Some linear algebra homework help, please

[TuxDave]

Originally posted by: eigen

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)

-3/2

Beat ya to it!

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.

Yeah, used the same method here. I've been trying to black out my days with linear algebra and my Nazi professor.

"What the determinant of the identity matrix? HMM?? HMM???? No one knows?!?! This would be a JOKE in the math department. If you asked them this, they would be laughing at you! They would go HA HA HA... what a stupid question!"

🙁 We didn't even learn what a determinant was...

 

[eigen]

Originally posted by: TuxDave

Originally posted by: eigen

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)

-3/2

Beat ya to it!

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.

Yeah, used the same method here. I've been trying to black out my days with linear algebra and my Nazi professor.

"What the determinant of the identity matrix? HMM?? HMM???? No one knows?!?! This would be a JOKE in the math department. If you asked them this, they would be laughing at you! They would go HA HA HA... what a stupid question!"

🙁 We didn't even learn what a determinant was...

Sucks for you huh...

When you have a matrix in triangular form the det is the product of the diagnol/

 

[Howard]

Originally posted by: RaynorWolfcastle

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)

-3/2?

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

maziwanka, is this what you did?

 

[maziwanka]

Originally posted by: Howard

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.

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?

yea, then expand the ||2u-3v|| part and you should see that after rearranging, its combinations of quantities for which you have values for. post more questions if you need some more help....

 

[TuxDave]

Originally posted by: eigen
Sucks for you huh...

When you have a matrix in triangular form the det is the product of the diagnol/

Yeah I know that NOW since I finished that class 4 years ago. But I think it's kinda rough to expect people to calculate a determinant of anything without knowing what a determinant was.

 

[Howard]

Originally posted by: TuxDave

Originally posted by: eigen

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)

-3/2

Beat ya to it!

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.

Yeah, used the same method here. I've been trying to black out my days with linear algebra and my Nazi professor.

"What the determinant of the identity matrix? HMM?? HMM???? No one knows?!?! This would be a JOKE in the math department. If you asked them this, they would be laughing at you! They would go HA HA HA... what a stupid question!"

🙁 We didn't even learn what a determinant was...

Says in my textbook that the det(A^-1) = 1/det(A)

 

[Howard]

Originally posted by: maziwanka

Originally posted by: Howard

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.

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?

yea, then expand the ||2u-3v|| part and you should see that after rearranging, its combinations of quantities for which you have values for. post more questions if you need some more help....

||(2x1 - 3x2, 2x2 - 3y2)|| is what I got after expanding ||2u - 3v||. Now I'd like to change x and y to u and v respectively.

||(2u1 - 3v1, 2u2 - 3v2)|| - phew. Now how do you suggest I expand this?

 

[TuxDave]

Originally posted by: Howard

Says in my textbook that the det(A^-1) = 1/det(A)

Let me fix it up to be absolutely correct.

You can get it to the point where.

2=3A
2inv(A)=3I (I is the identity matrix)
-inv(A)=-3/2I
det(-inv(A))=det(-3/2I)=-3/2

 

[Howard]

Originally posted by: TuxDave

Originally posted by: Howard

Says in my textbook that the det(A^-1) = 1/det(A)

Let me fix it up to be absolutely correct.

You can get it to the point where.

2=3A
2inv(A)=3I (I is the identity matrix)
-inv(A)=-3/2I
det(-inv(A))=det(-3/2I)=-3/2

Gracias, senor!

For the last step, are you taking the determinant of both sides?

 

[maziwanka]

no to raynor's answer. first, the answer cannot be negative...

here's how i did it:

u = {x1, y1}
v = {x2, y2}

||u||^2 = <u, u> = x1^2 + y1^2 = 9
||v||^2 = <v, v> = x2^2 + y2^2 = 16

<u, v> = x1x2 + y1y2 = 10

2u - 3v = {2x1 - 3x2, 2y1 - 3y2}

So,

||2u - 3v||^2 = <2u - 3v, 2u - 3v> = (2x1 - 3x2)^2 + (2y1 - 3y2)^2.

If you expand this and rearrange terms, you should be able to arrive at the solution.

 

[maziwanka]

Originally posted by: Howard

Originally posted by: TuxDave

Originally posted by: Howard

Says in my textbook that the det(A^-1) = 1/det(A)

Let me fix it up to be absolutely correct.

You can get it to the point where.

