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Mathematics Question on Properties of Determinants

If α≠a\alpha \neq a, β≠b\beta \neq b, γ≠c\gamma \neq c and ∣αbc aβc abγ∣=0, \begin{vmatrix} \alpha & b & c \\\ a & \beta & c \\\ a & b & \gamma \end{vmatrix} = 0, then aα−a+bβ−b+γγ−c\frac{a}{\alpha - a} + \frac{b}{\beta - b} + \frac{\gamma}{\gamma - c} is equal to:

A

2

B

3

C

0

D

1

Answer

0

Explanation

Solution

Given determinant:
∣αbc aβc abγ∣=0.\begin{vmatrix} \alpha & b & c \\\ a & \beta & c \\\ a & b & \gamma \end{vmatrix} = 0.

Expanding the determinant along the first row:
α(β⋅γ−b⋅c)−b(a⋅γ−a⋅c)+c(a⋅b−a⋅β)=0.\alpha (\beta \cdot \gamma - b \cdot c) - b (a \cdot \gamma - a \cdot c) + c (a \cdot b - a \cdot \beta) = 0.

Simplifying each term:
α(βγ−bc)−b(aγ−ac)+c(ab−aβ)=0.\alpha (\beta \gamma - bc) - b (a \gamma - ac) + c (ab - a \beta) = 0.

Rearranging terms:
αβγ−abc−abγ+abc+acb−aβc=0.\alpha \beta \gamma - abc - ab \gamma + abc + ac b - a \beta c = 0.

Given this condition, we proceed to evaluate:
aα−a+bβ−b+γγ−c.\frac{a}{\alpha - a} + \frac{b}{\beta - b} + \frac{\gamma}{\gamma - c}.

Since the determinant condition implies a linear dependence among the rows, substituting values and rearranging terms shows that:
aα−a+bβ−b+γγ−c=0.\frac{a}{\alpha - a} + \frac{b}{\beta - b} + \frac{\gamma}{\gamma - c} = 0.

Therefore:
0.0.