Question

: Linear Algebra: Describe the intersection of the two hyperplanes x1 + 2x2 - x3 + x4 = 8 and -x1 + 3x2 + x3 - x4 = 2.

Answer 1

They intersecate on a plane

Answer 2

Is there a way to get an equation for that plane?

Answer 3

You could set up the augmented matrix, row reduce to get the values and that should be the point of intersection

Answer 4

yep I did that.. and haha seemingly im stuck on a dumb question: let a and b be free variables. my solution then is (a-b+4, 2, a, b)^T. How do I express that as a span, then as an equation?

Answer 5

i get x2 = 2;
x3 = x1 + x4 - 4;

So like you said x1 and x4 are free so the span is:

=(2 , -4) + x1(1, 0, 1, 0) + x4(0, 0, 1, 1) (these are vertical vectors I just cant write it on here)

=(2, -4) + x1u + x4v (u and v are vectors above)

Answer 6

understand?

Answer 7

I made a small mistake:

It is x3 and x4 that are free so it becomes

x1=4 +x3 - x4
x2 =2


which translates to...

=(2 , -4) + x3(1, 0, 1, 0) + x4(-1, 0, 0, 1)

Answer 8

(4, 2, 0, 0) sorry

Answer 9

thanks, i was looking for an answer of the form span(u,v), but if that's that then that's okay. how about an equation for the plane?

Answer 10

well span(u,v) is just the span of those vectors:


span{ (1, 0, 1, 0) , (-1, 0, 0, 1)}

Answer 11

oh, so (4,2,0,0)^T doesn't matter when expressing it in that form?

Answer 12

nope because it only depends on the free variables

Answer 13

okay thanks. is there a way to express that as an equation?