Question

If $pqr\ne 0$ and the system of equations $(p+a)x+by+cz=0,$ $ax+(q+b)y+cz=0,$ $ax,by+(r+c)z=0$ has 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)

$-1$

$(p+a)x+by+cz=0,ax+(q+b)y+cz=0,ax+by+(r+c)z=0\left[\begin{array}{ccc}p+a& b& c\\ a& q+b& c\\ a& b& r+c\end{array}\right]=0\Rightarrow (p+a)((q+b)(r+c)-bc)-b(a(r+c)-ac)+c(ab-(q+b)a)=0\Rightarrow (p+a)(eq+qc+rb+bc+-bc)-b(ar+ac-ac)+c(ab-qa-ab)=0\Rightarrow (p+a)(rq+qc+rb)-abr-acq=0\Rightarrow aqr+bpr+cpq+cpq+pqr=0\Rightarrow aqr+bpr+cpq=-pqr$
$\Rightarrow \frac{a}{p}+\frac{b}{q}+\frac{c}{r}=-1$