)

- At3:36, Sal says that you have to draw the vector so that the orientation and magnitude are correct. What are orientation and magnitude?(3 votes)
- The 'magnitude' means the size. It's the length of the vector. The 'orientation' just means direction. That's the way the vector is pointed. Two vectors are equal if they have the same length, and they point in the same direction. They can be drawn anywhere on the graph because vectors only convey "magnitude" and "direction," NOT a starting point.

So, for example, a vector whose tail is at (1,1) and whose head is at (4,5) is the same as a vector whose tail is at (2,4) and whose head is at (5,8). They both have the same length. They are both 5 units long (from the pythagorean theorem... they both go up 3 units and to the right 4 units. So the square root of 3 squared plus 4 squared is 5). Also they both point the same direction, because they both have a slope of 3/4. Therefore, even though they are not in the same place on the graph, they are the same vector because their lengths and their directions are the same.(6 votes)

- Why would you want to find how to add two vectors to get a third?(2 votes)
- One application is physics when you want to find the direction of a force.

Imagine two horses pulling a wagon. Let's say the gray horse is pulling the wagon with a certain force which we will call G. (The standard unit of force is the Newton, but we'll just use letters for now) The brown horse (which is a bit stronger) is pulling with force B. The total force pulling the wagon forward is G + B.

However, and here is where it gets interesting: Suppose that we now attach the gray horse to the back of the wagon and have him pulling in the opposite direction. You have two forces acting on the wagon in opposite directions. B pulling forward and G pulling backward. Now there is still G and B force acting on the wagon, but they are acting against each other. The*resultant*force pulling the wagon forward is B-G.

Now, if we had two of the Brown horses, one attached to each end of the wagon, the resultant force would be B-B or 0. So, even though 2B of force is acting on the wagon, since each B is acting in the opposite direction, it is as though no force were acting on the wagon and it does not move. Think of that as adding Vector B to Vector -B and getting 0.

Now imagine that two brown horses are attached to the front of the wagon but one is pulling 45 degrees to the right and one is pulling 45 degrees to the left. What direction will the wagon go? Without getting into the equations, we can nevertheless see that the wagon will go forward because each horse is pulling at the same angle, but the*resultant*force on the wagon will not be 2B. It will be less, because the horses are pulling in different directions. And of course, if one horse decided to pull at 50 degrees, the wagon would start to veer in that direction. Or, if the weaker gray horse pulled right at 45 degrees with a stronger brown horse pulling left at 45 degrees, the wagon would veer left.

Now, think of the horses and their force as vectors. Vector G + Vector B = Vector W where G is the grey horse, B is the brown horse and W is the resultant force and its direction acting on the wagon.

I hope that example helps and makes sense.(5 votes)

- So, when I read about vectors in my geometry textbook and they say the notation for writing a vector is〈x,y〉it's the same as when Sal writes them as column vectors? (I suppose writing them as column vectors makes it easier to solve the problem.)(1 vote)
- At 12.53, Sal uses a word "vanier reality", can anyone please tell me what that means? I loved the way he connected philosophy and matrices. It would be nice if someone could tell me how to spell the word "vanier"? My english vocabulary is pretty weak and I wish to improve it.(2 votes)
- the word he uses is http://www.thefreedictionary.com/veneer

"A deceptive, superficial show; a façade"

[since there is no actual wood involved here ;) ](3 votes)

- At0:49, what is the origin of the word "isomorphism"?(3 votes)
- How to solve this:

x +6y -z =10

2x +3y -3z =17

3x -3y -2z =-19(1 vote)

