Let a,b,c∈R^(+) and the system of equations (1-a)x+y+z=0,x+(1-b)y+z=0 and x+y+(1-c)z=0 has infinitely many solutions, the minimum value of abc is

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MARKS Content Team · Accepted Answer

Correct answer: (C) 27. We have, 1-ax+y+z=0 x+1-by+z=0 x+y+1-cz=0 has many solutions ∴ 1-a 1 1 1 1-b 1 1 1 1-c =0 ⇒ ab+ac+bc=abc ∴ (ab+bc+ac)/3≥a2b2c2^(1/3)⇒abc^(1/3)≥3 ∴ Minimum of abc=3^3=27

Let $a, b, c \in R^{+}$ and the system of equations $(1 - a) x + y + z = 0, x + (1 - b) y + z = 0$ and $x + y + (1 - c) z = 0$ has infinitely many solutions, the minimum value of $a b c$ is

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Relevant Questions

Let a,b,c∈R+ and the system of equations (1-a)x+y+z=0,x+(1-b)y+z=0 and x+y+(1-c)z=0 has infinitely many solutions, the minimum value of abc is

Solution

Relevant Questions

Let $a, b, c \in R^{+}$ and the system of equations $(1 - a) x + y + z = 0, x + (1 - b) y + z = 0$ and $x + y + (1 - c) z = 0$ has infinitely many solutions, the minimum value of $a b c$ is