JEE MAIN - Mathematics (2019 - 10th April Evening Slot - No. 21)
Let $$\lambda $$ be a real number for which the system of linear equations x + y + z = 6, 4x + $$\lambda $$y – $$\lambda $$z = $$\lambda $$ – 2,
3x + 2y – 4z = – 5 has infinitely many solutions. Then $$\lambda $$ is a root of the quadratic equation:
$$\lambda $$2 + $$\lambda $$ - 6 = 0
$$\lambda $$2 - $$\lambda $$ - 6 = 0
$$\lambda $$2 - 3$$\lambda $$ - 4 = 0
$$\lambda $$2 + 3$$\lambda $$ - 4 = 0
Explanation
$$\Delta = 0$$
$$\left| {\matrix{ 1 & 1 & 1 \cr 4 & \lambda & { - \lambda } \cr 3 & 2 & { - 4} \cr } } \right| = 0$$
On solving we get $$\lambda $$ = 3
$$\left| {\matrix{ 1 & 1 & 1 \cr 4 & \lambda & { - \lambda } \cr 3 & 2 & { - 4} \cr } } \right| = 0$$
On solving we get $$\lambda $$ = 3
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