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my proof:- If $A$ and $B$ are two matrices such that $AB=BA=\det(A)I$, then $B=\operatorname{adj}(A)$. Therefore, $A\operatorname{adj}(A)=\operatorname{adj}(A)A=\det(A)I$. Now, for the matrix $\operatorname{adj}(A)$, there exists a matrix $A$ such that $\operatorname{adj}(A)A=A\operatorname{adj}(A)=\det(A)I$. So, $\operatorname{adj}\left(\operatorname{adj}(A)\right)=A$. But I know this result do not hold if the order of the square matrix is greater than $2$. Please clarify me-in which step I am wrong..

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  • $\begingroup$ Please use mathjax. $\endgroup$ – Chickenmancer Apr 26 '18 at 4:36
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    $\begingroup$ $\text{adj}(\text{adj}(A))$ is a scalar multiple of $A$. $\endgroup$ – Lord Shark the Unknown Apr 26 '18 at 4:41
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If A and B are two matrices such that AB=BA=detA.I [where I is an identity matrix]; then B =adj A.

This is not true. For example, if $A=0$, then $B$ could be any matrix. It is only true if $A$ is invertible.

But even ignoring that issue, your conclusion doesn't follow. You say

adjA. A=A.adjA=detA.I

but you can't conclude $\mbox{adj}(\mbox{adj} A) = A$ from this. You would need the right hand side to be $\det(\mbox{adj} A) I$ (which isn't true) to draw the conclusion $\mbox{adj}(\mbox{adj} A) = A$ from what you had before.

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