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Prove that the skew symmetric determinant of an odd order is zero.
\[\left| \left( \begin{matrix}
   0 & b & -c \\
   -b & 0 & a \\
   c & -a & 0 \\
\end{matrix} \right) \right|\]

Answer
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154.5k+ views
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 \[A\]=$0$.

“Complete step-by-step answer:”
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,
\[\begin{align}
  & {{a}_{ii}}=-{{a}_{ii}} \\
 & {{a}_{ii}}+{{a}_{ii}}=0 \\
 & \Rightarrow 2{{a}_{ii}}=0 \\
 & \therefore {{a}_{ii}}=0 \\
\end{align}\]