JEE MAIN - Mathematics (2019 - 8th April Morning Slot - No. 17)
The greatest value of c $$ \in $$ R for which the system
of linear equations
x – cy – cz = 0
cx – y + cz = 0
cx + cy – z = 0
has a non-trivial solution, is :
x – cy – cz = 0
cx – y + cz = 0
cx + cy – z = 0
has a non-trivial solution, is :
-1
0
1/2
2
Explanation
If the system of equations has non-trivial
solutions, then
D = 0
$$\left| {\matrix{ 1 & { - c} & { - c} \cr c & { - 1} & c \cr c & c & { - 1} \cr } } \right| = 0$$
$$ \Rightarrow $$ (1 - c2) + c(-c - c2) - c(c2 + c) = 0
$$ \Rightarrow $$ (1 + c)(1 - c - 2c2) = 0
$$ \Rightarrow $$ (1 + c)2 (1 - 2c) = 0
$$ \Rightarrow $$ c = -1 or $${1 \over 2}$$
$$ \therefore $$ Greatest value of c is $${1 \over 2}$$.
D = 0
$$\left| {\matrix{ 1 & { - c} & { - c} \cr c & { - 1} & c \cr c & c & { - 1} \cr } } \right| = 0$$
$$ \Rightarrow $$ (1 - c2) + c(-c - c2) - c(c2 + c) = 0
$$ \Rightarrow $$ (1 + c)(1 - c - 2c2) = 0
$$ \Rightarrow $$ (1 + c)2 (1 - 2c) = 0
$$ \Rightarrow $$ c = -1 or $${1 \over 2}$$
$$ \therefore $$ Greatest value of c is $${1 \over 2}$$.
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