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 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)

$-1$

Let $\Delta =\left|\begin{array}{ccc}p+a& b& c\\ a& q+b& c\\ a& b& r+c\end{array}\right|$

Applying $R_2\to R_2-R_1,R_3\to R_3-R_1$

$\Delta =\left|\begin{array}{ccc}p+a& b& c\\ -p& q& 0\\ -p& 0& r\end{array}\right|$

Expanding along with $C_3$

$c(+pq)+r(q(p+a)+bp)=0$

$\Rightarrow cqp+bpr+arq=-pqr$

Divided both sides by $pqr$

$\Rightarrow \frac{a}{p}+\frac{b}{q}+\frac{c}{r}=-1$