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Some linear algebra homework help, please

H

Howard

Lifer
Suppose that u and v are vectors in an inner product space V for which ||u|| = 3, ||v|| = 4, and <u,v> = 10. Find ||2u - 3v||.

I tried finding the angle between u and v so that I could find it geometrically, but no luck, since it wasn't a special angle.

Any ideas?

EDIT: Good point. Also, another matrices question down below.
 
OK, maybe you can help with this one.

Let A be a 3 x 3 invertible matrix, and assume that 2A^2 = 3A^3. Find |-A^-1|

(I assume that's the determinant of the negative of the inverse)
 
Originally posted by: Howard
OK, maybe you can help with this one.

Let A be a 3 x 3 invertible matrix, and assume that 2A^2 = 3A^3. Find |-A^-1|

(I assume that's the determinant of the negative of the inverse)
Click to expand...

I think you need to add the word 'linear' in front of 'algebra' in your topic title.
 
Originally posted by: Howard
OK, maybe you can help with this one.

Let A be a 3 x 3 invertible matrix, and assume that 2A^2 = 3A^3. Find |-A^-1|

(I assume that's the determinant of the negative of the inverse)
Click to expand...

been so long, sorry can't help. when you stop doing linear algebra for 3 years it kinda leaves you.
 
Originally posted by: Howard
OK, maybe you can help with this one.

Let A be a 3 x 3 invertible matrix, and assume that 2A^2 = 3A^3. Find |-A^-1|

(I assume that's the determinant of the negative of the inverse)
Click to expand...

-3/2?
 
Originally posted by: Howard
OK, maybe you can help with this one.

Let A be a 3 x 3 invertible matrix, and assume that 2A^2 = 3A^3. Find |-A^-1|

(I assume that's the determinant of the negative of the inverse)
Click to expand...
-3/2
 
Originally posted by: Howard
Suppose that u and v are vectors in an inner product space V for which ||u|| = 3, ||v|| = 4, and <u,v> = 10. Find ||2u - 3v||.

I tried finding the angle between u and v so that I could find it geometrically, but no luck, since it wasn't a special angle.

Any ideas?

EDIT: Good point. Also, another matrices question down below.
Click to expand...

Why can't you find it geometrically again? You have the magnitude and you have the dot product, so can't you find the angle really fast? And then with the angle, you can find the magnitude of 2u-3v pretty fast using geometry or trig.
 
hey dude. if you assume u={x1,y1} and v={x2,y2} you can figure out the answers to the first question. if you need more help ill post my answer

edit: i think i got sqrt(60) for the first problem.
 
Originally posted by: eigen
Originally posted by: Howard
OK, maybe you can help with this one.

Let A be a 3 x 3 invertible matrix, and assume that 2A^2 = 3A^3. Find |-A^-1|

(I assume that's the determinant of the negative of the inverse)
Click to expand...
-3/2
Click to expand...

Beat ya to it!
 
Originally posted by: TuxDave
Originally posted by: Howard
OK, maybe you can help with this one.

Let A be a 3 x 3 invertible matrix, and assume that 2A^2 = 3A^3. Find |-A^-1|

(I assume that's the determinant of the negative of the inverse)
Click to expand...

-3/2?
Click to expand...

yeah that would be my guess. For the other one are you using the standard inner product or is it some general inner product? I haven't looked at this stuff for awwhile but IIRC the inner product rules that apply are ||u|| = <u,u> and <u,v>=<v,u>. I'll have to think about it for a solution though

oh yeah and
||2u-3v|| = <2u-3v,2u-3v> which I think you were allowed to distribute as
<2u,2u> + <2u,-3v> + <-3v,2u> + <-3v,-3v>
I'm not sure about this last step, but I also think you can pull constants out of inner products by definition
so that would leave you with
4<u,u> - 12<u,v> +9<v,v> = 4*3 -12*10 + 9*4 = 12-120+36 = -72
Not positive about this, though
 
i thought an invertible matrix has 0 determinant.

edit: hahahaha, its the opposite thing: if the matrix is singular, its determinant is 0.
 
Originally posted by: TuxDave
Originally posted by: eigen
Originally posted by: Howard
OK, maybe you can help with this one.

Let A be a 3 x 3 invertible matrix, and assume that 2A^2 = 3A^3. Find |-A^-1|

(I assume that's the determinant of the negative of the inverse)
Click to expand...
-3/2
Click to expand...

Beat ya to it!
Click to expand...
A is non-singular so you can "cancel" the A's leaving 2=3A => 1=3/2A.Then since the Det(A^-1) = 1/A
you get the answer.
 
Originally posted by: maziwanka
hey dude. if you assume u={x1,y1} and v={x2,y2} you can figure out the answers to the first question. if you need more help ill post my answer

edit: i think i got sqrt(60) for the first problem.
Click to expand...
I get ||(2x1 - 3x2, 2x2 - 3y2)||. Is that what I'm supposed to get? If so, I can't find the relationship to <u,v> or the norms of u and v. <u,v> = x1x2 + y2y2, right? And ||u|| = (x1^2 + y1^2)^/2, correct?
 
tentative solution for the other problem in my other post. You'll have to verify the properties I use though, because it's been awhile since I've looked at it
 
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