## Video transcript

In the last video we saw how a
matrix and figuring out its inverse can be used to solve
a system of equations. And we did a 2 by 2. And in the future, we'll
do 3 by 3's. We won't do 4 by 4's because
those take too long. But you'll see it applies to
kind of an n by n matrix. And that is probably the
application of matrices that you learn if you learn this in
your Algebra 2 class, or your Algebra 1 class. And you often wonder,
well why even do the whole matrix thing? Now I will show you another
application of matrices that actually you're more likely to
see in your linear algebra class when you take
it in college. But the really neat thing here
is, and I think this will really hit the point home, that
the matrix representation is just one way of representing
multiple types of problems. And what's really
cool is that if different problems can be represented the
same way, it kind of tells you that they're the
same problem. And that's called an isomorphism
in math. That if you can reduce one
problem into another problem, then all the work you
did with one of them applies to the other. But anyway, let's figure
out a new way that matrices can be used. So I'm going to draw
some vectors. Let's say I have the vector--
Let's call this vector a. And I'm going to just write this
is as a column vector. And all of this is
just convention. These are just human
invented things. I could have written
this diagonally. I could have written
this however. But if I say vector a
is 3, negative 6. And I view this as the x
component of the vector, and this is equal to the y component
of the vector. And then I have vector b. Vector b is equal to 2, 6. And I want to know are there
some combinations of vectors a and b-- where you can say, 5
times vector a, plus 3 times vector b, or 10 times victor a
minus 6 times vector b-- some combination of vector a and b,
where I can get vector c. And vector c is the
vector 7, 6. So let me see if I can visually
draw this problem. So let me draw the
coordinate axes. Let's see this one. 3, negative 6. That'll be in quadrant--
these are both in the first quadrant. So I just want to figure
out how much of the axes I need to draw. So let's see-- Let me do
a different color. That's my y-axis. I'm not drawing the second or
third quadrants, because I don't think our vectors
show up there. And then this is the x-axis. Let me draw each of
these vectors. So first I'll do vector a. That's 3, negative 6. 1, 2, 3, and then negative 6. 1, 2, 3, 4, 5, 6. So it's there. So if I wanted to draw it
as a vector, usually start at the origin. And it doesn't have to start
at the origin like that. I'm just choosing to. You can move around a vector. It just has to have the
same orientation and the same magnitude. So that is vector
a for the green. Now let me do in magenta,
I'll do vector b. That is 2, 6. 1, 2, 3, 4, 5, 6. So 2, 6 is right over there. And that's vector b. So it'll look like this. That's vector b. And then let me write down
vector a down there. That's vector a. And I want to take some
combination of vectors a and b. And add them up and
get vector c. So what does vector
c look like? It's 7, 6. Let me do that in purple. So 1, 2, 3, 4, 5, 6, 7. Comma 6. So 7, 6 is right over there. That's vector c. Vector c looks like that. I'm going to draw
it like that. And that's vector c. So what was the original
problem I said? I said I want to add some
multiple of vector a to some multiple of vector b,
and get vector c. And I want to see what
those multiples are. So let's say the multiple
that I multiply times vector a is x. And the multiple of
vector b is y. So I essentially want to say
that-- let me do it in another neutral color-- that vector ax--
that's how much of vector a I'm contributing-- plus vector
by-- that's how much of vector b I'm contributing--
is equal to vector c. And you know, maybe I can't. Maybe there's no combinations of
vector a and b when you add them together equal vector c. But let's see if we
can solve this. So how do we solve? So let's expand out
vectors a and b. Vector a is what? 3, negative 6. So vector a, we could write
as 3, minus 6 times x. That just tells us how much
vector a we're contributing. Plus vector b, which is 2, 6. And then y is how much vector
b we're contributing. And that is equal to 7, 6. Vector c. Now this right here, this
problem can be rewritten just based on how we've defined
matrix multiplication, et cetera, et cetera, as this. As 3, minus 6, 2, 6, times
x, y, is equal to 7, 6. Now how does that work out? Well think about how matrix
multiplication works out. The way we learned matrix
multiplication, we said, 3 times x, plus 2 times
y is equal to 7. 3 times x plus 2 times
y is equal to 7. That's how we learned matrix
multiplication. That's the same thing here. 3 times x, plus 2 times y, is
going to be equal to 7. These x and y here are
just scalar numbers. So 3 times x plus 2 times
y is equal to 7. And then matrix multiplication
here, minus 6 times x plus 6 times y is equal to 6. That's just traditional matrix
multiplication that we learned several videos ago. That's the same thing here. Minus 6x plus 6y
is equal to 6. These x's and y's are
just numbers. They're just scalar numbers. They're not vectors
or anything. We would just multiply them
times both of these numbers. So hopefully you see that this
problem is the exact same thing as this problem. And you've maybe had an a-ha
