Question

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

Answer 1

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.
$$A=\left|\left(\begin{array}{ccc}0& b& -c\\ -b& 0& a\\ c& -a& 0\end{array}\right)\right|$$
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,
$$\begin{array}{rl}& det(A)=det(A^T)\\ & \Rightarrow det(cA)=cdet(A)\end{array}$$
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 $$A=\left|\left(\begin{array}{ccc}0& b& -c\\ -b& 0& a\\ c& -a& 0\end{array}\right)\right|$$
Let us take $$A^T$$, the rows become columns and vice-versa.
$$A^T=\left|\left(\begin{array}{ccc}0& -b& c\\ b& 0& -a\\ -c& a& 0\end{array}\right)\right|$$
Taking negative outside the determinant, we get,
$$A^T=-\left|\left(\begin{array}{ccc}0& b& -c\\ -b& 0& a\\ c& -a& 0\end{array}\right)\right|$$
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 $$c=-1.$$
$$\begin{array}{rl}& \Rightarrow det(A)=-det(A)\\ & det(A)+det(A)=0\\ & 2det(A)=0\\ & \Rightarrow det(A)=0\end{array}$$
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 $$\begin{array}{rl}& A^T=-A\\ & \Rightarrow ai{j}^T=-aji\\ & \Rightarrow aij=-aji\end{array}$$
For $$i,j=1,2,3,.....n$$
when$$i=j$$, we get that,
$$\begin{array}{rl}& a_{ii}=-a_{ii}\\ & a_{ii}+a_{ii}=0\\ & \Rightarrow 2a_{ii}=0\\ & \therefore a_{ii}=0\end{array}$$