[QUE/ME-02005] ME-PROBLEM

 Using the definition of Levi-Civita and Kronecker delta symbols to prove the following identity \begin{eqnarray}\label{EQ01} \epsilon_{i\,j\,k}\,\epsilon_{l\,m\,n}\ &=&\ \begin{vmatrix} \delta_{il}&\delta_{im}&\delta_{in}\\ \delta_{jl}&\delta_{jm}&\delta_{jn}\\ \delta_{kl}&\delta_{km}&\delta_{kn} \end{vmatrix} \end{eqnarray} 

Interchanging \(i\leftrightarrow j\) in the right hand side amounts to interchange of the first and second rows and therefore the right hand side changes sign. Similar statement holds for exchange of \(i\leftrightarrow k\) and \(i\leftrightarrow k\). Thus the hand side is completely antisymmetric in indices \(i,j,k\). Repeating the above argument we see that the right hand side is totally antisymmetrc in indices \(\ell, m, n \) also. Thus the right hand side must be proportional to \(\epsilon_{i\,j\,k}\,\epsilon_{l\,m\,n}\). Thus \begin{equation} \text{R.H.S. of \eqRef{EQ01}} = K \epsilon_{i\,j\,k}\,\epsilon_{l\,m\,n}\label{EQ02} \end{equation} The constant of proportionality \(K\) is fixed by substituting \(i=1,j=2,k=3, \ell=1, m=2, n=3\) in \eqRef{EQ02}. The right hand side of \eqRef{EQ01} becomes determinant of unit matix and hence equal to 1. The right hand side of \eqRef{EQ02} becomes \(K \epsilon_{123}\epsilon_{123}=K\). This gives \(K=1\) and the proof of \eqRef{EQ01} is complete.