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!!


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


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.


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