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Question: If \({{a}_{1}}\) , \({{a}_{2}}\) , \({{a}_{3}}\) , \({{b}_{1}}\) , \({{b}_{2}}\) , \({{b}_{3}}\) \(\...

If a1{{a}_{1}} , a2{{a}_{2}} , a3{{a}_{3}} , b1{{b}_{1}} , b2{{b}_{2}} , b3{{b}_{3}} ∈\in R and are such that aibj≠0{{a}_{i}}{{b}_{j}}\ne 0 for 1 ≤\le i,j ≤\le 3, then
∣1−a13b131−a1b11−a13b231−a1b21−a13b331−a1b3 1−a23b131−a2b11−a23b231−a2b21−a23b331−a2b3 1−a33b131−a3b11−a33b231−a3b21−a33b331−a3b3 ∣\left| \begin{matrix} \dfrac{1-{{a}_{1}}^{3}{{b}_{1}}^{3}}{1-{{a}_{1}}{{b}_{1}}} & \dfrac{1-{{a}_{1}}^{3}{{b}_{2}}^{3}}{1-{{a}_{1}}{{b}_{2}}} & \dfrac{1-{{a}_{1}}^{3}{{b}_{3}}^{3}}{1-{{a}_{1}}{{b}_{3}}} \\\ \dfrac{1-{{a}_{2}}^{3}{{b}_{1}}^{3}}{1-{{a}_{2}}{{b}_{1}}} & \dfrac{1-{{a}_{2}}^{3}{{b}_{2}}^{3}}{1-{{a}_{2}}{{b}_{2}}} & \dfrac{1-{{a}_{2}}^{3}{{b}_{3}}^{3}}{1-{{a}_{2}}{{b}_{3}}} \\\ \dfrac{1-{{a}_{3}}^{3}{{b}_{1}}^{3}}{1-{{a}_{3}}{{b}_{1}}} & \dfrac{1-{{a}_{3}}^{3}{{b}_{2}}^{3}}{1-{{a}_{3}}{{b}_{2}}} & \dfrac{1-{{a}_{3}}^{3}{{b}_{3}}^{3}}{1-{{a}_{3}}{{b}_{3}}} \\\ \end{matrix} \right| > 0 Provided either a1{{a}_{1}} < a2{{a}_{2}} < a3{{a}_{3}} and b1{{b}_{1}} < b2{{b}_{2}} < b3{{b}_{3}} or a1{{a}_{1}} > a2{{a}_{2}} a3{{a}_{3}} and b1>b2>b3{{b}_{1}}>{{b}_{2}}>{{b}_{3}} then show that (a1−a2)(a2−a3)(a3−a1)(b1−b2)(b2−b3)(b3−b1)<0\left( {{a}_{1}}-{{a}_{2}} \right)\left( {{a}_{2}}-{{a}_{3}} \right)\left( {{a}_{3}}-{{a}_{1}} \right)\left( {{b}_{1}}-{{b}_{2}} \right)\left( {{b}_{2}}-{{b}_{3}} \right)\left( {{b}_{3}}-{{b}_{1}} \right)<0

Explanation

Solution

We know that a3−b3a−b\dfrac{{{a}^{3}}-{{b}^{3}}}{a-b} is equal to a2+b2+ab{{a}^{2}}+{{b}^{2}}+ab where a is not equal to b. it is given that aibj≠0{{a}_{i}}{{b}_{j}}\ne 0 for 1 ≤\le i,j ≤\le 3 so we can write 1−a13b131−a1b1=1+a12b12+a1b1\dfrac{1-{{a}_{1}}^{3}{{b}_{1}}^{3}}{1-{{a}_{1}}{{b}_{1}}}=1+{{a}_{1}}^{2}{{b}_{1}}^{2}+{{a}_{1}}{{b}_{1}} . Then we can see the matrix is a product of 2 matrices. We can find the 2 matrices and find their determinant to solve the equation.

