Question

Prove the following
\left| {\begin{array}{*{20}{c}} {bc}&{bc' + b'c}&{b'c'} \\\ {ca}&{ca' + c'a}&{c'a'} \\\ {ab}&{ab' + a'b}&{a'b'} \end{array}} \right| = \left( {bc' - b'c} \right)\left( {ca' - c'a} \right)\left( {ab' - a'b} \right)

Answer 1

In this particular question use the concept of simplification first separate the determinant into two parts then form first and second determinant take abc and a’b’c’ common from first and third column respectively, so use these concepts to reach the solution of the question.

Complete step-by-step solution:
Given identity which we have to prove
\left| {\begin{array}{*{20}{c}} {bc}&{bc' + b'c}&{b'c'} \\\ {ca}&{ca' + c'a}&{c'a'} \\\ {ab}&{ab' + a'b}&{a'b'} \end{array}} \right| = \left( {bc' - b'c} \right)\left( {ca' - c'a} \right)\left( {ab' - a'b} \right)
Consider the LHS of the above identity we have,
\Rightarrow \left| {\begin{array}{*{20}{c}} {bc}&{bc' + b'c}&{b'c'} \\\ {ca}&{ca' + c'a}&{c'a'} \\\ {ab}&{ab' + a'b}&{a'b'} \end{array}} \right|
Now break this determinant into sum of two determinants we have,
\Rightarrow \left| {\begin{array}{*{20}{c}} {bc}&{bc'}&{b'c'} \\\ {ca}&{ca'}&{c'a'} \\\ {ab}&{ab'}&{a'b'} \end{array}} \right| + \left| {\begin{array}{*{20}{c}} {bc}&{b'c}&{b'c'} \\\ {ca}&{c'a}&{c'a'} \\\ {ab}&{a'b}&{a'b'} \end{array}} \right|
Now form first and second determinant take abc and a’b’c’ common from first and third column respectively we have,