moment now, if you watched the previous video. Because this matrix also
represented the problem, where do we find the intersection
of two lines? Where the two lines-- I'm just
going to do it on the side here-- the intersection of
the two lines, 3x plus 2y is equal to 7. And minus 6x plus 6y
is equal to 6. And so, I had drawn two lines. And we said, what's the
point of intersection, et cetera, et cetera. And it was represented
by this problem. But here, we have-- well I
won't say a completely different problem, because
we're learning they're actually very similar-- but here
I'm doing a problem of, I'm trying to find what
combination of the matrices a and b add up to the matrix c. But it got reduced to the same
matrix representation. And so we can solve this the
same exactly way we solved this problem. If we call this the matrix a,
let's figure out a inverse. So we get a inverse
is equal to what? It equals 1 over the
determinant of a. The determinant of
a is 3 times 6. 18 minus minus 12. So that's 18 plus 12,
which is 1/30. And we did this in the
previous video. You swap these two numbers. So you get 6 and 3. And then you make these
two negatives. So you get 6 and minus 2. That's a inverse. And now to solve for x and y, we
can multiply both sides of this equation by a inverse. If you multiply a inverse times
a, this cancels out. So you get x, y is equal to
a inverse times this. It's equal to 1/30 times
6, minus 2, 6, 3. Times 7, 6. And remember, with matrices,
the order that you multiply matters. So on this side, we multiplied
a inverse on this side of the equation. So we have to do a inverse on
the left side on this side of this equation. So that's why did it here. If we did it the other way,
all bets are off. So what does this equal? This is equal to 1/30 times--
and we did this the previous problem-- 6 times 7
is 42, minus 12. 30. 6 times 7, 42. Plus 18. 60. So that equals 1, 2. So what does this tell us? This tells us that if we have
1 times vector a, plus 2 times vector b. 1 times-- this is 1-- and
2 times vector b. So 1 times vector a plus
2 times vector b is equal to vector c. And let's confirm
that visually. So 1 times vector a. Well that's vector
a right there. So if we add 2 vector b's to
it, we should get vector c. So let's see if we
can do that. So if we just shift vector b
over this way, well vector let's see, vector b is
over 2 and up 6. So over 2 and up 6 would
get us there. So 1, vector b-- just doing
heads to tail visual method of adding vectors-- would
get us there. 1, 2, 3. Good. No, let me see. 1, 2, 3. And then vector b goes
over two more. two more. So it'll get us up 6. It's like that. So that's 1, vector b. And then if we add another-- but
we want 2 times vector b. We essentially need
two vector b's. So we had one, and then
we add another one. I think visually you see that
it does actually-- I didn't want to do it like that. I wanted to use the line
tool so it looks neat. So you add another vector b. And there you have it. That's a vector b. So it's 2 times vector b. So it's the same direction as
vector b, but it's two times the length. So we visually showed it. We solved it algabraically. But the real learning, and the
big real discovery of this whole video, is to show you that
the matrix representation can represent multiple different
problems. This was a finding the combinations
of a vector problem. And the previous one it
was figure out if two lines can intersect. But what it tells you is that
these two problems are connected in some deep way. That if we take the veneer of
reality, that underlying it, they are the same thing. And frankly, that's why math
is so interesting. Because when you realize that
two problems are really the same thing, it takes all of the
superficial human veneer away from things. Because our brains are kind of
wired to perceive the world in a certain way. But it tells us that there's
some fundamental truth, independent of our perception,
that is tying all of these different concepts together. But anyway, I don't want to
get all mystical on you. But if you do see
the mysticism in math, all the better. But hopefully you found that
pretty interesting. And actually, I know I'm going
over time, but I think this is-- A lot of people take linear
algebra, they learn how to do all of the things, and
they say, well what is the whole point of this? But this is kind of
an interesting thing to think about. We had this had vector a and
we had this vector b. And we were able to say, well
there's some combinations of the vectors a and b, that
when we added it up, we got vector c. So an interesting question is,
what are all the vectors that I could get to by adding
combinations of vectors a and b. Or adding or subtracting. Or you could say,
I could multiply them by negative numbers. But either way. What are all of the vectors
that I can get by taking linear combinations of
vectors a and b? And that's actually called the
vector space spanned by the vectors a and b. And we'll do more of that
in linear algebra. And here we're dealing
with a two dimensional Euclidean space. We could have had three
dimensional vectors. We could've had n dimensional
vectors. So it gets really, really,
really abstract. But this is, I think, a really
good toe dipping for linear algebra as well. So hopefully I haven't confused
or overwhelmed you. And I'll see you in
the next video.