Question

If ${{a}_{1}}$ , ${{a}_{2}}$ , ${{a}_{3}}$ , ${{b}_{1}}$ , ${{b}_{2}}$ , ${{b}_{3}}$ $\in $ R and are such that ${{a}_{i}}{{b}_{j}}\ne 0$ for 1 $\le $ i,j $\le $ 3, then
$\left| \begin{matrix}
   \dfrac{1-{{a}_{1}}^{3}{{b}_{1}}^{3}}{1-{{a}_{1}}{{b}_{1}}} & \dfrac{1-{{a}_{1}}^{3}{{b}_{2}}^{3}}{1-{{a}_{1}}{{b}_{2}}} & \dfrac{1-{{a}_{1}}^{3}{{b}_{3}}^{3}}{1-{{a}_{1}}{{b}_{3}}} \\
   \dfrac{1-{{a}_{2}}^{3}{{b}_{1}}^{3}}{1-{{a}_{2}}{{b}_{1}}} & \dfrac{1-{{a}_{2}}^{3}{{b}_{2}}^{3}}{1-{{a}_{2}}{{b}_{2}}} & \dfrac{1-{{a}_{2}}^{3}{{b}_{3}}^{3}}{1-{{a}_{2}}{{b}_{3}}} \\
   \dfrac{1-{{a}_{3}}^{3}{{b}_{1}}^{3}}{1-{{a}_{3}}{{b}_{1}}} & \dfrac{1-{{a}_{3}}^{3}{{b}_{2}}^{3}}{1-{{a}_{3}}{{b}_{2}}} & \dfrac{1-{{a}_{3}}^{3}{{b}_{3}}^{3}}{1-{{a}_{3}}{{b}_{3}}} \\
\end{matrix} \right|$ > 0 Provided either ${{a}_{1}}$ < ${{a}_{2}}$ < ${{a}_{3}}$ and ${{b}_{1}}$ < ${{b}_{2}}$ < ${{b}_{3}}$ or ${{a}_{1}}$ > ${{a}_{2}}$${{a}_{3}}$ and ${{b}_{1}}>{{b}_{2}}>{{b}_{3}}$ then show that $\left( {{a}_{1}}-{{a}_{2}} \right)\left( {{a}_{2}}-{{a}_{3}} \right)\left( {{a}_{3}}-{{a}_{1}} \right)\left( {{b}_{1}}-{{b}_{2}} \right)\left( {{b}_{2}}-{{b}_{3}} \right)\left( {{b}_{3}}-{{b}_{1}} \right)<0$

Answer 1

Hint: We know that $\dfrac{{{a}^{3}}-{{b}^{3}}}{a-b}$ is equal to ${{a}^{2}}+{{b}^{2}}+ab$ where a is not equal to b. it is given that ${{a}_{i}}{{b}_{j}}\ne 0$ for 1 $\le $ i,j $\le $ 3 so we can write $\dfrac{1-{{a}_{1}}^{3}{{b}_{1}}^{3}}{1-{{a}_{1}}{{b}_{1}}}=1+{{a}_{1}}^{2}{{b}_{1}}^{2}+{{a}_{1}}{{b}_{1}}$ . 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 $\left| \begin{matrix}
   \dfrac{1-{{a}_{1}}^{3}{{b}_{1}}^{3}}{1-{{a}_{1}}{{b}_{1}}} & \dfrac{1-{{a}_{1}}^{3}{{b}_{2}}^{3}}{1-{{a}_{1}}{{b}_{2}}} & \dfrac{1-{{a}_{1}}^{3}{{b}_{3}}^{3}}{1-{{a}_{1}}{{b}_{3}}} \\
   \dfrac{1-{{a}_{2}}^{3}{{b}_{1}}^{3}}{1-{{a}_{2}}{{b}_{1}}} & \dfrac{1-{{a}_{2}}^{3}{{b}_{2}}^{3}}{1-{{a}_{2}}{{b}_{2}}} & \dfrac{1-{{a}_{2}}^{3}{{b}_{3}}^{3}}{1-{{a}_{2}}{{b}_{3}}} \\
   \dfrac{1-{{a}_{3}}^{3}{{b}_{1}}^{3}}{1-{{a}_{3}}{{b}_{1}}} & \dfrac{1-{{a}_{3}}^{3}{{b}_{2}}^{3}}{1-{{a}_{3}}{{b}_{2}}} & \dfrac{1-{{a}_{3}}^{3}{{b}_{3}}^{3}}{1-{{a}_{3}}{{b}_{3}}} \\
\end{matrix} \right|$
