Question

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

Answer 1

Attachment......................

Answer 2

Answer:

It proved that the determinant of the skew-symmetric matrix of odd order is zero.

Step-by-step explanation:

Suppose that $n$ is an odd integer and let $A$ be an $n\times n$ skew-symmetric matrix.

Thus, we've $A^{T}=-A$ by definition of skew-symmetric.

Then we have to use the property $\det(A)=\det(A^T)$.

Since $A$ is skew-symmetric then $\det(A)=\det(-A)$

Now, we'll use the property $\det(cA)=c^n\det(A)$ then we get

$\det(A)=(-1)^n \det(A)$

It is as long as the skew-symmetric matrix is of odd order then $n$ is odd.

$\det(A)=-\det(A)$

Therefore, it yields that $2\det(A)=0$

Hence $\det(A)=0$.

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