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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