Q64 / 100 · BITSAT 2024 (Memory Based Paper 3) · Mathematics

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

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

Solution

Let $|A|=\left|\begin{array}{ccc}p+a & b & c \\ a & q+b & c \\ a & b & r+c\end{array}\right|$ Also, given that equations has non trivial solution $|A|=0$ $\begin{aligned} & \Rightarrow \quad\left|\begin{array}{ccc} p+a & b & c \\ a & q+b & c \\ a & b & r+c \end{array}\right|=0 \\ & \Rightarrow(p+a)[(q+b)(r+c)-b c]-b[a(r+c)-c a] \\ & +c[a b-a(q+b)]=0 \\ & \Rightarrow \quad(p+a)[q r+q c+b r]-b[a r]+c[-a q]=0 \\ & \Rightarrow \quad p q r+p q c+p b r+a q r=0 \end{aligned}$ Dividing whole equation by $p q r$ $\begin{aligned} \Rightarrow & \frac{p q r}{p q r}+\frac{p q c}{p q r}+\frac{p r b}{p q r}+\frac{q r a}{p q r} & =0 \\ \Rightarrow & 1+\frac{c}{r}+\frac{b}{q}+\frac{a}{p} & =0 \\ \Rightarrow & \frac{a}{p}+\frac{b}{q}+\frac{c}{r} & =-1 \end{aligned}$

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

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