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JEE Mains · Maths · STD 12 - 3 and 4 . metrices and determinant

જો રેખીય સમીકરણો \(x + y + z = 5\) ; \(x = 2y + 2z = 6\) ; \(x + 3y + \lambda z = u (\lambda \, \mu \in R)\) અનંત ઉકેલ ધરાવે છે તો  \(\lambda  + \mu \) ની કિમંત મેળવો.

  1. A \(12\)
  2. B \(7\)
  3. C \(10\)
  4. D \(9\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(10\)

Step-by-step Solution

Detailed explanation

\(x + 3y + \lambda z - u = a\left( {x + y + z - 5} \right) + b\left( {x + 2y + 2z - 6} \right)\) Comparing coefficients we get \(a+b=1\) and \(a+2b=3\) \((a,b)=(-1,2)\) So, \(x + 3y + \lambda z - u = x + 3y + 3z - \lambda \) \( \Rightarrow u = 7,\lambda = 3\)
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