JEE Main 2025 Jan 29 Shift 2, Mathematics Q11: Linear Equations And Cramer’s Rule

JEE Main2025Jan 29, Shift 2Mathematics

Let $α, β$ $(α \neq = β)$ be the values of $m$ , for which the equations $x + y + z = 1$ ; $x + 2 y + 4 z = m$ and $x + 4 y + 10 z = m^{2}$ have infinitely many solutions. Then the value of $n = 1 \sum 10 (n^{α} + n^{β})$ is equal to :

A
$440$
B
$3080$
C
$3410$
D
$560$

Solution

The coefficient determinant

$Δ = 1111241410 = 1 (20 - 16) - 1 (10 - 4) + 1 (4 - 2) = 4 - 6 + 2 = 0$ .

Since $Δ = 0$ , the system is consistent with infinitely many solutions only when every numerator determinant vanishes too. Solving $Δ_{x} = 0$ (or equivalently the consistency condition) gives $m^{2} - 3 m + 2 = 0 \Rightarrow m = 1, 2$ .

Hence $α = 1$ , $β = 2$ , and

$n = 1 \sum 10 (n + n^{2}) = \frac{10 \cdot 11}{2} + \frac{10 \cdot 11 \cdot 21}{6} = 55 + 385 = 440$ .

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JEE Main 2025 Jan 29 Shift 2, Mathematics Q11: Linear Equations And Cramer’s Rule

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