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I have to prove that determinant of skew- symmetric matrix of odd order is zero and also that its adjoint doesnt exist. I am sorry if the question is duplicate or already exists.I am not getting any start.I study in Class 11 so please give the proof accordingly. Thanks!

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    $\begingroup$ Did you mean inverse instead of adjoint? What is your definition for adjoint? $\endgroup$ – Element118 Nov 16 '15 at 6:01
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We know that eigenvalues of $A$ and $A^T$ are same and here $A^T=-A$ , that says eigenvalues of $A$ are symmetric about origin. i.e $\lambda$ is an evalue of $A$ iff $-\lambda$ is an evalue of $A$. Since order of matrix is odd. That proves the result.

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$A$ is skew-symmetric means $A^t=-A$. Taking determinant both sides $$\det(A^t)=\det(-A)\implies \det A =(-1)^n\det A \implies \det A =-\det A\implies \det A=0$$

I don't understand what do you mean by adjoint does not exist.

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