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If $p \neq a, q \neq b, r \neq c$ and the system of equations
$\begin{aligned}
& p x+a y+a z=0 \\
& b x+q y+b z=0 \\
& c x+c y+r z=0
\end{aligned}$
has a non-trivial solution, then the value of
$\frac{p}{p-a}+\frac{q}{q-b}+\frac{r}{r-c}$ is
Options:
$\begin{aligned}
& p x+a y+a z=0 \\
& b x+q y+b z=0 \\
& c x+c y+r z=0
\end{aligned}$
has a non-trivial solution, then the value of
$\frac{p}{p-a}+\frac{q}{q-b}+\frac{r}{r-c}$ is
Solution:
1989 Upvotes
Verified Answer
The correct answer is:
$2$
$2$
As the given system of equations has a non-trivial solution.
$\Delta=\left|\begin{array}{lll}p & a & a \\ b & q & b \\ c & c & r\end{array}\right|=0$
Applying $C_2 \rightarrow C_2-C_1$ and $C_3 \rightarrow C_3-C_1$
$\Delta=\left|\begin{array}{ccc}p & a-p & a-p \\ b & q-b & 0 \\ c & 0 & r-c\end{array}\right|=0$
Expanding along $C_3$, we get
$(a-p)\left|\begin{array}{cc}b & q-b \\ c & 0\end{array}\right|+(r-c)\left|\begin{array}{cc}p & a-p \\ b & q-b\end{array}\right|=0$
$\begin{aligned}
& \Rightarrow(a-p)(-c)(q-b)+(r-c) \\
& \{p(q-b)-b(a-p)\}=0 \\
& \Rightarrow \\
& (p-a)(q-b) c+p(r-c)(q-b)+b(r-c)(p-a)=0
\end{aligned}$
Dividing by $(p-a)(q-b)(r-c)$, we get
$\frac{c}{r-c}+\frac{p}{p-a}+\frac{b}{q-b}=0$
$\Rightarrow \frac{p}{p-a}+\frac{q}{q-b}+\frac{r}{r-c}=\frac{q-b}{q-b}+\frac{r-c}{r-c}=2$
$\Delta=\left|\begin{array}{lll}p & a & a \\ b & q & b \\ c & c & r\end{array}\right|=0$
Applying $C_2 \rightarrow C_2-C_1$ and $C_3 \rightarrow C_3-C_1$
$\Delta=\left|\begin{array}{ccc}p & a-p & a-p \\ b & q-b & 0 \\ c & 0 & r-c\end{array}\right|=0$
Expanding along $C_3$, we get
$(a-p)\left|\begin{array}{cc}b & q-b \\ c & 0\end{array}\right|+(r-c)\left|\begin{array}{cc}p & a-p \\ b & q-b\end{array}\right|=0$
$\begin{aligned}
& \Rightarrow(a-p)(-c)(q-b)+(r-c) \\
& \{p(q-b)-b(a-p)\}=0 \\
& \Rightarrow \\
& (p-a)(q-b) c+p(r-c)(q-b)+b(r-c)(p-a)=0
\end{aligned}$
Dividing by $(p-a)(q-b)(r-c)$, we get
$\frac{c}{r-c}+\frac{p}{p-a}+\frac{b}{q-b}=0$
$\Rightarrow \frac{p}{p-a}+\frac{q}{q-b}+\frac{r}{r-c}=\frac{q-b}{q-b}+\frac{r-c}{r-c}=2$
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