Math Is Fun Forum

basmah

Member

Registered: 2006-10-02

Posts: 18

Math Help required!!!

I need any kind of help or hints to find the answer to this:

adj (adj A) = ? (adj stands for adjoint and A is any matrix...please express the answer in terms of A and det(A).... det(A) is the determinant of A)

If you are able answer this question, then i would also like you to prove it aswell...if you are not able to prove it, its fine anything would be fine....also it is due tommorrow so as soon as possible please!!!

** the matrix is suppose to be for an n*n matrix

Last edited by basmah (2006-10-02 13:10:41)

WiZaRd

Real Member

Registered: 2006-09-22

Posts: 570

Re: Math Help required!!!

I know 2 things about it one is the Adjoint of A of Matrices of order 2*2 and the other i know is the Adjoint of MAtrices  of order 3*3

Which one do you want??

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Be Happy!

polylog

Member

Registered: 2006-09-28

Posts: 162

Re: Math Help required!!!

Well, writing out the definition of adj(A) for A nxn is long and annoying, and then taking the adjoint of that, would be painful. I'm not sure what the point of the question is, but perhaps this is an acceptable answer (since this is an expression for ajd(adj(A)):

A^(-1) = (1/det(A)) * adj(A)

det(A) * A^(-1) = adj(A)

adj(adj(A)) = adj(det(A) * A^(-1))

Though this may not be what is required.

basmah

Member

Registered: 2006-10-02

Posts: 18

Re: Math Help required!!!

i have this proof as a homework for my linear algebra class, im not sure if this will help or not but i will surely use it to derive my answer and make up a proof......so thx

Last edited by basmah (2006-10-02 13:40:27)

polylog

Member

Registered: 2006-09-28

Posts: 162

Re: Math Help required!!!

I found a result online that states that

if  n = 2 then adj(adj A) = A

But I guess we are talking about nxn, in which case it does not look like it holds.

I wish I could be of more concrete help, but this is kindof obscure.

polylog

Member

Registered: 2006-09-28

Posts: 162

Re: Math Help required!!!

I hope u are still awake since I did finally prove this! :

let A be nxn

We need to use these facts:

(1) (A^-1)^-1 = A
(2) (k A)^-1 = (1/k) A^-1
(3) det(adj A) = (det(A))^(n-1) where n is the dimension
(4) A^-1 = 1/det(A) adj(A)

begin with (4):

A^-1 = 1/det(A) adj(A)

(A^-1)^-1 = (1/det(A) adj(A))^-1

A = (det A)(adj A)^-1

A = (det A) [ 1/(det (adj A)) adj(adj A) ]

A = (det A)/(det (adj A)) adj(adj A)

A = (det A)/(det A)^(n-1) adj(adj A)

A = (det A)^(2-n) adj(adj A)

adj(adj A) = (det A)^(-(2-n)) A

adj(adj A) = (det A)^(n-2) A

and clearly as a special case for 3x3 matrices:

det(A) = 1 -> adj(adj A) = A

Finally !

Last edited by polylog (2006-10-02 16:47:19)

WiZaRd

Real Member

Registered: 2006-09-22

Posts: 570

Re: Math Help required!!!

Okay thank GoD you Find that

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Be Happy!

basmah

Member

Registered: 2006-10-02

Posts: 18

Re: Math Help required!!!

thanks polylog....i mean i really appretiate it....but its too little too late...i did not check late at night since i wanted to get it done....but your answer is the correct one.....if only you could finish on time.....or i checked early in the morning but i did not think anyone would get it, so i just turned it in....all not so very correct....hopefully i'll get some credit on it....but i really want to thank you....there is a new one that is due day after tommorrow that i will post now.....for this one you would have a lot more time....i will try to do it my self and then check my work with what ever responses i get on here.....

let A be nxn

We need to use these facts:

(1) (A^-1)^-1 = A
(2) (k A)^-1 = (1/k) A^-1
(3) det(adj A) = (det(A))^(n-1) where n is the dimension
(4) A^-1 = 1/det(A) adj(A)

begin with (4):

A^-1 = 1/det(A) adj(A)

(A^-1)^-1 = (1/det(A) adj(A))^-1

A = (det A)(adj A)^-1

A = (det A) [ 1/(det (adj A)) adj(adj A) ]

A = (det A)/(det (adj A)) adj(adj A)

A = (det A)/(det A)^(n-1) adj(adj A)

A = (det A)^(2-n) adj(adj A)

adj(adj A) = (det A)^(-(2-n)) A

adj(adj A) = (det A)^(n-2) A

and clearly as a special case for 3x3 matrices:

det(A) = 1 -> adj(adj A) = A

that is the correct answer!!!!....again thx for the effort:D