Verified Solution • 2025–26 Session moderate

QJEEM23Mab140: JEE Main 2023 Mathematics - Class 12 JEEM23Mab140 Solution: If the system of linear equations has infinitely many…

If the system of linear equations

amp;7x+11y+αz=13amp;5x+4y+7z=βamp;175x+194y+57z=361\begin{aligned}& 7 x+11 y+\alpha z=13 \\\\& 5 x+4 y+7 z=\beta \\\\& 175 x+194 y+57 z=361\end{aligned}

has infinitely many solutions, then α+β+2\alpha+\beta+2 is equal to :

Multiple Choice Options

  • A 6
  • B 4
  • C 5
  • D 3

Step-by-Step Verified Solution

Given,

amp;7x+11y+αz=13amp;5x+4y+7z=βamp;175x+194y+57z=361\begin{aligned}& 7 x+11 y+\alpha z=13 \\\\& 5 x+4 y+7 z=\beta \\\\& 175 x+194 y+57 z=361\end{aligned}

 For infinite solution, ∣7amp;11amp;α5amp;4amp;7175amp;194amp;57∣=0\text { For infinite solution, }\left|\begin{array}{ccc}7 & 11 & \alpha \\5 & 4 & 7 \\175 & 194 & 57\end{array}\right|=0

⇒∣7amp;11amp;α5amp;4amp;70amp;−81amp;57−25α∣=0\Rightarrow\left|\begin{array}{ccc}7 & 11 & \alpha \\5 & 4 & 7 \\0 & -81 & 57-25 \alpha\end{array}\right|=0

amp;⇒81(49−5α)+(57−25α)(−27)=0amp;⇒270α=−2430⇒α=−9\begin{aligned}& \Rightarrow 81(49-5 \alpha)+(57-25 \alpha)(-27)=0 \\\\& \Rightarrow 270 \alpha=-2430 \Rightarrow \alpha=-9\end{aligned}

And Δ1=0\Delta_1=0

amp;∣13amp;11amp;−9βamp;4amp;7361amp;194amp;57∣=0amp;⇒β=11\begin{aligned}& \left|\begin{array}{ccc}13 & 11 & -9 \\\beta & 4 & 7 \\361 & 194 & 57\end{array}\right|=0 \\\\& \Rightarrow \beta=11\end{aligned}

∴α+β+2=4\therefore \alpha+\beta+2=4

Related Learning Resources & Exam Preparation