3 and 4 .Determinants and Matrices
hard

જો સમીકરણો $x +y + z = 6$ ; $x + 2y + 3z= 10$ ; $x + 2y + \lambda z = 0$ એ એકાકી ઉકેલ ધરાવે છે તો  $\lambda $ ની કિમંત  . . .   શક્ય નથી.

A

$1$

B

$0$

C

$2$

D

$3$

(AIEEE-2012)

Solution

Given system of equations is 

$x + y + z = 6$

$x + 2y + 3z = 10$

$x + 2y + \lambda z = 0$

It has unique solution.

$\therefore \left| {\begin{array}{*{20}{c}}
1&1&1\\
1&2&3\\
1&2&\lambda 
\end{array}} \right| \ne 0$

$ \Rightarrow 1\left( {2\lambda  – 6} \right) – 1\left( {\lambda  – 3} \right) + 1\left( {2 – 2} \right) \ne 0$

$ \Rightarrow 2\lambda  – 6 – \lambda  + 3 \ne 0 \Rightarrow \lambda  – 3 \ne 0 \Rightarrow \lambda  \ne 3$

Standard 12
Mathematics

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