JEE MAIN - Mathematics (2021 - 27th August Evening Shift - No. 6)
Let [$$\lambda$$] be the greatest integer less than or equal to $$\lambda$$. The set of all values of $$\lambda$$ for which the system of linear equations
x + y + z = 4,
3x + 2y + 5z = 3,
9x + 4y + (28 + [$$\lambda$$])z = [$$\lambda$$] has a solution is :
x + y + z = 4,
3x + 2y + 5z = 3,
9x + 4y + (28 + [$$\lambda$$])z = [$$\lambda$$] has a solution is :
R
($$-$$$$\infty$$, $$-$$9) $$\cup$$ ($$-$$9, $$\infty$$)
[$$-$$9, $$-$$8)
($$-$$$$\infty$$, $$-$$9) $$\cup$$ [$$-$$8, $$\infty$$)
Explanation
$$D = \left| {\matrix{
1 & 1 & 1 \cr
3 & 2 & 5 \cr
9 & 4 & {28 + [\lambda ]} \cr
} } \right| = - 24 - [\lambda ] + 15 = - [\lambda ] - 9$$
if $$[\lambda ] + 9 \ne 0$$ then unique solution
if $$[\lambda ] + 9 = 0$$ then D1 = D2 = D3 = 0
so infinite solutions
Hence, $$\lambda$$ can be any red number.
if $$[\lambda ] + 9 \ne 0$$ then unique solution
if $$[\lambda ] + 9 = 0$$ then D1 = D2 = D3 = 0
so infinite solutions
Hence, $$\lambda$$ can be any red number.
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