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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