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# Prove that the skew symmetric determinant of an odd order is zero.$\left| \left( \begin{matrix} 0 & b & -c \\ -b & 0 & a \\ c & -a & 0 \\\end{matrix} \right) \right|$  Verified
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Hint:
Consider the matrix as $A$. Find ${{A}^{T}}$ of the matrix and prove that ${{A}^{T}}=-A$ as to prove that it is skew symmetric. Then using properties prove that the determinant of $A$=$0$.

Let us consider the given matrix as A.
$A=\left| \left( \begin{matrix} 0 & b & -c \\ -b & 0 & a \\ c & -a & 0 \\ \end{matrix} \right) \right|$
A matrix is called skew-symmetric if${{A}^{T}}=-A$, where ${{A}^{T}}$is the transpose of$A$.
We can use the properties of determinants to solve the expression.
The given matrix is $n\times n$ which is a $3\times 3$ matrix where $n=3$, which are the rows of the matrix and $n=3$, which are the columns of the matrix.
For any $n\times n$, matrix $A$ and a scalar $C$, we can say that,
\begin{align} & \det (A)=det({{A}^{T}}) \\ & \Rightarrow det(cA)=cdet(A) \\ \end{align}
Suppose that n is an odd integer and let A be a $n\times n$ skew-symmetric matrix. Thus we have${{A}^{T}}=-A$.
We have $A=\left| \left( \begin{matrix} 0 & b & -c \\ -b & 0 & a \\ c & -a & 0 \\ \end{matrix} \right) \right|$
Let us take ${{A}^{T}}$, the rows become columns and vice-versa.
${{A}^{T}}=\left| \left( \begin{matrix} 0 & -b & c \\ b & 0 & -a \\ -c & a & 0 \\ \end{matrix} \right) \right|$
Taking negative outside the determinant, we get,
${{A}^{T}}=-\left| \left( \begin{matrix} 0 & b & -c \\ -b & 0 & a \\ c & -a & 0 \\ \end{matrix} \right) \right|$
Hence the matrix is equal to $-A.$
$\therefore {{A}^{T}}=-A$.
By definition of skew-symmetric, we have,
$\det (A)=\det ({{A}^{T}})=\det (-A)$
Hence $A$ is a skew-symmetric matrix.
$\therefore \det (A)={{(-1)}^{n}}\det (A)=-\det (A)$, equal to ${{c}^{n}}\det (A),$n is odd, where $c=-1.$
\begin{align} & \Rightarrow \det (A)=-det(A) \\ & det(A)+det(A)=0 \\ & 2det(A)=0 \\ & \Rightarrow det(A)=0 \\ \end{align}
Hence we proved that the skew symmetric determinant of an odd order is zero.
Note:
We can prove it by supposing $An\times n=\left[ aij \right]$ which is a skew symmetric matrix and we can prove that $aii=0,$for $i=1,2,.......n$ and $aij=-aji.$
We can denote the ${{(i,j)}^{th}}$entry of ${{A}^{T}}=ai{{j}^{T}}.$
$\Rightarrow {{A}^{T}}=aij$, i.e. the transpose of $aij=aji.$
Together with \begin{align} & {{A}^{T}}=-A \\ & \Rightarrow ai{{j}^{T}}=-aji \\ & \Rightarrow aij=-aji \\ \end{align}
For $i,j=1,2,3,.....n$
when$i=j$, we get that,
\begin{align} & {{a}_{ii}}=-{{a}_{ii}} \\ & {{a}_{ii}}+{{a}_{ii}}=0 \\ & \Rightarrow 2{{a}_{ii}}=0 \\ & \therefore {{a}_{ii}}=0 \\ \end{align}