# What is value of determinent of adjoint of a matrix?

1. Feb 3, 2012

### vkash

1. The problem statement, all variables and given/known data

2. Relevant equations

AA-1=I

3. The attempt at a solution

pre multiplying both sides with A-1
using AA-1=I
taking determinant of both sides.
is it correct????
IF it is wrong then point out that spot.
thanks
--------------
vikash chandola

2. Feb 3, 2012

### Simon Bridge

Why did you drop the identity matrix on the RHS in line 3?

I have: for any nxn non-singular matrix:
I thnk you also want $|k\bm{A}|=k^n|\bm{A}|$ to go with $\bm{A}\adj{A}=\det{\bm{A}}\bm{I}$

Whatever people are prepared to pay for it - in a market economy. Have you considered putting it up for auction :)

3. Feb 3, 2012

### Fredrik

Staff Emeritus
What do you mean by |A|? The determinant of A? In that case, this equality doesn't hold for the 1×1 matrix (i), which has the adjoint (-i). Their product is (1), but the determinant of (i) is i.

4. Feb 3, 2012

### Simon Bridge

... seems a common enough identity to me.
ferinstance, used in https://www.physicsforums.com/archive/index.php/t-100824.html
Works for n > 1.

... which makes the relation work for n=1 as well doesn't it?

What do you make A.adj(A) to come to?

Last edited: Feb 3, 2012
5. Feb 3, 2012

### Fredrik

Staff Emeritus
The adjoint is the complex conjugate of the transpose. Let's try it with a 3×3 matrix:
$$A=\begin{pmatrix} i & 0 & 0\\ 0 & i & 0\\ 0 & 0 & i \end{pmatrix},\qquad A^\dagger= \begin{pmatrix} -i & 0 & 0\\ 0 & -i & 0\\ 0 & 0 & -i \end{pmatrix}=-A,\qquad \det A=i^3=-i$$ $$AA^\dagger=I,\qquad(\det A)I=-iI\neq I$$

6. Feb 3, 2012

### Ray Vickson

There is genuine confusion over the terminology, because the same word "adjoint" is used for two different concepts by different writers. Probably what the poster here means is the old-fashioned linear algebra concept with [adj(A)](i,j) = (-1)^(i+j)M(j,i), where M(u,v) = (u,v)-minor of A = determinant of the matrix obtained by deleting row u and column v of A. The Wikipedia article http://en.wikipedia.org/wiki/Adjugate_matrix refers to this as the "adjugate or classical adjoint" and points out:
"The adjugate has sometimes been called the 'adjoint', but that terminology is ambiguous. Today, 'adjoint' of a matrix normally refers to its corresponding adjoint operator, which is its conjugate transpose." I don't think the use of the term 'adjugate' is nearly as widespread as the Wiki article claims.

RGV

7. Feb 3, 2012

### Fredrik

Staff Emeritus
OK, thanks. I don't think I've ever heard anyone call that matrix the "adjoint". "Adjugate" sounds vaguely familiar, but I'm not sure I've heard that term either. I think I've heard it called "matrix of cofactors", but I could be mistaken. I haven't had to use this concept many times. Edit: I checked the Wikipedia article. The adjugate matrix is the transpose of the cofactor matrix.

Last edited: Feb 3, 2012
8. Feb 3, 2012

### Simon Bridge

I think one of those links I gave you explains this too.

The one you refer to is normally called the conjugate transpose... and searching wiki for "adjoint matrix" will take you to the conjugate transpose page.
The one I've mostly seen is the hermitian adjoint.
The one you get when you google for "determinant of adjoint" (i.e. in context) is the classical adjoint.

This makes sense - after all, what's the determinant of the conjugate transpose?

All of which tends to support Ray - term 'adjugate' is nearly as not as widespread as the Wiki article makes out.

Need to hear from OP.

9. Feb 3, 2012

### vkash

my third line was
so here i am multiplying Identity matrix(I) with a matrix so it is self matrix.
AI=A; I think this statement is correct.(A and I are of same order)

10. Feb 4, 2012

### Simon Bridge


I had noticed you went from $\A^{-1}\A \adj{\A}=\A^{-1}\det{\A}\I$ to $\I \adj{\A}=\A^{-1}\det{\A}$ using $\A \A^{-1}=\I$
vis:
You kept the I on the LHS but not on the RHS. That's all. This kind of thing alerts your marker to look more closely at this part of the reasoning... if you have made an arror it is likely to be close to that line. Let's see...

The strategy seems to be to get rid of an inconvenient A on the LHS at the expense of the losing the diagonalization on the RHS.
So I was wondering what you gained by this compared with just taking the determinant of both sides, right at the beginning, and exploiting the distributive property of determinants?
Wot-ho - continuing....

Since $\det{k\A}=k^n\det{\A}$ for an nxn matrix $\A$ ... when you do the next step: $\det{\adj{\A}}=\det{\A^{-1}\det{\A}}=\det{\A}^n\det{\A^{-1}}$ ... which is different from what you got. So when you put $\det{\A}\det{\A^{-1}}=1$ ...