{\dfrac{{bc}}{{abc}}}&{bc'}&{\dfrac{{b'c'}}{{a'b'c'}}} \\\ {\dfrac{{ca}}{{abc}}}&{ca'}&{\dfrac{{c'a'}}{{a'b'c'}}} \\\ {\dfrac{{ab}}{{abc}}}&{ab'}&{\dfrac{{a'b'}}{{a'b'c'}}} \end{array}} \right| + \left( {abc} \right)\left( {a'b'c'} \right)\left| {\begin{array}{*{20}{c}} {\dfrac{{bc}}{{abc}}}&{b'c}&{\dfrac{{b'c'}}{{a'b'c'}}} \\\ {\dfrac{{ca}}{{abc}}}&{c'a}&{\dfrac{{c'a'}}{{a'b'c'}}} \\\ {\dfrac{{ab}}{{abc}}}&{a'b}&{\dfrac{{a'b'}}{{a'b'c'}}} \end{array}} \right|$$ Now simplify we have, $$ \Rightarrow \left( {abc} \right)\left( {a'b'c'} \right)\left| {\begin{array}{*{20}{c}} {\dfrac{1}{a}}&{bc'}&{\dfrac{1}{{a'}}} \\\ {\dfrac{1}{b}}&{ca'}&{\dfrac{1}{{b'}}} \\\ {\dfrac{1}{c}}&{ab'}&{\dfrac{1}{{c'}}} \end{array}} \right| + \left( {abc} \right)\left( {a'b'c'} \right)\left| {\begin{array}{*{20}{c}} {\dfrac{1}{a}}&{b'c}&{\dfrac{1}{{a'}}} \\\ {\dfrac{1}{b}}&{c'a}&{\dfrac{1}{{b'}}} \\\ {\dfrac{1}{c}}&{a'b}&{\dfrac{1}{{c'}}} \end{array}} \right|$$ $$ \Rightarrow \left( {abc} \right)\left( {a'b'c'} \right)\left[ {\left| {\begin{array}{*{20}{c}} {\dfrac{1}{a}}&{bc'}&{\dfrac{1}{{a'}}} \\\ {\dfrac{1}{b}}&{ca'}&{\dfrac{1}{{b'}}} \\\ {\dfrac{1}{c}}&{ab'}&{\dfrac{1}{{c'}}} \end{array}} \right| + \left| {\begin{array}{*{20}{c}} {\dfrac{1}{a}}&{b'c}&{\dfrac{1}{{a'}}} \\\ {\dfrac{1}{b}}&{c'a}&{\dfrac{1}{{b'}}} \\\ {\dfrac{1}{c}}&{a'b}&{\dfrac{1}{{c'}}} \end{array}} \right|} \right]$$ Now expand the determinant we have, $$ \Rightarrow \left( {abc} \right)\left( {a'b'c'} \right)\left[ {\dfrac{1}{a}\left| {\begin{array}{*{20}{c}} {ca'}&{\dfrac{1}{{b'}}} \\\ {ab'}&{\dfrac{1}{{c'}}} \end{array}} \right| - bc'\left| {\begin{array}{*{20}{c}} {\dfrac{1}{b}}&{\dfrac{1}{{b'}}} \\\ {\dfrac{1}{c}}&{\dfrac{1}{{c'}}} \end{array}} \right| + \dfrac{1}{{a'}}\left| {\begin{array}{*{20}{c}} {\dfrac{1}{b}}&{ca'} \\\ {\dfrac{1}{c}}&{ab'} \end{array}} \right| + \dfrac{1}{a}\left| {\begin{array}{*{20}{c}} {c'a}&{\dfrac{1}{{b'}}} \\\ {a'b}&{\dfrac{1}{{c'}}} \end{array}} \right| - b'c\left| {\begin{array}{*{20}{c}} {\dfrac{1}{b}}&{\dfrac{1}{{b'}}} \\\ {\dfrac{1}{c}}&{\dfrac{1}{{c'}}} \end{array}} \right| + \dfrac{1}{{a'}}\left| {\begin{array}{*{20}{c}} {\dfrac{1}{b}}&{c'a} \\\ {\dfrac{1}{c}}&{a'b} \end{array}} \right|} \right]$$ Now expand the all mini determinant we have, $$ \Rightarrow \left( {abc} \right)\left( {a'b'c'} \right)\left[ {\dfrac{1}{a}\left( {\dfrac{{ca'}}{{c'}} - \dfrac{{ab'}}{{b'}}} \right) - bc'\left( {\dfrac{1}{{bc'}} - \dfrac{1}{{b'c}}} \right) + \dfrac{1}{{a'}}\left( {\dfrac{{ab'}}{b} - \dfrac{{ca'}}{c}} \right) + \dfrac{1}{a}\left( {\dfrac{{c'a}}{{c'}} - \dfrac{{a'b}}{{b'}}} \right) - b'c\left( {\dfrac{1}{{bc'}} - \dfrac{1}{{b'c}}} \right) + \dfrac{1}{{a'}}\left( {\dfrac{{a'b}}{b} - \dfrac{{c'a}}{c}} \right)} \right]$$Now simplify we have, $$ \Rightarrow \left( {abc} \right)\left( {a'b'c'} \right)\left[ {\dfrac{{ca'}}{{ac'}} - 1 - 1 + \dfrac{{bc'}}{{b'c}} + \dfrac{{ab'}}{{a'b}} - 1 + 1 - \dfrac{{a'b}}{{ab'}} - \dfrac{{b'c}}{{bc'}} + 1 + 1 - \dfrac{{c'a}}{{a'c}}} \right]$$ $$ \Rightarrow \left( {abc} \right)\left( {a'b'c'} \right)\left[ {\dfrac{{ca'}}{{ac'}} + \dfrac{{bc'}}{{b'c}} + \dfrac{{ab'}}{{a'b}} - \dfrac{{a'b}}{{ab'}} - \dfrac{{b'c}}{{bc'}} - \dfrac{{c'a}}{{a'c}} + 3 - 3} \right]$$ $$ \Rightarrow \left( {abc} \right)\left( {a'b'c'} \right)\left[ {\left( {\dfrac{{ca'}}{{ac'}} - \dfrac{{c'a}}{{a'c}}} \right) + \left( {\dfrac{{bc'}}{{b'c}} - \dfrac{{b'c}}{{bc'}}} \right) + \left( {\dfrac{{ab'}}{{a'b}} - \dfrac{{a'b}}{{ab'}}} \right)} \right]$$ $$ \Rightarrow \left( {abc} \right)\left( {a'b'c'} \right)\left[ {\left( {\dfrac{{{{\left( {ca'} \right)}^2} - {{\left( {c'a} \right)}^2}}}{{aa'cc'}}} \right) + \left( {\dfrac{{{{\left( {bc'} \right)}^2} - {{\left( {b'c} \right)}^2}}}{{bb'cc'}}} \right) + \left( {\dfrac{{{{\left( {ab'} \right)}^2} - {{\left( {a'b} \right)}^2}}}{{aa'bb'}}} \right)} \right]$$ $$ \Rightarrow \left( {abc} \right)\left( {a'b'c'} \right)\left[ {\dfrac{{bb'{{\left( {ca'} \right)}^2} - bb'{{\left( {c'a} \right)}^2} + aa'{{\left( {bc'} \right)}^2} - aa'{{\left( {b'c} \right)}^2} + cc'{{\left( {ab'} \right)}^2} - cc'{{\left( {a'b} \right)}^2}}}{{aa'bb'cc'}}} \right]$$ $$ \Rightarrow bb'{\left( {ca'} \right)^2} - bb'{\left( {c'a} \right)^2} + aa'{\left( {bc'} \right)^2} - aa'{\left( {b'c} \right)^2} + cc'{\left( {ab'} \right)^2} - cc'{\left( {a'b} \right)^2}$$................. (1) Now consider the RHS of the given equation we have, $ \Rightarrow \left( {bc' - b'c} \right)\left( {ca' - c'a} \right)\left( {ab' - a'b} \right)$ Now multiply first two terms we have, $ \Rightarrow \left( {bc'ca' - bc'c'a - b'cca' + b'cc'a} \right)\left( {ab' - a'b} \right)$ $ \Rightarrow \left( {bc'ca'ab' - bc'c'aab' - b'cca'ab' + b'cc'aab' - a'bbc'ca' + a'bbc'c'a + a'bb'cca' - a'bb'cc'a} \right)$ Now cancel out the same terms with positive and negative sign we have, $ \Rightarrow \left( { - bc'c'aab' - b'cca'ab' + b'cc'aab' - a'bbc'ca' + a'bbc'c'a + a'bb'cca'} \right)$ $$ \Rightarrow bb'{\left( {ca'} \right)^2} - bb'{\left( {c'a} \right)^2} + aa'{\left( {bc'} \right)^2} - aa'{\left( {b'c} \right)^2} + cc'{\left( {ab'} \right)^2} - cc'{\left( {a'b} \right)^2}$$............... (2) So as we see that equation (1) and (2) both are the same. Therefore, LHS = RHS Hence proved. **Note:** In such types of questions it is advised to simplify the LHS or the RHS according to their complexity of the functions. Sometimes proving LHS = RHS needs simplification on both sides of the equation. Remember to convert dissimilar functions to get to the final result, and check whether R.H.S is equal to L.H.S or not if yes then it is the required answer.