Complete step by step solution:
The given determinant is ∣1−a13b131−a1b11−a13b231−a1b21−a13b331−a1b3 1−a23b131−a2b11−a23b231−a2b21−a23b331−a2b3 1−a33b131−a3b11−a33b231−a3b21−a33b331−a3b3 ∣\left| \begin{matrix} \dfrac{1-{{a}_{1}}^{3}{{b}_{1}}^{3}}{1-{{a}_{1}}{{b}_{1}}} & \dfrac{1-{{a}_{1}}^{3}{{b}_{2}}^{3}}{1-{{a}_{1}}{{b}_{2}}} & \dfrac{1-{{a}_{1}}^{3}{{b}_{3}}^{3}}{1-{{a}_{1}}{{b}_{3}}} \\\ \dfrac{1-{{a}_{2}}^{3}{{b}_{1}}^{3}}{1-{{a}_{2}}{{b}_{1}}} & \dfrac{1-{{a}_{2}}^{3}{{b}_{2}}^{3}}{1-{{a}_{2}}{{b}_{2}}} & \dfrac{1-{{a}_{2}}^{3}{{b}_{3}}^{3}}{1-{{a}_{2}}{{b}_{3}}} \\\ \dfrac{1-{{a}_{3}}^{3}{{b}_{1}}^{3}}{1-{{a}_{3}}{{b}_{1}}} & \dfrac{1-{{a}_{3}}^{3}{{b}_{2}}^{3}}{1-{{a}_{3}}{{b}_{2}}} & \dfrac{1-{{a}_{3}}^{3}{{b}_{3}}^{3}}{1-{{a}_{3}}{{b}_{3}}} \\\ \end{matrix} \right|
We know that a3−b3a−b\dfrac{{{a}^{3}}-{{b}^{3}}}{a-b} is equal to a2+b2+ab{{a}^{2}}+{{b}^{2}}+ab where a is not equal to b
So we can write ∣1−a13b131−a1b11−a13b231−a1b21−a13b331−a1b3 1−a23b131−a2b11−a23b231−a2b21−a23b331−a2b3 1−a33b131−a3b11−a33b231−a3b21−a33b331−a3b3 ∣\left| \begin{matrix} \dfrac{1-{{a}_{1}}^{3}{{b}_{1}}^{3}}{1-{{a}_{1}}{{b}_{1}}} & \dfrac{1-{{a}_{1}}^{3}{{b}_{2}}^{3}}{1-{{a}_{1}}{{b}_{2}}} & \dfrac{1-{{a}_{1}}^{3}{{b}_{3}}^{3}}{1-{{a}_{1}}{{b}_{3}}} \\\ \dfrac{1-{{a}_{2}}^{3}{{b}_{1}}^{3}}{1-{{a}_{2}}{{b}_{1}}} & \dfrac{1-{{a}_{2}}^{3}{{b}_{2}}^{3}}{1-{{a}_{2}}{{b}_{2}}} & \dfrac{1-{{a}_{2}}^{3}{{b}_{3}}^{3}}{1-{{a}_{2}}{{b}_{3}}} \\\ \dfrac{1-{{a}_{3}}^{3}{{b}_{1}}^{3}}{1-{{a}_{3}}{{b}_{1}}} & \dfrac{1-{{a}_{3}}^{3}{{b}_{2}}^{3}}{1-{{a}_{3}}{{b}_{2}}} & \dfrac{1-{{a}_{3}}^{3}{{b}_{3}}^{3}}{1-{{a}_{3}}{{b}_{3}}} \\\ \end{matrix} \right| as ∣1+a1b1+a12b121+a1b2+a12b221+a1b3+a12b32 1+a2b1+a22b121+a2b2+a22b221+a2b3+a22b32 1+a3b1+a32b121+a3b2+a32b221+a3b3+a32b32 ∣\left| \begin{matrix} 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{matrix} \right|
If we observe we can see the above matrix is the product of 2 matrix.
