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

Question

If pqr ≠ 0 and the system of equations (p+a) x+b y+c z=0 a x+(q+b) y+c z=0 a x+by+(r+c) z=0 has a non-trivial solution, then value of (a/p)+(b/q)+(c/r) is (A) -1 (B) 0 (C) 1 (D) 2

Tardigrade Admin · Accepted Answer

Δ=|p+a&b&c a&q+b&c a&b&r+c|=0 Applying R1 arrow R2-R1 and R3 arrow R3-R1, we get |p+a&b&c -p&q&0 -p&0&r|=0 ⇒ p q c+[q(p+a)+b p] r=0 Divide by p q r to obtain (a/p)+(b/q)+(c/r)=-1

Q. If $pq r \neq = 0$ and the system of equations
$(p + a) x + b y + cz = 0$
$a x + (q + b) y + cz = 0$
$a x + b y + (r + c) z = 0$ has a non-trivial solution, then value of $\frac{a}{p} + \frac{b}{q} + \frac{c}{r}$ is

-1

0

1

2

$Δ = ∣ ∣ p + a a a b q + b b c c r + c ∣ ∣ = 0$
Applying $R_{1} \to R_{2} - R_{1}$ and $R_{3} \to R_{3} - R_{1},$ we get
$∣ ∣ p + a - p - p b q 0 c 0 r ∣ ∣ = 0$
$\Rightarrow pq c + [q (p + a) + b p] r = 0$
Divide by $pq r$ to obtain $\frac{a}{p} + \frac{b}{q} + \frac{c}{r} = - 1$

Q. If pqr=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 pa​+qb​+rc​ is

-1

0

1

2

Δ=∣∣​p+aaa​bq+bb​ccr+c​∣∣​=0 Applying R1​→R2​−R1​ and R3​→R3​−R1​, we get ∣∣​p+a−p−p​bq0​c0r​∣∣​=0 ⇒pqc+[q(p+a)+bp]r=0 Divide by pqr to obtain pa​+qb​+rc​=−1

Q. If $pq r \neq = 0$ and the system of equations
$(p + a) x + b y + cz = 0$
$a x + (q + b) y + cz = 0$
$a x + b y + (r + c) z = 0$ has a non-trivial solution, then value of $\frac{a}{p} + \frac{b}{q} + \frac{c}{r}$ is

$Δ = ∣ ∣ p + a a a b q + b b c c r + c ∣ ∣ = 0$
Applying $R_{1} \to R_{2} - R_{1}$ and $R_{3} \to R_{3} - R_{1},$ we get
$∣ ∣ p + a - p - p b q 0 c 0 r ∣ ∣ = 0$
$\Rightarrow pq c + [q (p + a) + b p] r = 0$
Divide by $pq r$ to obtain $\frac{a}{p} + \frac{b}{q} + \frac{c}{r} = - 1$