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QJEEM24Mab8ff: JEE Main 2024 Mathematics - Class 12 JEEM24Mab8ff Solution: If the system of equations has infinitely many solutions, then…

If the system of equations

amp;2x+3y−z=5amp;x+αy+3z=−4amp;3x−y+βz=7\begin{aligned}& 2 x+3 y-z=5 \\\\& x+\alpha y+3 z=-4 \\\\& 3 x-y+\beta z=7\end{aligned}

has infinitely many solutions, then 13αβ13 \alpha \beta is equal to :

Multiple Choice Options

  • A 1110
  • B 1120
  • C 1210
  • D 1220

Step-by-Step Verified Solution

amp; Given 2x+3y−z=5amp;x+αy+3z=−4amp;3x−y+βz=7amp;Δ2=∣2amp;−1amp;51amp;3amp;−43amp;βamp;7∣amp;Δ2=2(21+4β)+1(7+12)+5(β−9)amp;Δ2=42+8β+19+5β−45amp;Δ2=13β+16amp;Δ2=0\begin{aligned} & \text { Given } 2 x+3 y-z=5 \\\\ & x+\alpha y+3 z=-4 \\\\ & 3 x-y+\beta z=7 \\\\ & \Delta_2=\left|\begin{array}{ccc}2 & -1 & 5 \\ 1 & 3 & -4 \\ 3 & \beta & 7\end{array}\right| \\\\ & \Delta_2=2(21+4 \beta)+1(7+12)+5(\beta-9) \\\\& \Delta_2=42+8 \beta+19+5 \beta-45 \\\\ & \Delta_2=13 \beta+16 \\\\ & \Delta_2=0\end{aligned}

amp;∴β=−1613amp;Δ3=∣2amp;3amp;51amp;αamp;−43amp;−1amp;7∣amp;Δ3=2(7α−4)−3(7+12)+5(−1−3α)amp;Δ3=14α−8−57−5−15αamp;Δ3=−α−70\begin{aligned} & \therefore \beta=-\frac{16}{13} \\\\ & \Delta_3=\left|\begin{array}{lll}2 & 3 & 5 \\ 1 & \alpha & -4 \\ 3 & -1 & 7\end{array}\right| \\\\ & \Delta_3=2(7 \alpha-4)-3(7+12)+5(-1-3 \alpha) \\\\ & \Delta_3=14 \alpha-8-57-5-15 \alpha \\\\ & \Delta_3=-\alpha-70\end{aligned}

amp;Δ3=0amp;α=−70amp;13αβ=(13)(−70)(−1613)amp;=+1120\begin{aligned} & \Delta_3=0 \\\\ & \alpha=-70 \\\\ & 13 \alpha \beta=(13)(-70)\left(-\frac{16}{13}\right) \\\\ & =+1120\end{aligned}

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