Determinant: Let lambda be a real number for which the system of linear equations… | JEE Main 2019

JEE Main 2019 (Online) 10th April Evening ShiftQuestion 1

Let $\lambda$ be a real number for which the system of linear equations $x+y+z=6$, $4x+\lambda y-\lambda z=\lambda-2$ and $3x+2y-4z=-5$ has infinitely many solutions. Then $\lambda$ is a root of the quadratic equation :

Determinant

$\lambda^{2}+3\lambda-4=0$$\lambda^{2}-\lambda-6=0$✓ Correct answer$\lambda^{2}-3\lambda-4=0$$\lambda^{2}+\lambda-6=0$

Solution

$\Delta=\begin{vmatrix}1&1&1\\4&\lambda&-\lambda\\3&2&-4\end{vmatrix}=24-8\lambda=0$ gives $\lambda=3$, and then the system is consistent (infinitely many solutions). $\lambda=3$ satisfies $\lambda^2-\lambda-6=0$.

Determinant - JEE Mathematics question (JEE Main 2019 (Online) 10th April Evening Shift)

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