JEE Main 2022 · 28 Jun · Shift 1 · Mathematics Q3

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MathematicsSingle correct+4 / -1Question 3 of 90

If the system of linear equations
$2x+3y-z=-2$
$x+y+z=4$
$x-y+|\lambda|z=4\lambda-4$
where $\lambda\in\mathbb R$ , has no solution, then

\lambda=7

\lambda=-7

\lambda=8

\lambda^2=1

Answer and worked solution

Official answer: B

Determinant. $\begin{vmatrix}2&3&-1\\1&1&1\\1&-1&|\lambda|\end{vmatrix}=7-|\lambda|$ , zero when $\lambda=\pm7$ .

Consistency. The third row equals $-2(\text{row }1)+5(\text{row }2)$ , which needs right side $4+20=24$ . $\lambda=7$ gives 24 (infinitely many solutions); $\lambda=-7$ gives $-32$ (no solution).

(B)

Topics:Systems of linear equations Inequalities and modulus

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JEE Main 2022 · 28 Jun · Shift 1, Mathematics question 3