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Statement

If a1, a2, ...

Question

If $a_{1}, a_{2}, a_{3}, b_{1}, b_{2}, b_{3} \in R$ and are such that $a_{i} b_{j} \neq 1$ for $1 \leq i, j \leq 3$ , then

$∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} \frac{1 - a_{1}^{3} b_{1}^{3}}{1 - a_{1} b_{1}} & \frac{1 - a_{1}^{3} b_{2}^{3}}{1 - a_{1} b_{2}} & \frac{1 - a_{1}^{3} b_{3}^{3}}{1 - a_{1} b_{3}} \frac{1 - a_{2}^{3} b_{1}^{3}}{1 - a_{2} b_{1}} & \frac{1 - a_{2}^{3} b_{2}^{3}}{1 - a_{2} b_{2}} & \frac{1 - a_{2}^{3} b_{3}^{3}}{1 - a_{2} b_{3}} \frac{1 - a_{3}^{3} b_{1}^{3}}{1 - a_{3} b_{1}} & \frac{1 - a_{3}^{3} b_{2}^{3}}{1 - a_{3} b_{2}} & \frac{1 - a_{3}^{3} b_{3}^{3}}{1 - a_{3} b_{3}} \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣$ > 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 $(a_{1} - a_{2}) (a_{2} - a_{3}) (a_{3} - a_{1}) (b_{1} - b_{2}) (b_{2} - b_{3}) (b_{3} - b_{1}) < 0$ ,

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Solution

Since $\frac{x^{3} - y^{3}}{x - y} = \frac{(x - y) (x^{2} + x y + y^{2})}{(x - y)} = x^{2} + x y + y^{2}$
Hence the given determinant becomes

$∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 + a_{1} b_{1} + a_{1}^{2} b_{1}^{2} & 1 + a_{1} b_{2} + a_{1}^{2} b_{2}^{2} & 1 + a_{1} b_{3} + a_{1}^{2} b_{3}^{2} 1 + a_{2} b_{1} + a_{2}^{2} b_{1}^{2} & 1 + a_{2} b_{2} + a_{2}^{2} b_{2}^{2} & 1 + a_{2} b_{3} + a_{2}^{2} b_{3}^{2} 1 + a_{3} b_{1} + a_{3}^{2} b_{1}^{2} & 1 + a_{3} b_{2} + a_{3}^{2} b_{2}^{2} & 1 + a_{3} b_{3} + a_{3}^{2} b_{3}^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣ > 0$

$\Rightarrow ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & a_{1} & a_{1}^{2} 1 & a_{2} & a_{2}^{2} 1 & a_{3} & a_{3}^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣ \times ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & b_{1} & b_{1}^{2} 1 & b_{2} & b_{2}^{2} 1 & b_{3} & b_{3}^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣ > 0$

$\Rightarrow (a_{1} - a_{2}) (a_{2} - a_{3}) (a_{3} - a_{1}) (b_{1} - b_{2}) (b_{2} - b_{3}) (b_{3} - b_{1}) > 0$
$⎧ ⎪ ⎨ ⎪ ⎩ ∵ ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & a & a^{2} 1 & b & b^{2} 1 & c & c^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣ = (a - b) (b - c) (c - a) ⎫ ⎪ ⎬ ⎪ ⎭$
Case I
If $a_{1} < a_{2} < a_{3}$ and $b_{1} < b_{2} < b_{3}$
then $(a_{1} - a_{2}) < 0, (a_{2} - a_{3}) < 0$ and $(b_{1} - b_{2}) < 0, (b_{2} - b_{3}) < 0$
$(a_{1} - a_{3}) < 0$ $(b_{1} - b_{3}) < 0$
$∴ (a_{3} - a_{1}) > 0$ $∴ b_{3} - b_{1} > 0$
then $(a_{1} - a_{2}) (a_{2} - a_{3}) (a_{3} - a_{1}) > 0$ and
$(b_{1} - b_{2}) (b_{2} - b_{3}) (b_{3} - b_{1}) > 0$
$∴ (a_{1} - a_{2}) (a_{2} - a_{3}) (a_{3} - a_{1}) (b_{1} - b_{2}) (b_{2} - b_{3}) (b_{3} - b_{1}) > 0$
which is true.
Case II
If $a_{1} ‘ > a_{2} > a_{3}$ and $b_{1} > b_{2} > b_{3}$
$∴ a_{1} - a_{2} > 0, a_{2} - a_{3} > 0$ and $b_{1} - b_{2} > 0, b_{2} - b_{3} > 0$
$a_{1} - a_{3} > 0 \Rightarrow a_{3} - a_{1} < 0$ $b_{1} - b_{3} > 0 \Rightarrow b_{3} - b_{1} < 0$
Hence $(a_{1} - a_{2}) (a_{2} - a_{3}) (a_{3} - a_{1}) < 0$ and $(b_{1} - b_{2}) (b_{2} - b_{3}) (b_{3} - b_{1}) < 0$
$∴ (a_{1} - a_{2}) (a_{2} - a_{3}) (a_{3} - a_{1}) (b_{1} - b_{2}) (b_{2} - b_{3}) (b_{3} - b_{1}) > 0$

