Answer
Verified
435.6k+ views
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.
\[\left| {\begin{array}{*{20}{c}}
1&a&{{a^2}} \\
1&b&{{b^2}} \\
1&c&{{c^2}}
\end{array}} \right| = (a - b)(b - c)(c - a)\]
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.
Recently Updated Pages
what is the correct chronological order of the following class 10 social science CBSE
Which of the following was not the actual cause for class 10 social science CBSE
Which of the following statements is not correct A class 10 social science CBSE
Which of the following leaders was not present in the class 10 social science CBSE
Garampani Sanctuary is located at A Diphu Assam B Gangtok class 10 social science CBSE
Which one of the following places is not covered by class 10 social science CBSE
Trending doubts
Which are the Top 10 Largest Countries of the World?
The states of India which do not have an International class 10 social science CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
How do you graph the function fx 4x class 9 maths CBSE
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE
Difference Between Plant Cell and Animal Cell
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Why is there a time difference of about 5 hours between class 10 social science CBSE
Name the three parallel ranges of the Himalayas Describe class 9 social science CBSE