If the system of equation $x - 2 y + z = a, 2 x + y - 2 z = b$ and $x + 3 y - 3 z = c$ have atleast one solution, then the relationship between a, b and c is

a + b + c = 0

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a – b + c = 0

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–a + b + c = 0

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a + b – c = 0

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Solution

The correct option is B
a – b + c = 0

$D = ∣ ∣ ∣ ∣ \begin{matrix} 1 & - 2 & 1 2 & 1 & - 2 a & - 2 & - 3 \end{matrix} ∣ ∣ ∣ ∣$ which vanishes; hence for atleast one solution $D_{1} = D_{1} = D_{3} = 0$
$D_{1} = ∣ ∣ ∣ ∣ \begin{matrix} a & - 2 & 1 b & 1 & - 2 c & 3 & - 3 \end{matrix} ∣ ∣ ∣ ∣ = 0 \Rightarrow a - b + c = 0$

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Similar questions

x - 2 y + z = a

2 x + y - 2 z = b

x + 3 y - 3 z = c

a, b

c

- 2 x + y + z = a, x - 2 y + z = b, x + y - 2 z = c

2 x + 2 y + 3 z = a

3 x - y + 5 z = b

x - 3 y + 2 z = c

a, b, c

\frac{a}{x - 2 y + 3 z} = \frac{b}{y - 2 z + 3 x} = \frac{c}{z - 2 x + 3 y}

x + y + z \neq 0

\frac{x - a + d}{α - δ} = \frac{y - a}{α} = \frac{z - a - d}{α + δ}

\frac{x - b + c}{β - γ} = \frac{y - b}{β} = \frac{z - b - c}{β + γ}

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