We know that $\dfrac{{{a}^{3}}-{{b}^{3}}}{a-b}$ is equal to ${{a}^{2}}+{{b}^{2}}+ab$ where a is not equal to b
So we can write $\left| \begin{matrix}
   \dfrac{1-{{a}_{1}}^{3}{{b}_{1}}^{3}}{1-{{a}_{1}}{{b}_{1}}} & \dfrac{1-{{a}_{1}}^{3}{{b}_{2}}^{3}}{1-{{a}_{1}}{{b}_{2}}} & \dfrac{1-{{a}_{1}}^{3}{{b}_{3}}^{3}}{1-{{a}_{1}}{{b}_{3}}} \\
   \dfrac{1-{{a}_{2}}^{3}{{b}_{1}}^{3}}{1-{{a}_{2}}{{b}_{1}}} & \dfrac{1-{{a}_{2}}^{3}{{b}_{2}}^{3}}{1-{{a}_{2}}{{b}_{2}}} & \dfrac{1-{{a}_{2}}^{3}{{b}_{3}}^{3}}{1-{{a}_{2}}{{b}_{3}}} \\
   \dfrac{1-{{a}_{3}}^{3}{{b}_{1}}^{3}}{1-{{a}_{3}}{{b}_{1}}} & \dfrac{1-{{a}_{3}}^{3}{{b}_{2}}^{3}}{1-{{a}_{3}}{{b}_{2}}} & \dfrac{1-{{a}_{3}}^{3}{{b}_{3}}^{3}}{1-{{a}_{3}}{{b}_{3}}} \\
\end{matrix} \right|$ as $\left| \begin{matrix}
   1+{{a}_{1}}{{b}_{1}}+{{a}_{1}}^{2}{{b}_{1}}^{2} & 1+{{a}_{1}}{{b}_{2}}+{{a}_{1}}^{2}{{b}_{2}}^{2} & 1+{{a}_{1}}{{b}_{3}}+{{a}_{1}}^{2}{{b}_{3}}^{2} \\
   1+{{a}_{2}}{{b}_{1}}+{{a}_{2}}^{2}{{b}_{1}}^{2} & 1+{{a}_{2}}{{b}_{2}}+{{a}_{2}}^{2}{{b}_{2}}^{2} & 1+{{a}_{2}}{{b}_{3}}+{{a}_{2}}^{2}{{b}_{3}}^{2} \\
   1+{{a}_{3}}{{b}_{1}}+{{a}_{3}}^{2}{{b}_{1}}^{2} & 1+{{a}_{3}}{{b}_{2}}+{{a}_{3}}^{2}{{b}_{2}}^{2} & 1+{{a}_{3}}{{b}_{3}}+{{a}_{3}}^{2}{{b}_{3}}^{2} \\
\end{matrix} \right|$
If we observe we can see the above matrix is the product of 2 matrix.
We can write
$\Rightarrow \left| \begin{matrix}
   1+{{a}_{1}}{{b}_{1}}+{{a}_{1}}^{2}{{b}_{1}}^{2} & 1+{{a}_{1}}{{b}_{2}}+{{a}_{1}}^{2}{{b}_{2}}^{2} & 1+{{a}_{1}}{{b}_{3}}+{{a}_{1}}^{2}{{b}_{3}}^{2} \\
   1+{{a}_{2}}{{b}_{1}}+{{a}_{2}}^{2}{{b}_{1}}^{2} & 1+{{a}_{2}}{{b}_{2}}+{{a}_{2}}^{2}{{b}_{2}}^{2} & 1+{{a}_{2}}{{b}_{3}}+{{a}_{2}}^{2}{{b}_{3}}^{2} \\
   1+{{a}_{3}}{{b}_{1}}+{{a}_{3}}^{2}{{b}_{1}}^{2} & 1+{{a}_{3}}{{b}_{2}}+{{a}_{3}}^{2}{{b}_{2}}^{2} & 1+{{a}_{3}}{{b}_{3}}+{{a}_{3}}^{2}{{b}_{3}}^{2} \\
\end{matrix} \right|=\left| \begin{matrix}
   1 & {{a}_{1}} & {{a}_{1}}^{2} \\
   1 & {{a}_{2}} & {{a}_{2}}^{2} \\
   1 & {{a}_{3}} & {{a}_{3}}^{2} \\
\end{matrix} \right|\left| \begin{matrix}
   1 & 1 & 1 \\
   {{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\
   {{b}_{1}}^{2} & {{b}_{2}}^{2} & {{b}_{3}}^{2} \\
\end{matrix} \right|$
