3 and 4 .Determinants and Matrices
hard

સમીકરણની સંહતિ $x + y + z = 6$, $x + 2y + 3z = 10,x + 2y + \lambda z = \mu $ નો એકપણ ઉકેલ શક્ય ન હોય તો . . .

A

$\lambda \ne 3,\mu = 10$

B

$\lambda = 3,\mu \ne 10$

C

$\lambda \ne 3,\mu \ne 10$

D

એકપણ નહી.

Solution

(b) The value of the determinant will vanish if $\lambda = 3$ and ${\Delta _1} \ne 0$, $\therefore \,\,\mu \ne 10$.

Standard 12
Mathematics

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