The following problems will be used to study for a quiz from, since I
have already completed most of these problems, no explanations in
words are necessary, but I will need to see step-by-step how each
problem has been solved.  Also, I know that $3.00 is not a huge amount
of money for these problems, but they are simple and should only take
a few minutes each, and I will be putting up many more difficult(more
expensive?) problems as the year progesses, and I will keep in mind
who provides good answers that are helpful and who answers
quickly.(P.S. calebu2-ga has been a great researcher thus far):)

1.  A^-1=[3,2;1,3] b^-1=[2,5;3,-2]
    Find A and B
2.  Find the area of the right triangle with vertices (0, 0),
(0,3),(4,0).  Verify by using the formula A=1/2(base)(height)
3.  Which of the vectors u1=(1,2), u2=(0,1),
u3=(-2,-4),u4=(-2,1),u5=(2,4),u6=(-6,3) are:
(a)Orthogonal?
(b)In the same direction?
(c)In opposite directions?
4.  Show that if w is orthogonal to u and v, then w is orthogonal to
ru+sv, where r and s are the scalars.
5.  For questions 2, 3, and 4:  In addition to solving the given
problems, please provide a similar problem to the one given, except 3
dimensional, and provide the respective answer.

Thanks for all of your time,
Bildy-ga

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Clarification of Question by bildy-ga on 08 Oct 2002 19:22 PDT

I was told everything was done the same in 2D as it is done in 3D

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Clarification of Question by bildy-ga on 08 Oct 2002 20:04 PDT

If you can complete everything except for the 3D area of a triangle
question then please do so

Forgive me for being naïve, but I don't recollect, and my search did
not yield, any formula for finding the area of a triangle in 3-D space
using linear algebra.
Would you enlighten me about the formula that you learned? (I know the
2-D formula, but not the 3-D one.)
Thanks
secret901

If I am not mistaken, the general equation for finding an area of triangle is

A = 1/2 | v x w | 

where v and w are vectors defining the triangle. These work in 3 space too, I think.

Just for the heck of it, I'm going to outline some principles...
numerical answers can easily be gotten using Mathematica or Maple.

1. for 2-space, matrix inversion is done using:
   where A = [a,b;c,d] 
   the inverse A-1 = 1/(ad-bc)[a,-b;-c,a]

    in 3-space, A-1 = 1/|A| adj (A)T. 

That's the reciprocal of the determinant of A multiplied by adjunct
matrix of the transposed A.

2. Use Cramer's rule:
A = 1/2 | x1 x2 x3 x1 |
        | y1 y2 y3 y1 |
and verify with 
A = 1/2 b h
using a transformation. Or use A = 1/2 | v x w | (the x is a cross
product, not a multiplication)

3. The test for orthogonality is always the dot product. Just dot the
pairs and see if they're zero.. if yes, then they're orthogonal. As
for the others, it's a matter of simple vector algebra. if a vector A
is parallel to a vector B, then A = kB where k is a constant. if k is
negative, then the vectors are in opposite directions.

4. Using vector rules

w.u = 0
w.v = 0

to show that w.(ru + sv) = 0

L.H.S. = w.(ru + sv) 
       = w.(ru) + w.(sv)
       = r w.u + s w.v
       = 0 + 0
       = R.H.S.

Disclaimer: I haven't done linear algebra for years, so I might be
mistaken about some things

Oh what the heck... let me just do it. Errors and Omissions Excepted.

1) A = 1/7 [3,-2;-1,3]
   B = 1/-19 [-2,-5;-3,2]

2) A = 1/2 bh = 1/2 * 4 * 3 = 6 
   A = 1/2 | 0 0 4 0 |
           | 0 3 0 0 |
     = 1/2 ( 0 - 0 + 0 - 12 + 0 - 0 )
     = -6 (signed) so the unsigned area is 6

3) there are 6 vectors to test, but by inspection, I managed to pick out:
   a) u1.u4 = 0
   b) u1 and u5 
   c) u1 and u3, u3 and u5

4) as shown above

5) Heh heh... A-1 = I (3x3) B-1 = I-1 (3x3)
   therefore A = I, B = I. No, seriously, use the other relationship.

   for a 3D triangle, A(1,0,0) B(0,1,0) C(0,0,1), the area is given by
   let v = B - A = <-1,1,0>
       w = C - A = <-1,0,1>
   v x w = <1,1,1>
   A = 1/2 | v x w | = 1/2 sqrt( 1 + 1 + 1) = sqrt(3)/2

   I'm not sure if this absolutely right though... someone please confirm this.