Now we can perform row and column operation
$\Rightarrow \left| \begin{matrix}
   1+{{a}_{1}}{{b}_{1}}+{{a}_{1}}^{2}{{b}_{1}}^{2} & 1+{{a}_{1}}{{b}_{2}}+{{a}_{1}}^{2}{{b}_{2}}^{2} & 1+{{a}_{1}}{{b}_{3}}+{{a}_{1}}^{2}{{b}_{3}}^{2} \\
   1+{{a}_{2}}{{b}_{1}}+{{a}_{2}}^{2}{{b}_{1}}^{2} & 1+{{a}_{2}}{{b}_{2}}+{{a}_{2}}^{2}{{b}_{2}}^{2} & 1+{{a}_{2}}{{b}_{3}}+{{a}_{2}}^{2}{{b}_{3}}^{2} \\
   1+{{a}_{3}}{{b}_{1}}+{{a}_{3}}^{2}{{b}_{1}}^{2} & 1+{{a}_{3}}{{b}_{2}}+{{a}_{3}}^{2}{{b}_{2}}^{2} & 1+{{a}_{3}}{{b}_{3}}+{{a}_{3}}^{2}{{b}_{3}}^{2} \\
\end{matrix} \right|=\left| \begin{matrix}
   1 & {{a}_{1}} & {{a}_{1}}^{2} \\
   0 & {{a}_{2}}-{{a}_{1}} & {{a}_{2}}^{2}-{{a}_{1}}^{2} \\
   0 & {{a}_{3}}-{{a}_{1}} & {{a}_{3}}^{2}-{{a}_{1}}^{2} \\
\end{matrix} \right|\left| \begin{matrix}
   1 & 0 & 0 \\
   {{b}_{1}} & {{b}_{2}}-{{b}_{1}} & {{b}_{3}}-{{b}_{1}} \\
   {{b}_{1}}^{2} & {{b}_{2}}^{2}-{{b}_{1}}^{2} & {{b}_{3}}^{2}-{{b}_{1}}^{2} \\
\end{matrix} \right|$
Now evaluating the determinant, we get
$\begin{align}
  & \Rightarrow \left| \begin{matrix}
   1+{{a}_{1}}{{b}_{1}}+{{a}_{1}}^{2}{{b}_{1}}^{2} & 1+{{a}_{1}}{{b}_{2}}+{{a}_{1}}^{2}{{b}_{2}}^{2} & 1+{{a}_{1}}{{b}_{3}}+{{a}_{1}}^{2}{{b}_{3}}^{2} \\
   1+{{a}_{2}}{{b}_{1}}+{{a}_{2}}^{2}{{b}_{1}}^{2} & 1+{{a}_{2}}{{b}_{2}}+{{a}_{2}}^{2}{{b}_{2}}^{2} & 1+{{a}_{2}}{{b}_{3}}+{{a}_{2}}^{2}{{b}_{3}}^{2} \\
   1+{{a}_{3}}{{b}_{1}}+{{a}_{3}}^{2}{{b}_{1}}^{2} & 1+{{a}_{3}}{{b}_{2}}+{{a}_{3}}^{2}{{b}_{2}}^{2} & 1+{{a}_{3}}{{b}_{3}}+{{a}_{3}}^{2}{{b}_{3}}^{2} \\
\end{matrix} \right|= \\
 & \left[ \left( {{a}_{2}}-{{a}_{1}} \right)\left( {{a}_{3}}^{2}-{{a}_{1}}^{2} \right)-\left( {{a}_{3}}-{{a}_{1}} \right)\left( {{a}_{2}}^{2}-{{a}_{1}}^{2} \right) \right]\left[ \left( {{b}_{2}}-{{b}_{1}} \right)\left( {{b}_{3}}^{2}-{{b}_{1}}^{2} \right)-\left( {{b}_{3}}-{{b}_{1}} \right)\left( {{b}_{2}}^{2}-{{b}_{1}}^{2} \right) \right] \\
\end{align}$
 Further solving it we get
$\begin{align}
  & \Rightarrow \left| \begin{matrix}
   1+{{a}_{1}}{{b}_{1}}+{{a}_{1}}^{2}{{b}_{1}}^{2} & 1+{{a}_{1}}{{b}_{2}}+{{a}_{1}}^{2}{{b}_{2}}^{2} & 1+{{a}_{1}}{{b}_{3}}+{{a}_{1}}^{2}{{b}_{3}}^{2} \\
