JEE MAIN - Mathematics (2021 - 17th March Morning Shift - No. 10)
The system of equations kx + y + z = 1, x + ky + z = k and x + y + zk = k2 has no solution if k is equal to :
0
$$-$$1
$$-$$2
1
Explanation
$$D = \left| {\matrix{
k & 1 & 1 \cr
1 & k & 1 \cr
1 & 1 & k \cr
} } \right| = 0$$
$$ \Rightarrow k({k^2} - 1) - (k - 1) + (1 - k) = 0$$
$$ \Rightarrow (k - 1)({k^2} + k - 1 - 1) = 0$$
$$ \Rightarrow (k - 1)({k^2} + k - 2) = 0$$
$$ \Rightarrow (k - 1)(k - 1)(k + 2) = 0$$
$$ \Rightarrow k = 1,k = -2$$
for k = 1 equation identical so k = $$-$$2 for no solution.
$$ \Rightarrow k({k^2} - 1) - (k - 1) + (1 - k) = 0$$
$$ \Rightarrow (k - 1)({k^2} + k - 1 - 1) = 0$$
$$ \Rightarrow (k - 1)({k^2} + k - 2) = 0$$
$$ \Rightarrow (k - 1)(k - 1)(k + 2) = 0$$
$$ \Rightarrow k = 1,k = -2$$
for k = 1 equation identical so k = $$-$$2 for no solution.
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