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3 and 4 .Determinants and Matrices
hard
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.
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