Q76 / 125 · BITSAT 2020 · Mathematics

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

  • Correct answer
    A
    -1
  • B
    0
  • C
    1
  • D
    2

Solution

Since the system is homogeneous and has a non-trivial solution $0=\left|\begin{array}{ccc}p+a & b & c \\ a & q+b & c \\ a & b & r+c\end{array}\right|=\left|\begin{array}{ccc}p & -q & 0 \\ 0 & q & -r \\ a & b & r+c\end{array}\right|$ using $\mathrm{R}_{1} \rightarrow \mathrm{R}_{1}-\mathrm{R}_{2}$ and $\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{3}$ Expanding along $\mathrm{C}$, we get $\Rightarrow \mathrm{p}[\mathrm{q}(\mathrm{r}+\mathrm{c})+\mathrm{br}]+\mathrm{aqr}=0$ $\Rightarrow \mathrm{pqr}+\mathrm{pqc}+\mathrm{prb}+\mathrm{qra}=0$ $\Rightarrow \frac{a}{p}+\frac{b}{q}+\frac{c}{r}=-1$

Hence, the correct answer is option: 'A'.

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