We can write
⇒∣1+a1b1+a12b121+a1b2+a12b221+a1b3+a12b32 1+a2b1+a22b121+a2b2+a22b221+a2b3+a22b32 1+a3b1+a32b121+a3b2+a32b221+a3b3+a32b32 ∣=∣1a1a12 1a2a22 1a3a32 ∣∣111 b1b2b3 b12b22b32 ∣\Rightarrow \left| \begin{matrix} 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{matrix} \right|=\left| \begin{matrix} 1 & {{a}_{1}} & {{a}_{1}}^{2} \\\ 1 & {{a}_{2}} & {{a}_{2}}^{2} \\\ 1 & {{a}_{3}} & {{a}_{3}}^{2} \\\ \end{matrix} \right|\left| \begin{matrix} 1 & 1 & 1 \\\ {{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\\ {{b}_{1}}^{2} & {{b}_{2}}^{2} & {{b}_{3}}^{2} \\\ \end{matrix} \right|
Now we can perform row and column operation
⇒∣1+a1b1+a12b121+a1b2+a12b221+a1b3+a12b32 1+a2b1+a22b121+a2b2+a22b221+a2b3+a22b32 1+a3b1+a32b121+a3b2+a32b221+a3b3+a32b32 ∣=∣1a1a12 0a2−a1a22−a12 0a3−a1a32−a12 ∣∣100 b1b2−b1b3−b1 b12b22−b12b32−b12 ∣\Rightarrow \left| \begin{matrix} 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{matrix} \right|=\left| \begin{matrix} 1 & {{a}_{1}} & {{a}_{1}}^{2} \\\ 0 & {{a}_{2}}-{{a}_{1}} & {{a}_{2}}^{2}-{{a}_{1}}^{2} \\\ 0 & {{a}_{3}}-{{a}_{1}} & {{a}_{3}}^{2}-{{a}_{1}}^{2} \\\ \end{matrix} \right|\left| \begin{matrix} 1 & 0 & 0 \\\ {{b}_{1}} & {{b}_{2}}-{{b}_{1}} & {{b}_{3}}-{{b}_{1}} \\\ {{b}_{1}}^{2} & {{b}_{2}}^{2}-{{b}_{1}}^{2} & {{b}_{3}}^{2}-{{b}_{1}}^{2} \\\ \end{matrix} \right|
Now evaluating the determinant, we get
⇒∣1+a1b1+a12b121+a1b2+a12b221+a1b3+a12b32 1+a2b1+a22b121+a2b2+a22b221+a2b3+a22b32 1+a3b1+a32b121+a3b2+a32b221+a3b3+a32b32 ∣= [(a2−a1)(a32−a12)−(a3−a1)(a22−a12)][(b2−b1)(b32−b12)−(b3−b1)(b22−b12)] \begin{aligned} & \Rightarrow \left| \begin{matrix} 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{matrix} \right|= \\\ & \left[ \left( {{a}_{2}}-{{a}_{1}} \right)\left( {{a}_{3}}^{2}-{{a}_{1}}^{2} \right)-\left( {{a}_{3}}-{{a}_{1}} \right)\left( {{a}_{2}}^{2}-{{a}_{1}}^{2} \right) \right]\left[ \left( {{b}_{2}}-{{b}_{1}} \right)\left( {{b}_{3}}^{2}-{{b}_{1}}^{2} \right)-\left( {{b}_{3}}-{{b}_{1}} \right)\left( {{b}_{2}}^{2}-{{b}_{1}}^{2} \right) \right] \\\ \end{aligned}
Further solving it we get
