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If ${a_1},{a_2},{a_3},{b_1},{b_2},{b_3} \in R$ and are such that ${a_i}{b_j} \ne 1$ $1 \leqslant i,j \leqslant 3$ , then
$\left| {\begin{array}{*{20}{c}}
  {\dfrac{{1 - a_1^3b_1^3}}{{1 - {a_1}{b_1}}}}&{\dfrac{{1 - a_1^3b_2^3}}{{1 - {a_1}{b_2}}}}&{\dfrac{{1 - a_1^3b_3^3}}{{1 - {a_1}{b_3}}}} \\
  {\dfrac{{1 - a_2^3b_1^3}}{{1 - {a_2}{b_1}}}}&{\dfrac{{1 - a_2^3b_2^3}}{{1 - {a_2}{b_2}}}}&{\dfrac{{1 - a_2^3b_3^3}}{{1 - {a_2}{b_3}}}} \\
  {\dfrac{{1 - a_3^3b_1^3}}{{1 - {a_3}{b_1}}}}&{\dfrac{{1 - a_3^3b_2^3}}{{1 - {a_3}{b_2}}}}&{\dfrac{{1 - a_3^3b_3^3}}{{1 - {a_3}{b_3}}}}
\end{array}} \right| > 0$ provided either ${a_1} < {a_2} < {a_3}$ and ${b_1} < {b_2} < {b_3}$ or
${a_1} > {a_2} > {a_3}$ and ${b_1} > {b_2} > {b_3}$ then show that $\left( {{a_1} - {a_2}} \right)({a_2} - {a_3})({a_3} - {a_1})({b_1} - {b_2})({b_2} - {b_3})({b_3} - {b_1}) > 0$.

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Last updated date: 13th Jun 2024
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Answer
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Hint: As the given question is based on determinant, we will use some of the properties of matrices and determinant like splitting determinants to solve the problem such as:
\[\left| {\begin{array}{*{20}{c}}
  1&a&{{a^2}} \\
  1&b&{{b^2}} \\
  1&c&{{c^2}}
\end{array}} \right| = (a - b)(b - c)(c - a)\]
And also the given determinant includes terms of the form ${a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right)$ which can be used in between to simplify the given complex form.

Complete step-by-step solution:
Step 1: As we have in the determinant the term $\dfrac{{1 - a_1^3b_1^3}}{{1 - {a_1}{b_1}}}$where the numerator is of the form ${a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right)$. Thus substituting this in $\dfrac{{1 - a_1^3b_1^3}}{{1 - {a_1}{b_1}}}$ we get
\[\dfrac{{1 - a_1^3b_1^3}}{{1 - {a_1}{b_1}}} = \dfrac{{(1 - {a_1}{b_1})(1 + {a_1}{b_1} + {a_1}^2{b_1}^2)}}{{1 - {a_1}{b_1}}} = 1 + {a_1}{b_1} + {a_1}^2{b_1}^2\]
Using the same procedure and rewriting all the terms in the determinant we get,
$\Rightarrow$\[\left| {\begin{array}{*{20}{c}}
  {1 + {a_1}{b_1} + {a_1}^2{b_1}^2}&{1 + {a_1}{b_2} + {a_1}^2{b_2}^2}&{1 + {a_1}{b_3} + {a_1}^2{b_3}^2} \\
  {1 + {a_2}{b_1} + {a_2}^2{b_1}^2}&{1 + {a_2}{b_2} + {a_2}^2{b_2}^2}&{1 + {a_2}{b_3} + {a_2}^2{b_3}^2} \\
  {1 + {a_3}{b_1} + {a_3}^2{b_1}^2}&{1 + {a_3}{b_2} + {a_3}^2{b_2}^2}&{1 + {a_3}{b_3} + {a_3}^2{b_3}^2}
\end{array}} \right| > 0\]
Step 2: Now splitting this determinant we get,
$\Rightarrow$\[\left| {\begin{array}{*{20}{c}}
  1&{{a_1}}&{{a_1}^2} \\
  1&{{a_2}}&{{a_2}^2} \\
  1&{{a_3}}&{{a_3}^2}
\end{array}} \right|\left| {\begin{array}{*{20}{c}}
  1&{{b_1}}&{{b_1}^2} \\
  1&{{b_2}}&{{b_2}^2} \\
  1&{{b_3}}&{{b_3}^2}
\end{array}} \right| > 0\] ………………..…Formula1
Step 3: Now applying the property, \[\left| {\begin{array}{*{20}{c}}
  1&a&{{a^2}} \\
  1&b&{{b^2}} \\
  1&c&{{c^2}}
\end{array}} \right| = (a - b)(b - c)(c - a)\] we can rewrite formula1 as,
$\left( {{a_1} - {a_2}} \right)({a_2} - {a_3})({a_3} - {a_1})({b_1} - {b_2})({b_2} - {b_3})({b_3} - {b_1}) > 0$ which is the required solution.
 Hence the proof.
Step 4: Now we need to check for each case provided.
For case 1, ${a_1} < {a_2} < {a_3}$ and ${b_1} < {b_2} < {b_3}$
Thus taking out each term in the product we need and checking whether it is positive or negative.
$\left( {{a_1} - {a_2}} \right) < 0,({a_2} - {a_3}) < 0,({a_3} - {a_1}) > 0$
And $({b_1} - {b_2}) < 0,({b_2} - {b_3}) < 0,({b_3} - {b_1}) > 0$
Now combining partially,
$\Rightarrow$$\left( {{a_1} - {a_2}} \right)({a_2} - {a_3})({a_3} - {a_1}) > 0,({b_1} - {b_2})({b_2} - {b_3})({b_3} - {b_1}) > 0$
Combining them in total, $\left( {{a_1} - {a_2}} \right)({a_2} - {a_3})({a_3} - {a_1})({b_1} - {b_2})({b_2} - {b_3})({b_3} - {b_1}) > 0$
Step 5: Now applying case 2, ${a_1} > {a_2} > {a_3}$ and ${b_1} > {b_2} > {b_3}$
Thus taking out each term in the product we need and checking whether it is positive or negative.
$\left( {{a_1} - {a_2}} \right) > 0,({a_2} - {a_3}) > 0,({a_3} - {a_1}) < 0$
And $({b_1} - {b_2}) > 0,({b_2} - {b_3}) > 0,({b_3} - {b_1}) < 0$
Now combining partially,
$\Rightarrow$$\left( {{a_1} - {a_2}} \right)({a_2} - {a_3})({a_3} - {a_1}) > 0,({b_1} - {b_2})({b_2} - {b_3})({b_3} - {b_1}) > 0$
Combining them in total, $\left( {{a_1} - {a_2}} \right)({a_2} - {a_3})({a_3} - {a_1})({b_1} - {b_2})({b_2} - {b_3})({b_3} - {b_1}) > 0$

Final answer : For each of the cases, $\left( {{a_1} - {a_2}} \right)({a_2} - {a_3})({a_3} - {a_1})({b_1} - {b_2})({b_2} - {b_3})({b_3} - {b_1}) > 0$

Note: Here we need to be careful while splitting the determinant and while applying the cases.
Splitting the determinant can be reverted to check whether both the split and original determinants yield the same value. Applying cases is completely the application of algebraic property in real numbers named as ordering. That is if a,b are real numbers with a0.