If system of linear equations x + y + z = 5 ; x = 2y + 2z = 6 ; x + 3y + λ z = u

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3 and 4 .Determinants and Matrices

hard

If the system of linear equations $x + y + z = 5$ ; $x = 2y + 2z = 6$ ; $x + 3y + \lambda z = u (\lambda \, \mu \in R)$, has infinitely many solutions then the value of $\lambda + \mu $ is

A

$12$

B

$7$

C

$10$

D

$9$

(JEE MAIN-2019)

Solution

$x + 3y + \lambda z – u = a\left( {x + y + z – 5} \right) + b\left( {x + 2y + 2z – 6} \right)$

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$

Standard 12

Mathematics

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