Verified Solution • 2025–26 Session moderate

QJEEM23Mab11a: JEE Main 2023 Mathematics - Class 11 JEEM23Mab11a Solution: For the system of equations , which one of the following is NOT …

For the system of equations

x+y+z=6x+y+z=6

x+2y+αz=10x+2 y+\alpha z=10

x+3y+5z=βx+3 y+5 z=\beta , which one of the following is NOT true?

Multiple Choice Options

  • A System has a unique solution for α=3,β≠14\alpha=3,\beta\ne14 .
  • B System has infinitely many solutions for α=3,β=14\alpha=3, \beta=14 .
  • C System has no solution for α=3,β=24\alpha=3, \beta=24 .
  • D System has a unique solution for α=−3,β=14\alpha=-3, \beta=14 .

Step-by-Step Verified Solution

Given system of equations,

x+y+zamp;=6........(i)x+2y+αzamp;=10........(ii)x+3y+5zamp;=β........(iii)\begin{aligned}x+y+z & =6 ........(i)\\\\x+2 y+\alpha z & =10 ........(ii)\\\\x+3 y+5 z & =\beta ........(iii)\end{aligned}

Here,

Δamp;=∣1amp;1amp;11amp;2amp;α1amp;3amp;5∣amp;=1(10−3α)−1(5−α)+1(3−2)amp;=10−3α−5+α+1amp;=6−2α\begin{aligned}\Delta & =\left|\begin{array}{lll}1 & 1 & 1 \\1 & 2 & \alpha \\1 & 3 & 5\end{array}\right| \\\\& =1(10-3 \alpha)-1(5-\alpha)+1(3-2) \\\\& =10-3 \alpha-5+\alpha+1 \\\\& =6-2 \alpha\end{aligned}

For unique solution, Δ≠0\Delta \neq 0

⇒6−2α≠0⇒α≠3\Rightarrow 6-2 \alpha \neq 0 \Rightarrow \alpha \neq 3

When, α=3\alpha=3

Δ1amp;=∣6amp;1amp;110amp;2amp;3βamp;3amp;5∣amp;=∣0amp;1amp;0−2amp;2amp;1β−18amp;3amp;2∣=−1(−4−β+18)amp;=β−14\begin{aligned}\Delta_1 & =\left|\begin{array}{ccc}6 & 1 & 1 \\10 & 2 & 3 \\\beta & 3 & 5\end{array}\right| \\\\& =\left|\begin{array}{ccc}0 & 1 & 0 \\-2 & 2 & 1 \\\beta-18 & 3 & 2\end{array}\right|=-1(-4-\beta+18) \\\\& =\beta-14\end{aligned}

and

Δ2amp;=∣1amp;6amp;11amp;10amp;31amp;βamp;5∣amp;=1(50−3β)−6(5−3)+1(β−10)amp;=50−3β−12+β−10amp;=28−2β=2(14−β)\begin{aligned} \Delta_2 & =\left|\begin{array}{ccc}1 & 6 & 1 \\ 1 & 10 & 3 \\ 1 & \beta & 5\end{array}\right| \\\\ & =1(50-3 \beta)-6(5-3)+1(\beta-10) \\\\ & =50-3 \beta-12+\beta-10 \\\\ & =28-2 \beta=2(14-\beta)\end{aligned}

and

Δ3amp;=∣1amp;1amp;61amp;2amp;101amp;3amp;β∣amp;=1(2β−30)−1(β−10)+6(3−2)amp;=2β−30−β+10+6amp;=β−14\begin{aligned}\Delta_3 & =\left|\begin{array}{ccc}1 & 1 & 6 \\1 & 2 & 10 \\1 & 3 & \beta\end{array}\right| \\& =1(2 \beta-30)-1(\beta-10)+6(3-2) \\\\& =2 \beta-30-\beta+10+6 \\\\& =\beta-14\end{aligned}

Thus, at β=14,Δ1=Δ2=Δ3=0\beta=14, \Delta_1=\Delta_2=\Delta_3=0

⇒α=3,β=14\Rightarrow \alpha=3, \beta=14

So, system has infinite solutions.

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