MHB Proving the Inverse of the Adjoint Matrix Property for nxn Matrices

  • Thread starter Thread starter Yankel
  • Start date Start date
  • Tags Tags
    Inverse
Yankel
Messages
390
Reaction score
0
Hello

I need some help proving the next thing, I can't seem to be able to work it out..

Let A be an nxn matrix.

Prove that:

(adj A)^{-1} = adj(A^{-1})

Thanks...
 
Physics news on Phys.org
$A = IA = A^*(A^*)^{-1}A$

so:

$A^* = (A^*(A^*)^{-1}A)^* = A^*((A^*)^{-1})^*A$

therefore:

$A^*A^{-1} = A^*((A^*)^{-1})^*$

and multiplying on the left by $(A^*)^{-1}$ we get:

$A^{-1} = ((A^*)^{-1})^*$

so

$(A^{-1})^* = ((A^*)^{-1})^{**} = (A^*)^{-1}$
 
thanks, took me some time to understand your proof, but now I see it, nice one !
(Yes)
 
Thread 'How to define a vector field?'
Hello! In one book I saw that function ##V## of 3 variables ##V_x, V_y, V_z## (vector field in 3D) can be decomposed in a Taylor series without higher-order terms (partial derivative of second power and higher) at point ##(0,0,0)## such way: I think so: higher-order terms can be neglected because partial derivative of second power and higher are equal to 0. Is this true? And how to define vector field correctly for this case? (In the book I found nothing and my attempt was wrong...

Similar threads

  • · Replies 15 ·
Replies
15
Views
3K
  • · Replies 1 ·
Replies
1
Views
9K
Replies
2
Views
1K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 14 ·
Replies
14
Views
3K
Replies
34
Views
3K
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K