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← JEE Main 2022 · 26 Jun · Shift 2
MathematicsSingle correct+4 / -1Question 2 of 90

JEE Main 2022 · 26 Jun · Shift 2, Mathematics question 2

If the system of equations
αx+y+z=5,x+2y+3z=4,x+3y+5z=β\alpha x+y+z=5,\quad x+2y+3z=4,\quad x+3y+5z=\beta
has infinitely many solutions, then the ordered pair (α,β)(\alpha,\beta) is equal to :

Choose one option
Answer and worked solution

Official answer: C

Determinant. α(10−9)−(5−3)+(3−2)=α−1=0\alpha(10-9)-(5-3)+(3-2)=\alpha-1=0, so α=1\alpha=1.

Consistency. Equation 2 − equation 1 gives y+2z=−1y+2z=-1; equation 3 − equation 2 gives y+2z=β−4y+2z=\beta-4. So β=3\beta=3.

(C)

Topics:Systems of linear equationsDeterminants and properties
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