If the system of linear equations

2x + 2y + 3z = a

3x – y + 5z = b

x – 3y + 2z = c

Where a, b, c are non-zero real numbers, has more than one solution, then:

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JEE Mains Previous Paper 1 (Held On: 11 Jan 2019 Shift 1)
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  1. b – c + a = 0
  2. b – c – a = 0
  3. a + b + c = 0
  4. b + c – a = 0

Answer (Detailed Solution Below)

Option 2 : b – c – a = 0
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JEE Main 04 April 2024 Shift 1
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Detailed Solution

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If the system of equations has more than one solution, then D = 0 and D1 = D2 = D3 = 0.

From the given set of equation, we form,

D1 = 0

\(\Rightarrow \left| \begin{matrix} a & 2 & 3 \\ b & -1 & 5 \\ c & -3 & 2 \\ \end{matrix} \right|=0\)

Now,

⇒ a(-2 + 15) – 2(2b – 5c) + 3(-3b + c) = 0

⇒ 13a – 4b + 10c – 9b + 3c = 0

⇒ 13a – 13b + 13c = 0

⇒ a – b + c = 0

∴ b – a – c = 0
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