Prove that the determinant of skew-symmetric matrices of odd order is zero

Question

I have to prove that determinant of skew-symmetric matrix of odd order is zero and also that its adjoint doesn't 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!

Answer 1

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.

Answer 2

$A$ 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.

Answer 3

Let the skew- symmetric matrix be, \begin{aligned}A=-A^{T}\\ A^{T}=\left( -A^{T}\right) ^{T}\\ A^{T}=-A\\ \left| A^{T}\right| =\left| -A\right| \\ \left| A^{T}\right| =-1 ^{2k+1}\left| A\right| \end{aligned} \begin{aligned}\left| A^{T}\right| =-\left| A\right| \\ \left| A\right| ^{T}=-\left| A\right| \\ \left| A\right| =-\left| A\right| \end{aligned} \begin{aligned}2\left| A\right| =0\\ \left| A\right| =0\end{aligned}