   1+{{a}_{2}}{{b}_{1}}+{{a}_{2}}^{2}{{b}_{1}}^{2} & 1+{{a}_{2}}{{b}_{2}}+{{a}_{2}}^{2}{{b}_{2}}^{2} & 1+{{a}_{2}}{{b}_{3}}+{{a}_{2}}^{2}{{b}_{3}}^{2} \\
   1+{{a}_{3}}{{b}_{1}}+{{a}_{3}}^{2}{{b}_{1}}^{2} & 1+{{a}_{3}}{{b}_{2}}+{{a}_{3}}^{2}{{b}_{2}}^{2} & 1+{{a}_{3}}{{b}_{3}}+{{a}_{3}}^{2}{{b}_{3}}^{2} \\
\end{matrix} \right|=\left( {{a}_{2}}-{{a}_{1}} \right)\left( {{a}_{3}}-{{a}_{2}} \right)\left( {{a}_{3}}-{{a}_{1}} \right)\left( {{b}_{2}}-{{b}_{1}} \right)\left( {{b}_{3}}-{{b}_{1}} \right)\left( {{b}_{3}}-{{b}_{2}} \right) \\
 & \\
\end{align}$
$\begin{align}
  & \Rightarrow \left| \begin{matrix}
   1+{{a}_{1}}{{b}_{1}}+{{a}_{1}}^{2}{{b}_{1}}^{2} & 1+{{a}_{1}}{{b}_{2}}+{{a}_{1}}^{2}{{b}_{2}}^{2} & 1+{{a}_{1}}{{b}_{3}}+{{a}_{1}}^{2}{{b}_{3}}^{2} \\
   1+{{a}_{2}}{{b}_{1}}+{{a}_{2}}^{2}{{b}_{1}}^{2} & 1+{{a}_{2}}{{b}_{2}}+{{a}_{2}}^{2}{{b}_{2}}^{2} & 1+{{a}_{2}}{{b}_{3}}+{{a}_{2}}^{2}{{b}_{3}}^{2} \\
   1+{{a}_{3}}{{b}_{1}}+{{a}_{3}}^{2}{{b}_{1}}^{2} & 1+{{a}_{3}}{{b}_{2}}+{{a}_{3}}^{2}{{b}_{2}}^{2} & 1+{{a}_{3}}{{b}_{3}}+{{a}_{3}}^{2}{{b}_{3}}^{2} \\
\end{matrix} \right|=\left( {{a}_{1}}-{{a}_{2}} \right)\left( {{a}_{2}}-{{a}_{3}} \right)\left( {{a}_{3}}-{{a}_{1}} \right)\left( {{b}_{1}}-{{b}_{2}} \right)\left( {{b}_{2}}-{{b}_{3}} \right)\left( {{b}_{3}}-{{b}_{1}} \right) \\
 & \\
\end{align}$so if $\left| \begin{matrix}
   \dfrac{1-{{a}_{1}}^{3}{{b}_{1}}^{3}}{1-{{a}_{1}}{{b}_{1}}} & \dfrac{1-{{a}_{1}}^{3}{{b}_{2}}^{3}}{1-{{a}_{1}}{{b}_{2}}} & \dfrac{1-{{a}_{1}}^{3}{{b}_{3}}^{3}}{1-{{a}_{1}}{{b}_{3}}} \\
   \dfrac{1-{{a}_{2}}^{3}{{b}_{1}}^{3}}{1-{{a}_{2}}{{b}_{1}}} & \dfrac{1-{{a}_{2}}^{3}{{b}_{2}}^{3}}{1-{{a}_{2}}{{b}_{2}}} & \dfrac{1-{{a}_{2}}^{3}{{b}_{3}}^{3}}{1-{{a}_{2}}{{b}_{3}}} \\
   \dfrac{1-{{a}_{3}}^{3}{{b}_{1}}^{3}}{1-{{a}_{3}}{{b}_{1}}} & \dfrac{1-{{a}_{3}}^{3}{{b}_{2}}^{3}}{1-{{a}_{3}}{{b}_{2}}} & \dfrac{1-{{a}_{3}}^{3}{{b}_{3}}^{3}}{1-{{a}_{3}}{{b}_{3}}} \\
\end{matrix} \right|$ > 0 then $\left( {{a}_{1}}-{{a}_{2}} \right)\left( {{a}_{2}}-{{a}_{3}} \right)\left( {{a}_{3}}-{{a}_{1}} \right)\left( {{b}_{1}}-{{b}_{2}} \right)\left( {{b}_{2}}-{{b}_{3}} \right)\left( {{b}_{3}}-{{b}_{1}} \right)$ > 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.