Question

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

Hint: We will first interchange the values as provided in the question. Then, we will substitute the values of x, y and z back into the given equations of $\xi $,$\eta $and $\zeta $respectively. After that, we will simplify the equations into terms of $\xi $,$\eta $and $\zeta $, then we will compare both sides for the coefficients of $\xi $,$\eta $and $\zeta $for the required proof.

Complete step-by-step answer:
We are required to prove that:
${l}^{2}$ + 2mn = 1, ${m}^{2}$ + 2ln = 1, ${n}^{2}$ + 2lm = 1
We are given three equations as: $\xi $= l x + m y + n z, $\eta $= n x + l y + m z, $\zeta $= m x + n y + l z
It is said that if $\xi $,$\eta $, $\zeta $ are interchanged with x, y, z respectively, then the same equations will still hold true.
$ \Rightarrow x = l\xi + m\eta + n\zeta ,y = n\xi + l\eta + m\zeta ,z = m\xi + n\eta + l\zeta $
Now, substituting these values of x, y and z in the given equation, we get
$ \Rightarrow \xi = l\left( {l\xi + m\eta + n\zeta } \right) + m\left( {n\xi + l\eta + m\zeta } \right) + n\left( {m\xi + n\eta + l\zeta } \right)$
$ \Rightarrow \xi = {l^2}\xi + lm\eta + \ln \zeta + mn\xi + ml\eta + {m^2}\zeta + nm\xi + {n^2}\eta + nl\zeta $
We get an identity upon substitution and now simplifying this equation in terms of $\xi $,$\eta $and $\zeta $, we get
$ \Rightarrow \xi = \left( {{l^2} + 2mn} \right)\xi + \left( {{m^2} + 2\ln } \right)\zeta + \left( {{n^2} + 2ml} \right)\eta $
On equating coefficients both sides, we get
$ \Rightarrow $ ${l}^{2}$ + 2mn = 1, ${m}^{2}$ + 2ln = 1, ${n}^{2}$ + 2lm = 1

Note: In such questions, you may get confused with language provided in the question. Be careful while solving after interchanging the variables $\xi $,$\eta $ and $\zeta $with x, y and z. You can also prove this question by putting the value of x in any of the remaining equations.