Verified Solution • 2025–26 Session moderate

QJEEM21Ma9baa: JEE Main 2021 Mathematics - Class 12 JEEM21Ma9baa Solution: Consider the system of linear equations x + y + 2z = 0 3x ay +…

Consider the system of linear equations

−- x + y + 2z = 0

3x −- ay + 5z = 1

2x −- 2y −- az = 7

Let S1 be the set of all a ∈\in R for which the system is inconsistent and S2 be the set of all a ∈\in R for which the system has infinitely many solutions. If n(S1) and n(S2) denote the number of elements in S1 and S2 respectively, then

Multiple Choice Options

  • A n(S1) = 2, n(S2) = 2
  • B n(S1) = 1, n(S2) = 0
  • C n(S1) = 2, n(S2) = 0
  • D n(S1) = 0, n(S2) = 2

Step-by-Step Verified Solution

Δ=|−1123−a52−2−a|

=−1(a2+10)−1(−3a−10)+2(−6+2a)= - 1({a^2} + 10) - 1( - 3a - 10) + 2( - 6 + 2a)

=−a2−10+3a+10−12+4a= - {a^2} - 10 + 3a + 10 - 12 + 4a

Δ=−a2+7a−12\Delta = - {a^2} + 7a - 12

Δ=−[a2−7a+12]\Delta = - [{a^2} - 7a + 12]

Δ=−[(a−3)(a−4)]\Delta = - [(a - 3)(a - 4)]

Δ1=|0121−a57−2−a|


=a+35−4+14a= a + 35 - 4 + 14a

= 15a+3115a + 31

Now, Δ1=15a+31{\Delta _1} = 15a + 31

For inconsistent Δ\Delta = 0 ∴\therefore a = 3, a = 4 and for a = 3 and 4, Δ\Delta 1 ≠\ne 0

n(S1) = 2

For infinite solution : Δ\Delta = 0 and Δ\Delta 1 = Δ\Delta 2 = Δ\Delta 3 = 0

Not possible

∴\therefore n(S2) = 0

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