Question

Prove that the skew symmetric determinant of an odd order is zero.

0 & b & -c \\\ -b & 0 & a \\\ c & -a & 0 \\\ \end{matrix} \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.

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{aligned} & \det (A)=det({{A}^{T}}) \\\ & \Rightarrow det(cA)=cdet(A) \\\ \end{aligned}$$ Suppose that n is an odd integer and let A be a $$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{aligned} & \Rightarrow \det (A)=-det(A) \\\ & det(A)+det(A)=0 \\\ & 2det(A)=0 \\\ & \Rightarrow det(A)=0 \\\ \end{aligned}$$ Hence we proved that the skew symmetric determinant of an odd order is zero. Note: We can prove it by supposing $$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{aligned} & {{A}^{T}}=-A \\\ & \Rightarrow ai{{j}^{T}}=-aji \\\ & \Rightarrow aij=-aji \\\ \end{aligned}$$ For $$i,j=1,2,3,.....n$$ when$$i=j$$, we get that, $$\begin{aligned} & {{a}_{ii}}=-{{a}_{ii}} \\\ & {{a}_{ii}}+{{a}_{ii}}=0 \\\ & \Rightarrow 2{{a}_{ii}}=0 \\\ & \therefore {{a}_{ii}}=0 \\\ \end{aligned}$$