Suppose p, q, r ≠ 0 and system of equation gathered (p+a) x+b y+c z=0 a x+(q+b) y+c z=0 gathered a x+b y+(r+c) z=0 , has a non-trivial solution, then the value of (a)/(p)+(b)/(q)+(c)/(r) is

Correct answer: A. -1 Let |A|=| array ccc p+a & b & c a & q+b & c a & b & r+c array | Also, given that equations has non trivial solution |A|=0 aligned & | array ccc p+a & b & c a & q+b & c a & b & r+c array |=0 & (p+a)[(q+b)(r+c)-b c]-b[a(r+c)-c a] & +c[a b-a(q+b)]=0 & (p+a)[q r+q c+b r]-b[a r]+c[-a q]=0 & p q r+p q c+p b r+a q r=0 aligned Dividing whole equation by p q r aligned & (p q r)/(p q r)+(p q c)/(p q r)+(p r b)/(p q r)+(q r a)/(p q r) & =0 & 1+(c)/(r)+(b)/(q)+(a)/(p) & =0 & (a)/(p)+(b)/(q)+(c)/(r) & =-1 aligned

BITSAT 2024

Exam: BITSAT
Year: 2024
Subject: Mathematics
Chapter: Determinants

BITSAT 2024 · Mathematics · Determinants

Suppose p, q, r ≠ 0 and system of equation gathered (p+a) x+b y+c...

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}

Options

Correct answer

A. -1

Step-by-step 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}

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