Visual SolutionPYQ 2024 · Apr 8 Shift 2Standard
Question
If αa,βb,γc\alpha \neq a, \beta \neq b, \gamma \neq c and αbcaβcabγ=0\begin{vmatrix} \alpha & b & c \\ a & \beta & c \\ a & b & \gamma \end{vmatrix} = 0, then aαa+bβb+γγc\frac{a}{\alpha - a} + \frac{b}{\beta - b} + \frac{\gamma}{\gamma - c} is equal to:
(A)33
(B)00
(C)11
(D)22
Solution Path
Row operations and expansion yield aαa+bβb+γγc=0\dfrac{a}{\alpha - a} + \dfrac{b}{\beta - b} + \dfrac{\gamma}{\gamma - c} = 0.
01Question Setup
1/4
Given determinant =0= 0 with αa,βb,γc\alpha \neq a, \beta \neq b, \gamma \neq c. Find aαa+bβb+γγc\dfrac{a}{\alpha - a} + \dfrac{b}{\beta - b} + \dfrac{\gamma}{\gamma - c}.
Expression=  ?\text{Expression} = \;?
02Row Operations
2/4
Apply R1R1R2R_1 \to R_1 - R_2 and R2R2R3R_2 \to R_2 - R_3 to simplify the determinant.
Simplified determinant
03SimplifyKEY INSIGHT
3/4
Expand and divide by (αa)(βb)(γc)(\alpha - a)(\beta - b)(\gamma - c) to get the expression equals 00.
Expression=0\text{Expression} = 0
04Final Answer
4/4
aαa+bβb+γγc=0\dfrac{a}{\alpha - a} + \dfrac{b}{\beta - b} + \dfrac{\gamma}{\gamma - c} = 0.
0\boxed{0}