For consistent solutions Δ=0
⇒∣−k−15−4341−14−k3∣=0
⇒−k(12+k)−3(−45−4k)−14(−15+16)=0
⇒−12k−k2+135+12k−14=0
⇒k2=121
⇒k=11,−11
JEE Main 2022 — Mathematics Algebra
The system of equations
−kx+3y−14z=25
−15x+4y−kz=3
−4x+y+3z=4
Question: is consistent for all k in the set
Held on 25 Jun 2022 · Verified 6 Jul 2026.
R
R−−11,13
R−−13
R−−11,11
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