The Determinant of a Skew-Symmetric Matrix of odd order is Zero.

In this post we are going to show that Determinent of a Skew-Symmetric Matrix of odd order is Zero.First of all lets see what is a skew-symmetric matrix

A matrix \(\;M\;\) of order \(\; n \times n\; \) is said to be a skew-symmetric matirx iff \( M^{t}= -M\)

see the followig example

Let \[M=\begin{bmatrix} 0& 2 & 3\\ -2 & 0& 5\\ -3 & -5 & 0 \end{bmatrix} \;\;\;\;\;\;\;\;\;\;\;\;.....1 \] now if we take the transpose of this matrix i.e \(M^t \) will be
\[M^t=\begin{bmatrix} 0& -2 & -3\\ 2 & 0& -5\\ 3 & 5 & 0 \end{bmatrix}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\]
we can see that if you take out \(-1\) form the above matrix it will be of form \[M^t=-\begin{bmatrix} 0& 2 & 3\\ -2 & 0& 5\\ -3 & -5 & 0 \end{bmatrix} \;\;\;\;\;\;\;\;\;\;\;\;.....2 \]

From above we can see that matrix on right is original matrix \(M\) i.e form \(1\) and \(2\) we have .

\[M^t=-M\] so the above given matrix is skew-symmetric

Now lets move on to Proff of "The Determinant of a Skew-Symmetric Matrix of odd order is Zero." In this proff we are going to use two important properties of Determinent which are :-

\(1). \det(M)=\det(M^{t})\)

\(2). \det(kM)=k^n \det(M)\;\;\ \text{ n is order of matrix}\)

Given- A skew-symmetric matrix \(M\) of odd order \(n\)

To prove-The Determinant of a Matrix \(M\) is Zero.

Proff- \[\det(M)=\det(M^t)\;\;\; (\text{form above})\]
\[\det(M)=\det(−M)\;\; (\text{because M is skew-symmetric})\]
\[\det(M)=(−1)^n \det(M)\;\;\;(\text{by property} )\]
\[\det(M)=−\det(M) \;\;\;\;\; \text{because n is odd so (\(-1)^n =-1\)}\]
\(\text{taking \(-\det(M)\) on other side we will get }\) \[2\det(M)=0\], \[=\det(M)=0.\] \[\textbf{HENCE PROVED}\]