2=3A
2inv(A)=3I (I is the identity matrix)
-inv(A)=-3/2I
det(-inv(A))=det(-3/2I)=-3/2

Gracias, senor!

hey. if you look at the mathworld website, i think this solution is wrong.

 

[Howard]

Originally posted by: maziwanka

Originally posted by: Howard

Originally posted by: TuxDave

Originally posted by: Howard

Says in my textbook that the det(A^-1) = 1/det(A)

Let me fix it up to be absolutely correct.

You can get it to the point where.

2=3A
2inv(A)=3I (I is the identity matrix)
-inv(A)=-3/2I
det(-inv(A))=det(-3/2I)=-3/2

Gracias, senor!

hey. if you look at the mathworld website, i think this solution is wrong.

Great...

 

[TuxDave]

Sorry... sorry.. you're right, I looked at my solution again.

The det(-3/2I)=(-3/2)^3 for a 3x3 identity matrix... right?

 

[TuxDave]

too many double posts.

 

[maziwanka]

From mathworld....

If 2A^2 = 3A^3, then

det(2A^2) = det(3A^3)

Since A is a 3x3 matrix,

(2^3)*det(A^2) = (3^3)det(A^3)
(2/3)^3*det(A)^2=det(A)^3*

Since det(A) is just a number,

det(A)=(2/3)^3

Again, A is a 3x3 matrix, so,

det(-A)=(-1^3)*det(A)=-det(A)=-(2/3)^3

And, det(A^-1)=1/det(A), so,

det(-A^-1)=-1/det(A)=-(3/2)^3

*This comes from the property: det(A^n)=det(A)^n b/c determinants are distributive.

 

[TuxDave]

Originally posted by: maziwanka
From mathworld....

If 2A^2 = 3A^3, then

det(2A^2) = det(3A^3)

Since A is a 3x3 matrix,

(2^3)*det(A^2) = (3^3)det(A^3)
(2/3)^3*det(A)^2=det(A)^3

Since det(A) is just a number,

det(A)=(2/3)^3

Again, A is a 3x3 matrix, so,

det(-A)=(-1^3)*det(A)=-det(A)=-(2/3)^3

And, det(A^-1)=1/det(A), so,

det(-A^-1)=-1/det(A)=-(3/2)^3

Yeah... I caught my error when I took a second glance at it. The methodology is correct all the way to the end where I messed up the computation of the determinant... doh!!

 

[maziwanka]

in 40 posts your entire hwk assignment was completed.

 

[Howard]

Originally posted by: maziwanka
From mathworld....

If 2A^2 = 3A^3, then

det(2A^2) = det(3A^3)

Since A is a 3x3 matrix,

(2^3)*det(A^2) = (3^3)det(A^3)
(2/3)^3*det(A)^2=det(A)^3*

Since det(A) is just a number,

det(A)=(2/3)^3

Again, A is a 3x3 matrix, so,

det(-A)=(-1^3)*det(A)=-det(A)=-(2/3)^3

And, det(A^-1)=1/det(A), so,

det(-A^-1)=-1/det(A)=-(3/2)^3

*This comes from the property: det(A^n)=det(A)^n b/c determinants are distributive.

Thanks, now I remember it all. I should have looked over my old stuff. 🙁

 

[Howard]

Originally posted by: maziwanka
in 40 posts your entire hwk assignment was completed.

I'm actually still working on the product question. Those x's and y's are really confusing me. (I'm used to working with u1 u2 and v1 v2)

 

[Howard]

nm

 

[maziwanka]

If you expand this,

(2x1 - 3x2)^2 + (2y1 - 3y2)^2

= 4x1^2 + 9x2^2 - 12x1x2 + 4y1^2 + 9y2^2 - 12y1y2
= 4(x1^2 + y1^2) + 9(x2^2 + y2^2) - 12(x1x2 + y1y2)

AND, you have all the values for the sums in the paranthesis.

 

[Howard]

Originally posted by: maziwanka
If you expand this,

(2x1 - 3x2)^2 + (2y1 - 3y2)^2

= 4x1^2 + 9x2^2 - 12x1x2 + 4y1^2 + 9y2^2 - 12y1y2
= 4(x1^2 + y1^2) + 9(x2^2 + y2^2) - 12(x1x2 + y1y2)

AND, you have all the values for the sums in the paranthesis.

I could expand it if I understood how you ended up with (2u1 - 3v1)^2 + (2u2 - 3v2)^2.

I understand now.