Question

On page 2, Partial Proof (23), equation 26 located at:
http://ocw.mit.edu/courses/mathematics/18-02sc-multivariable-calculus-fall-2010/1.-vectors-and-matrices/part-b-matrices-and-systems-of-equations/session-14-solutions-to-square-systems/MIT18_02SC_MNotes_m3.pdf

Answer 1

On page 2, Partial Proof (23), equation 26 located at: http://ocw.mit.edu/courses/mathematics/18-02sc-multivariable-calculus-fall-2010/1.-vectors-and-matrices/part-b-matrices-and-systems-of-equations/session-14-solutions-to-square-systems/MIT18_02SC_MNotes_m3.pdf

How is it we can express vector c in terms of vector a & b here? What if, for instance, vector c points in a different direction to a & b and they can't be added?

Answer 2

if that was the case, then |A| would not equal to zero

Answer 3

But they are still in the same plane though? If we multiply a or b by some number, wouldn't that just scale them in the wrong direction to vector c? Even if they share the same plane.

Answer 4

because |A|=0, the row vectors a, b, c lie in a plane.

Answer 5

c is a linear combination of a,b

Answer 6

or b is a linear combination of a,c

Answer 7

there, c is a linear combination of a,b

Answer 8

lets use an example, suppose that vector `a` is (1,0,0) and `b` is (0,1,0).
Now `c` could be (3,4,0)

Answer 9

Ok

Answer 10

So this is applicable to all c, for any vectors a and b, we can multiply a and b to get any vector c on the same plane?

Answer 11

vector c is in the 'span' of vectors a,b

Answer 12

What if for example, a is (1,2,0), we can't multiply by r to get (3,4,0)? For instance if r = 3, c = (3,6,0)

Answer 13

right, but theres another vector `b`, that is independent of `a`

Answer 14

for instance, if a = (1,2,0) and b= (1,1,0) , now you can get (3,4,0)

Answer 15

But what if b is (2,1,0)?

Answer 16

yes that works , any vector that is not parallel to (1,2,0) and we want the third coordinate to be zero

Answer 17

So if a = (1,2,0) * r
b = (2,1,0) * s
What values r and s would give c = (3,4,0)?

Answer 18

ok lets see

Answer 19

r*(1,2,0) + s*(2,1,0) = (3,4,0)

multiply the r, and s

Answer 20

( r, 2r, 0) + (2s, s, 0) = (3,4,0)

( r + 2s , 2r + s , 0 ) = (3,4,0)

equate coefficients

r + 2s = 3
2r + s = 4

Answer 21

Ohh interesting. So I can do that for any vectors on the same plane.

Answer 22

right, we get a system of equations , which we can solve

Answer 23

Great, that explains it, thanks for your help.

Answer 24

see if you can solve for r and s

Answer 25

i get s = 2/3 , r = 5/3

Answer 26

\[\left(\begin{matrix}r \\ s\end{matrix}\right) = \left[\begin{matrix}1 & -2 \\ -2 & 1\end{matrix}\right] /3 \left(\begin{matrix}3 \\ 4\end{matrix}\right) = \left(\begin{matrix}5/3 \\ 2/3\end{matrix}\right)\]

Answer 27

i see, you solved the matrix equation
A* (r,s) = (3,4)

Answer 28

A* (r,s) = (3,4)
(r,s) = A^-1 (3,4)

Answer 29

Yep. That's interesting. How could I visualize that geometrically? Because I thought if we multiplied any vector by a scalar, we just change its magnitude and not the direction. And if the directions of the two vectors do not add to the third, I didn't think it's possible. But it seems it is.

Answer 30

right, it helps to look at the case in 2 dimensions

Answer 31

two linearly independent vectors fill up the entire x y plane

Answer 32

the linear combinations of the two vectors fill up the plane, like so

Answer 33

pick any point on the plane, and i will show you how to make a linear combination to get that point

Answer 34

Ohh, so you're travelling a certain amount in the direction of one vector, then another amount in the other? Or in the opposite direction. And by doing that you can reach any point?