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

Correct answer: A. -1 Since the system is homogeneous and has a non-trivial solution 0=| array ccc p+a & b & c a & q+b & c a & b & r+c array |=| array ccc p & -q & 0 0 & q & -r a & b & r+c array | using R_ 1 arrow R_ 1 -R_ 2 and R_ 2 arrow R_ 2 -R_ 3 Expanding along C , we get p[q(r+c)+br]+aqr=0 pqr+pqc+prb+qra=0 (a)/(p)+(b)/(q)+(c)/(r)=-1

BITSAT 2020

Exam: BITSAT
Year: 2020
Subject: Mathematics
Chapter: Determinants

BITSAT 2020 · Mathematics · Determinants

Suppose p, q, r ≠ 0 and system of equation (p+a) x+by+cz=0a x+(q+b)...

Suppose

\mathrm{p}, \mathrm{q}, \mathrm{r} \neq 0

and system of equation

(\mathrm{p}+\mathrm{a}) \mathrm{x}+\mathrm{by}+\mathrm{cz}=0a x+(q+b) y+c z=0a x+b y+(r+c) z=0

has a non-trivial solution, then value of

\frac{a}{p}+\frac{b}{q}+\frac{c}{r}

Options

Correct answer

A. -1

Step-by-step 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

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