Verified Solution • 2025–26 Session moderate

QJEEM23Mad41c: JEE Main 2023 Mathematics - Class 12 JEEM23Mad41c Solution: For the system of linear equations , which one of the following …

For the system of linear equations αx+y+z=1,x+αy+z=1,x+y+αz=β\alpha x+y+z=1,x+\alpha y+z=1,x+y+\alpha z=\beta , which one of the following statements is NOT correct?

Multiple Choice Options

  • A It has infinitely many solutions if α=1\alpha=1 and β=1\beta=1
  • B It has infinitely many solutions if α=2\alpha=2 and β=−1\beta=-1
  • C x+y+z=34x+y+z=\frac{3}{4} if α=2\alpha=2 and β=1\beta=1
  • D It has no solution if α=−2\alpha=-2 and β=1\beta=1

Step-by-Step Verified Solution

For infinite solution Δ=Δx=Δy=Δz=0\Delta=\Delta_x=\Delta_y=\Delta_z=0

Δ=∣αamp;1amp;11amp;αamp;11amp;1amp;α∣=0⇒(α3−3α+2)=0⇒α=1,−2\Delta=\left|\begin{array}{lll}\alpha & 1 & 1 \\1 & \alpha & 1 \\1 & 1 & \alpha\end{array}\right|=0 \Rightarrow\left(\alpha^3-3 \alpha+2\right)=0 \Rightarrow \alpha=1,-2

If β=1\beta=1 , then all planes are overlapping

∴\therefore Option (A) is correct.

Option (B) :

If α=2⇒Δ≠0\alpha=2 \Rightarrow \Delta \neq 0

∴\therefore Unique solution exist

∴\therefore Option (B) is incorrect.

Option (C) :

amp;α=2,β=1amp;2x+y+z=1amp;x+2y+z=1amp;x+y+2z=1\begin{aligned}& \alpha=2, \beta=1 \\\\& 2 x+y+z=1 \\\\& x+2 y+z=1 \\\\& x+y+2 z=1\end{aligned}

Adding all three equations,

x+y+z=34x+y+z=\frac{3}{4}

∴\therefore option (C) is correct.

Option (D) :

If α=−2\alpha=-2 and β=1\beta=1 , then Δ=0,Δx≠0\Delta=0, \Delta_x \neq 0

∴\therefore No solution.

∴\therefore Option (D) is correct.

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