Matrices and Determinants 2 Question 37
41. If $a \neq p, b \neq q, c \neq r$ and $\begin{vmatrix}p & b & c \\ a & q & c \\ a & b & r\end{vmatrix}$
Then, find the value of $\frac{p}{p-a}+\frac{q}{q-b}+\frac{r}{r-c}$.
$(1991,4 M)$
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Answer:
Correct Answer: 41. (2)
Solution:
- Let $\Delta=\begin{vmatrix}p & b & c \\ a & q & c \\ a & b & r\end{vmatrix}$
Applying $R_{1} \rightarrow R_{2}-R_{1}$ and $R_{3} \rightarrow R_{3}-R_{1}$, we get
$ \begin{aligned} \Delta & =\begin{vmatrix} p & b & c \\ a-p & q-b & 0 \\ a-p & 0 & r-c \end{vmatrix} \\ & =c\begin{vmatrix} a-p & q-b \\ a-p & 0 \end{vmatrix}+(r-c)\begin{vmatrix} p & b \\ a-p & q-b \end{vmatrix} \\ & =-c(a-p)(q-b)+(r-c)[p(q-b)-b(a-p)] \\ & =-c(a-p)(q-b)+p(r-c)(q-b)-b(r-c)(a-p) \end{aligned} $
Since, $\Delta=0$
$ \begin{aligned} & \Rightarrow-c(a-p)(q-b)+p(r-c)(q-b)-b(r-c)(a-p)=0 \\ & \Rightarrow \quad \frac{c}{r-c}+\frac{p}{p-a}+\frac{b}{q-b}=0 \end{aligned} $
[on dividing both sides by $(a-p)(q-b)(r-c)$ ]
$ \begin{aligned} \Rightarrow & \frac{p}{p-a}+\frac{b}{q-b}+1+\frac{c}{r-c}+1 & =2 \\ \Rightarrow & \frac{p}{p-a}+\frac{q}{q-b}+\frac{r}{r-c} =2 \end{aligned} $
