Question

Eliminate l,m between the equations \[{\text{lx + my = a, mx - ly = b, }}{{\text{l}}^2}{\text{ + }}{{\text{m}}^2}{\text{ = 1}}\] .

Answer 1

Hint:- Square both the equations given in terms of x & y and add.

Let the given equation be
$
  {\text{lx + my = a }} \cdots \left( 1 \right) \\
  {\text{mx - ly = b }} \cdots \left( 2 \right) \\
  {{\text{l}}^2}{\text{ + }}{{\text{m}}^2}{\text{ = 1 }} \cdots \left( 3 \right) \\
 $
For eliminating l and m, we need to perform squaring on both sides of equation (1) and (2).
$
  {\left( {{\text{lx + my}}} \right)^2} = {\text{ }}{{\text{a}}^2}{\text{ }} \cdots \left( 4 \right) \\
  {\left( {{\text{mx - ly}}} \right)^2}{\text{ = }}{{\text{b}}^2}{\text{ }} \cdots \left( 5 \right) \\
 $
On adding equations (4) and (5) and simplifying the equation, we get
$
  {\left( {{\text{lx + my}}} \right)^2}{\text{ + }}{\left( {{\text{mx - ly}}} \right)^2}{\text{ = }}{{\text{a}}^2}{\text{ + }}{{\text{b}}^2}{\text{ }} \cdots \left( 6 \right) \\
  {{\text{l}}^2}{{\text{x}}^2}{\text{ + }}{{\text{m}}^2}{{\text{y}}^2}{\text{ + 2(lx)(my) + }}{{\text{m}}^2}{{\text{x}}^2}{\text{ + }}{{\text{l}}^2}{{\text{y}}^2}{\text{ - 2(mx)(ly) = }}{{\text{a}}^2}{\text{ + }}{{\text{b}}^2} \\
$
Cancelling (2lmxy) terms and taking ${{\text{l}}^2}{\text{ + }}{{\text{m}}^2}$ common , we get
$\left( {{{\text{l}}^2}{\text{ + }}{{\text{m}}^2}} \right){{\text{x}}^2}{\text{ + }}\left( {{{\text{l}}^2}{\text{ + }}{{\text{m}}^2}} \right){{\text{y}}^2} = {\text{ }}{{\text{a}}^2}{\text{ + }}{{\text{b}}^2}{\text{ }} \cdots \left( 7 \right)$
Putting the value of ${{\text{l}}^2}{\text{ + }}{{\text{m}}^2}$ from equation (3), we get
${{\text{x}}^2}{\text{ + }}{{\text{y}}^2}{\text{ = }}{{\text{a}}^2}{\text{ + }}{{\text{b}}^2}$
It is the required answer.

Note:- There are two well known methods to solve the linear algebraic equations. They are (i) by substitution and (ii) by elimination by multiplication. These methods can easily be applied when the equations have two unknowns. But here we have four unknowns. Assuming a and b as constant.So, these types of problems are solved differently.