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How do matrices work

by okkvlt
Tags: matrices, work
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okkvlt
#1
Jul24-08, 02:40 PM
P: 53
Say i want to find [xa,ya,za]*[xb,yb,zb]

I could use pythagoras to find |a| and |b|. Then use pythagoras again to find the distance between the endpoints of a and b. Then use law of cosines to find the angle formed by a and b, then do |a||b|cos(angle).
Or, i could just do xa*xb+ya*yb+za*zb.
How are all these operations compressed into such a simple form?

(By the way, the dot product in 2 dimensions is the area of the parallelogram formed by the endpoints of a+b, a, b, and 0, right?) Then the dot product of vectors in 3 dimensions is the area of the parallelogram squared to get a 3 dimensional shape, right? Whats the signifigance of a negative dot product?


And if i want to find the cross product, all i have to do is arrange them like this
xa ya za
xb yb zb

And for each coordinate of the cross product i just remove that column and find the determinant of the remaining 2x2 matrix, reversing the sign for y. Doing it otherwise, i would have to do alot of complicated things, especially finding that perpendicular unit vector.


Also, how do matrix determinants work in finding the solution to systems of equations?
How do matrices and vectors work?
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Defennder
#2
Jul24-08, 10:27 PM
HW Helper
P: 2,616
Quote Quote by okkvlt View Post
Say i want to find [xa,ya,za]*[xb,yb,zb]

I could use pythagoras to find |a| and |b|. Then use pythagoras again to find the distance between the endpoints of a and b. Then use law of cosines to find the angle formed by a and b, then do |a||b|cos(angle).
Or, i could just do xa*xb+ya*yb+za*zb.
How are all these operations compressed into such a simple form?
You could "prove" it by letting defining b with respect to a. Use a as the reference vector a_x i and try to find another way to express axbx. of course this prove doesn't work in [tex]R^n[/tex] where [tex]n > 3[/tex] and it becomes impossible to visualise the vectors geometrically.

(By the way, the dot product in 2 dimensions is the area of the parallelogram formed by the endpoints of a+b, a, b, and 0, right?) Then the dot product of vectors in 3 dimensions is the area of the parallelogram squared to get a 3 dimensional shape, right?
Actually the area of the parallelogram is given by the determinant of the matrix consisting of the column vectors of a,b. In three dimensions, it is also the determinant of 3 (linearly independent) vectors which gives the volume of the parallelopiped.

Whats the signifigance of a negative dot product?
Well it just means that the the projection of one vector onto another is pointing in the opposite direction.

And if i want to find the cross product, all i have to do is arrange them like this
xa ya za
xb yb zb

And for each coordinate of the cross product i just remove that column and find the determinant of the remaining 2x2 matrix, reversing the sign for y. Doing it otherwise, i would have to do alot of complicated things, especially finding that perpendicular unit vector.
The cross product of [tex]\textbf{a}, \textbf{b}[/tex] is [tex]\textbf{a} \times \textbf{b} = \left| \begin{array}{ccc}i&j&k\\a_x&a_y&a_z\\b_x&b_y&b_z \end{array} \right|[/tex].

Also, how do matrix determinants work in finding the solution to systems of equations? How do matrices and vectors work?
You can pick up any introductory linear algebra textbook for a good introduction. I read Elementary Linear Algebra by Anton, 9th edn.


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