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Cramer's Rule

1 + a1 + b1 ...

Question

$∣ ∣ ∣ ∣ \begin{matrix} 1 + a_{1} + b_{1} & a_{1} + b_{2} & a_{1} + b_{3} a_{2} + b_{1} & 1 + a_{2} + b_{2} & a_{2} + b_{3} a_{3} + b_{1} & a_{3} + b_{2} & 1 + a_{3} + b_{3} \end{matrix} ∣ ∣ ∣ ∣$ . Evaluate the determinant.

A

1 + 3 \sum i = 1 (a_{i} + b_{i}) + \sum \sum 1 \leq i < j \leq 3 (a_{i} - a_{j}) (b_{j} - b_{i})

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B

1 + 3 \sum i = 1 (a_{i} + b_{i}) + \sum \sum 1 \leq i < j \leq 3 (a_{i} + a_{j}) (b_{j} + b_{i})

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C

1 + 3 \sum i = 1 (a_{i} - b_{i}) + \sum \sum 1 \leq i < j \leq 3 (a_{i} - a_{j}) (b_{j} - b_{i})

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D

None of these.

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Solution

The correct option is A $1 + 3 \sum i = 1 (a_{i} + b_{i}) + \sum \sum 1 \leq i < j \leq 3 (a_{i} - a_{j}) (b_{j} - b_{i})$
Let $Δ = ∣ ∣ ∣ ∣ \begin{matrix} 1 + a_{1} + b_{1} & a_{1} + b_{2} & a_{1} + b_{3} a_{2} + b_{1} & 1 + a_{2} + b_{2} & a_{2} + b_{3} a_{3} + b_{1} & a_{3} + b_{2} & 1 + a_{3} + b_{3} \end{matrix} ∣ ∣ ∣ ∣$
$= ∣ ∣ ∣ ∣ \begin{matrix} 1 & a_{1} + b_{2} & a_{1} + b_{3} 0 & 1 + a_{2} + b_{2} & a_{2} + b_{3} 0 & a_{3} + b_{2} & 1 + a_{3} + b_{3} \end{matrix} ∣ ∣ ∣ ∣ + ∣ ∣ ∣ ∣ \begin{matrix} a_{1} & a_{1} + b_{2} & a_{1} + b_{3} a_{2} & 1 + a_{2} + b_{2} & a_{2} + b_{3} a_{3} & a_{3} + b_{2} & 1 + a_{3} + b_{3} \end{matrix} ∣ ∣ ∣ ∣ + ∣ ∣ ∣ ∣ \begin{matrix} b_{1} & a_{1} + b_{2} & a_{1} + b_{3} b_{1} & 1 + a_{2} + b_{2} & a_{2} + b_{3} b_{1} & a_{3} + b_{2} & 1 + a_{3} + b_{3} \end{matrix} ∣ ∣ ∣ ∣$
$Δ = Δ_{1} + Δ_{2} + Δ_{3}$ say (1)
Expand $Δ_{1}$ with respect to $C_{1}$ , we get
$Δ_{1} = 1 + a_{2} + a_{3} + b_{2} + b_{3} + (a_{2} - a_{3}) (b_{3} - b_{2})$
and
$Δ_{2} = ∣ ∣ ∣ ∣ \begin{matrix} a_{1} & a_{1} + b_{2} & a_{1} + b_{3} a_{2} & 1 + a_{2} + b_{2} & a_{2} + b_{3} a_{3} & a_{3} + b_{2} & 1 + a_{3} + b_{3} \end{matrix} ∣ ∣ ∣ ∣$
Applying $C_{2} \to C_{2} - C_{1}; C_{3} \to C_{3} - C_{1}$
$= ∣ ∣ ∣ ∣ \begin{matrix} a_{1} & b_{2} & b_{3} a_{2} & 1 + b_{2} & b_{3} a_{3} & b_{2} & 1 + b_{3} \end{matrix} ∣ ∣ ∣ ∣$
Applying $R_{2} \to R) 2 - R_{1}; R_{3} \to R_{3} - R_{1}$
$= ∣ ∣ ∣ ∣ \begin{matrix} a_{1} & b_{2} & b_{3} a_{2} - a_{1} & 1 & 0 a_{3} - a_{1} & 0 & 1 \end{matrix} ∣ ∣ ∣ ∣$
$∴ Δ_{2} = a_{1} + b_{2} (a_{1} - a_{2}) + b_{3} (a_{1} - a_{3})$
$Δ_{3} = ∣ ∣ ∣ ∣ \begin{matrix} b_{1} & a_{1} + b_{2} & a_{1} + b_{3} b_{1} & 1 + a_{2} + b_{2} & a_{2} + b_{3} b_{1} & a_{3} + b_{2} & 1 + a_{3} + b_{3} \end{matrix} ∣ ∣ ∣ ∣$
Applying $R_{2} \to R_{2} - R_{1}; R_{3} \to R_{3} - R_{1}$
$∣ ∣ ∣ ∣ \begin{matrix} b_{1} & a_{1} + b_{+} 2 & a_{1} + b_{3} 0 & 1 + a_{2} - a_{1} & a_{2} - a_{1} 0 & a_{3} - a_{1} & 1 + a_{3} - a_{1} \end{matrix} ∣ ∣ ∣ ∣$
$Δ_{3} = b_{1} - b_{1} (a_{1} - a_{2}) - b_{1} (a_{1} - a_{3})$
Therefore from (1), we have
$Δ = Δ_{1} + Δ_{2} + Δ_{3}$
$= 1 + 3 \sum i = 1 (a_{i} + b_{i}) + \sum \sum 1 \leq i < j \leq 3 (a_{i} - a_{j}) (b_{j} - b_{i})$

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