Linear Algebra – Prove 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

$A$ 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.