Question

If $pqr \neq 0$ and the system of equations
$(p+a) x+b y+c z=0 $
$a x+(q+b) y+c z=0 $
$a x+by+(r+c) z=0$ has a non-trivial solution, then value of $\frac{a}{p}+\frac{b}{q}+\frac{c}{r}$ is

• A. -1
• B. 0
• C. 1
• D. 2

Answer 1

Answer: (A)

$\Delta=\begin{vmatrix}p+a&b&c\\ a&q+b&c\\ a&b&r+c\end{vmatrix}=0 $
Applying $R_{1} \rightarrow R_{2}-R_{1}$ and $R_{3} \rightarrow R_{3}-R_{1},$ we get
$\begin{vmatrix}p+a&b&c\\ -p&q&0\\ -p&0&r\end{vmatrix}=0 $
$\Rightarrow p q c+[q(p+a)+b p] r=0$
Divide by $p q r$ to obtain $\frac{a}{p}+\frac{b}{q}+\frac{c}{r}=-1$