Determinants Matrices Question 65
Question: If $ a\ne p, $
$ b\ne q, $
$ c\ne r $ and $ \begin{vmatrix} p & b & c \\ a & q & c \\ a & b & r \\ \end{vmatrix}=0 $ then the value of $ \frac{p}{p-a}+\frac{q}{q-b}+\frac{r}{r-c} $ is equal to
Options:
A) $ -1 $
B) $ 1 $
C) $ -2 $
D) $ 2 $
Show Answer
Answer:
Correct Answer: D
Solution:
- [d] $ Given \begin{vmatrix} p & b & c \\ a & q & c \\ a & b & r \\ \end{vmatrix}=0 $
$ R_1\to R_1-R_2,R_2\to R_2-R_3 $ reduces the determinant to $ \begin{vmatrix} p-a & b-q & 0 \\ 0 & q-b & c-r \\ a & b & r \\ \end{vmatrix}=0 $
$ \Rightarrow (p-a)(q-b)r+a(b-q)(c-r)-b(p-a)(c-r)=0 $
$ \Rightarrow $ Dividing throughout by $ (p-a)(q-b)(r-c), $ we get
$ \Rightarrow \frac{r}{r-c}+\frac{a}{p-a}+\frac{b}{q-b}=0 $
$ \Rightarrow \frac{r}{r-c}+\frac{a}{p-a}+\frac{b}{q-b}=2 $
BETA