⇒∣1+a1b1+a12b121+a1b2+a12b221+a1b3+a12b32 1+a2b1+a22b121+a2b2+a22b221+a2b3+a22b32 1+a3b1+a32b121+a3b2+a32b221+a3b3+a32b32 ∣=(a2−a1)(a3−a2)(a3−a1)(b2−b1)(b3−b1)(b3−b2)  \begin{aligned} & \Rightarrow \left| \begin{matrix} 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{matrix} \right|=\left( {{a}_{2}}-{{a}_{1}} \right)\left( {{a}_{3}}-{{a}_{2}} \right)\left( {{a}_{3}}-{{a}_{1}} \right)\left( {{b}_{2}}-{{b}_{1}} \right)\left( {{b}_{3}}-{{b}_{1}} \right)\left( {{b}_{3}}-{{b}_{2}} \right) \\\ & \\\ \end{aligned}
⇒∣1+a1b1+a12b121+a1b2+a12b221+a1b3+a12b32 1+a2b1+a22b121+a2b2+a22b221+a2b3+a22b32 1+a3b1+a32b121+a3b2+a32b221+a3b3+a32b32 ∣=(a1−a2)(a2−a3)(a3−a1)(b1−b2)(b2−b3)(b3−b1)  \begin{aligned} & \Rightarrow \left| \begin{matrix} 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{matrix} \right|=\left( {{a}_{1}}-{{a}_{2}} \right)\left( {{a}_{2}}-{{a}_{3}} \right)\left( {{a}_{3}}-{{a}_{1}} \right)\left( {{b}_{1}}-{{b}_{2}} \right)\left( {{b}_{2}}-{{b}_{3}} \right)\left( {{b}_{3}}-{{b}_{1}} \right) \\\ & \\\ \end{aligned}so if ∣1−a13b131−a1b11−a13b231−a1b21−a13b331−a1b3 1−a23b131−a2b11−a23b231−a2b21−a23b331−a2b3 1−a33b131−a3b11−a33b231−a3b21−a33b331−a3b3 ∣\left| \begin{matrix} \dfrac{1-{{a}_{1}}^{3}{{b}_{1}}^{3}}{1-{{a}_{1}}{{b}_{1}}} & \dfrac{1-{{a}_{1}}^{3}{{b}_{2}}^{3}}{1-{{a}_{1}}{{b}_{2}}} & \dfrac{1-{{a}_{1}}^{3}{{b}_{3}}^{3}}{1-{{a}_{1}}{{b}_{3}}} \\\ \dfrac{1-{{a}_{2}}^{3}{{b}_{1}}^{3}}{1-{{a}_{2}}{{b}_{1}}} & \dfrac{1-{{a}_{2}}^{3}{{b}_{2}}^{3}}{1-{{a}_{2}}{{b}_{2}}} & \dfrac{1-{{a}_{2}}^{3}{{b}_{3}}^{3}}{1-{{a}_{2}}{{b}_{3}}} \\\ \dfrac{1-{{a}_{3}}^{3}{{b}_{1}}^{3}}{1-{{a}_{3}}{{b}_{1}}} & \dfrac{1-{{a}_{3}}^{3}{{b}_{2}}^{3}}{1-{{a}_{3}}{{b}_{2}}} & \dfrac{1-{{a}_{3}}^{3}{{b}_{3}}^{3}}{1-{{a}_{3}}{{b}_{3}}} \\\ \end{matrix} \right| > 0 then (a1−a2)(a2−a3)(a3−a1)(b1−b2)(b2−b3)(b3−b1)\left( {{a}_{1}}-{{a}_{2}} \right)\left( {{a}_{2}}-{{a}_{3}} \right)\left( {{a}_{3}}-{{a}_{1}} \right)\left( {{b}_{1}}-{{b}_{2}} \right)\left( {{b}_{2}}-{{b}_{3}} \right)\left( {{b}_{3}}-{{b}_{1}} \right) > 0

Note: If product of matrix A and B is equal to matrix c, then product of determinant of A and B is equal to determinant of C. The determinant of matrix A is equal to determinant of transpose of A. If the determinant of A is equal to 0, Then we can not find the inverse of the matrix A. The determinant of the inverse of matrix A is equal to the reciprocal of determinant of matrix A.