Question

If $\xi =lx+my+nz$, $\eta =nx+ly+mz$, $\zeta =mx+ny+lz$, and if the same equations are true for all values of $x,y,z$ when $\xi ,\eta ,\zeta$ are interchanged with $x,y,z$ respectively, show that
$l^2+2mn=1$, $m^2+2ln=1$, $n^2+2lm=1$.

Answer 1

To show that $l^2+2mn=1,m^2+2ln=1,n^2+2lm=1$

We have,

$\xi =lx+my+nz,$

Since, same equations are true for all values of $x,y,z$

Then if $\xi ,\eta ,\zeta$ are interchanged with $x,y,z$respectively, we have

$x=l\xi +m\eta +n\zeta ,y=n\xi +l\eta +m\zeta ,z=m\xi +n\eta +l\zeta$

$\therefore$ By substitution, we have the identity

$\xi =l(l\xi +m\eta +n\zeta )+m(n\xi +l\eta +m\zeta )+n(m\xi +n\eta +l\zeta )$

On simplifying this we have

$\xi =\xi (l^2+2mn)+\eta (m^2+2ln)+\zeta (n^2+2lm)$

On equating the coefficients of $\xi ,\eta ,\zeta$ on both sides, we obtain the required relations

$l^2+2mn=1,m^2+2ln=1,n^2+2lm=1$