1
$\begingroup$

1) If A is a symmetric matrix , then Adj(A) is symmetric.

A . Adj(A) = Adj(A) . A =|A| I (where 'I' is an indentity matrix).

What i did was take tranpose of the original equation giving me :-

A$^t$ . [Adj(A)]$^t$ = [Adj(a)]$^t$ . A$^t$ = |A| I. (Also A$^t$ = A as A is symmetric).

Therefore:- A . [Adj(A)]$^t$ = [Adj(a)]$^t$ . A = |A| I.We also know adjoint of a matrix is unique therefore Adj(A)=[Adj(A)]$^t$.

But in above case we have assumed |A| $\not=$ 0. So if we define Adjoint of a matrix such A . Adj (A) = Adj (A) A = |A| I , then adjoint of matrix A is not unique if |A|=0. And if we define adjoint as transpose of cofactor matrix then Adj (A) is unique.

So what is the correct defination of adjoint of a matrix? Is adjoint defined for singular matrix? If 2nd defination is correct then is our statement valid for singular matrix ? If yes how to prove it?


2) If A is skew symmetric matrix of order n(where n is even) then Adj(A) is also skew symmetric . I used above method to prove it . Then again i have same doubts as in first question. How to prove it when |A|=0. And if Adj(A) is not defined for singular matrix then solve simply for transpose of cofactor matrix.(for the first question also)


3) transpose of cofactor matrix of a matrix of order n (odd) is symmetric if matrix is skew symmetric.

I did it for n=3 , but i don't how do it for general 'n'.Please provide a solution.

A = $$ \begin{bmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \\ \end{bmatrix} $$

Transpose of cofactor matrix:- $$ \begin{bmatrix} c^2 & -bc & ca \\ -bc & b^2 & -ab\\ ca & -ab & a^2 \\ \end{bmatrix} $$


Please tell me if i made any mistake in my proof or if any of the given statement is wrong. thankyou for help!!

$\endgroup$
0
$\begingroup$

I was just making things complicated , it's actually very simple.

Adj(A$^t$)=[Adj(A)]$^t$

If A is symmetric [Adj(A)]$^t$ = Adj(A$^t$) =Adj(A)

If A is skew symmetric [Adj(A)]$^t$ = Adj(A$^t$) = Adj(-A) (We know that Adj(kA)= k$^{n-1}$ A) = -Adj(A) if n is even and Adj(A) if n is odd.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.