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Similar questions

a_{1} > a_{2} > a_{3}

b_{1} > b_{2} > b_{3}

a_{i} b_{j} \neq 1

1 \leq i

j \leq 3

, then the determinant

Δ = ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} \frac{1 - a_{1}^{3} b_{1}^{3}}{1 - a_{1} b_{1}} & \frac{1 - a_{1}^{3} b_{2}^{3}}{1 - a_{1} b_{2}} & \frac{1 - a_{1}^{3} b_{3}^{3}}{1 - a_{1} b_{3}} \frac{1 - a_{2}^{3} b_{1}^{3}}{1 - a_{2} b_{1}} & \frac{1 - a_{2}^{3} b_{2}^{3}}{1 - a_{2} b_{2}} & \frac{1 - a_{2}^{3} b_{3}^{3}}{1 - a_{2} b_{3}} \frac{1 - a_{3}^{3} b_{1}^{3}}{1 - a_{1} b_{1}} & \frac{1 - a_{3}^{3} b_{2}^{3}}{1 - a_{3} b_{2}} & \frac{1 - a_{3}^{3} b_{3}^{3}}{1 - a_{3} b_{3}} \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣

is

2 y - x = 8; 5 y - x = 14, y - 2 x = 1

(a_{1}, b_{1}), (a_{2}, b_{2}), (a_{3}, b_{3})

(a_{1} + a_{2} + a_{3} + b_{1} + b_{2} + b_{3})

Q. Assertion :Suppose four distinct positive numbers

a_{1}, a_{2}, a_{3}, a_{4}

b_{1} = a_{1}, b_{2} = b_{1} + a_{2}, b_{3} = b_{2} + a_{3}

b_{4} = b_{3} + a_{4} .

STATEMENT- 1: The numbers

b_{1}, b_{2}, b_{3}, b_{4}

are neither in AP nor in GP, and Reason: STATEMENT- 1: The numbers

b_{1}, b_{2}, b_{3}, b_{4}

Q. Suppose four distinct positive numbers

a_{1}, a_{2}, a_{3}, a_{4}

G . P

b_{1} = a_{1}, b_{2} = b_{1} + a_{2}, b_{3} = b_{2} + a_{3}

b_{4} = b_{3} + a_{4}

.
STATEMENT-1 : The numbers

b_{1}, b_{2}, b_{3}, b_{4}

A . P

G . P

.
STATEMENT-2 : The numbers

b_{1}, b_{2}, b_{3}, b_{4}

H . P

.

Q. Suppose four distinct positive number

a_{1}, a_{2}, a_{3}, a_{4}

are in G.P. Let

b_{1} = a_{1}, b_{2} = b_{1} + a_{2}, a_{3} = b_{2} + a_{3}

b_{4} = b_{3} + a_{4}

Statement 1 - The numbers

b_{1}, b_{2}, b_{3}, b_{4}

are neither in A.P. nor in G.P.
Statement 2 - The numaber

b_{1}, b_{2}, b_{3}, b_{4}

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