Verified Solution • 2025–26 Session moderate

QJEEAdv23Ma00b0: JEE Advanced 2023 Mathematics - Class 12 JEEAdv23Ma00b0 Solution: Let and be real numbers. Consider the following system of…

Let α,β\alpha, \beta and γ\gamma be real numbers. Consider the following system of linear equations

amp;x+2y+z=7amp;x+αz=11amp;2x−3y+βz=γ\begin{aligned}& x+2 y+z=7 \\\\& x+\alpha z=11 \\\\& 2 x-3 y+\beta z=\gamma\end{aligned}

Match each entry in List-I to the correct entries in List-II.

List - I List - II
(P) If β=12(7α−3)\beta=\frac{1}{2}(7 \alpha-3) and γ=28\gamma=28 , then the system has (1) a unique solution
(Q) If β=12(7α−3)\beta=\frac{1}{2}(7 \alpha-3) and γ≠28\gamma \neq 28 , then the system has (2) no solution
(R) If β≠12(7α−3)\beta \neq \frac{1}{2}(7 \alpha-3) where α=1\alpha=1 and γ≠28\gamma \neq 28 , then the system has (3) infinitely many solutions
(S) If β≠12(7α−3)\beta \neq \frac{1}{2}(7 \alpha-3) where α=1\alpha=1 and γ=28\gamma=28 , then the system has (4) x=11,y=−2x=11, y=-2 and z=0z=0 as a solution
(5) x=−15,y=4x=-15, y=4 and z=0z=0 as a solution

The correct option is:

Multiple Choice Options

  • A (P)→(3)  (Q)→(2)  (R)→(1)  (S)→(4)(P) \rightarrow(3) ~~ (Q) \rightarrow(2) ~~(R) \rightarrow(1)~~ (S) \rightarrow(4)
  • B (P)→(3)  (Q)→(2)  (R)→(5)  (S)→(4)(P) \rightarrow(3) ~~(Q) \rightarrow(2) ~~(R) \rightarrow(5)~~ (S) \rightarrow(4)
  • C (P)→(2)  (Q)→(1)  (R)→(4)  (S)→(5)(P) \rightarrow(2)~~ (Q) \rightarrow(1) ~~ (R) \rightarrow(4) ~~ (S) \rightarrow(5)
  • D (P)→(2)  (Q)→(1)  (R)→(1)  (S)→(3)(P) \rightarrow(2) ~~ (Q) \rightarrow(1) ~~ (R) \rightarrow(1) ~~ (S) \rightarrow(3)

Step-by-Step Verified Solution

Given

x+2y+z=7x+αz=112x−3y+βz=γ\begin{gathered}x+2 y+z=7 \\\\ x+\alpha z=11 \\\\ 2 x-3 y+\beta z=\gamma\end{gathered}

Using Cramer's rule

Δamp;=∣1amp;2amp;11amp;0amp;α2amp;−3amp;β∣amp;=1(3α)−2(β−2α)+1(−3)amp;=3α−2β+4α−3amp;=7α−2β−3\begin{aligned}\Delta & =\left|\begin{array}{ccc}1 & 2 & 1 \\1 & 0 & \alpha \\2 & -3 & \beta\end{array}\right| \\\\& =1(3 \alpha)-2(\beta-2 \alpha)+1(-3) \\\\& =3 \alpha-2 \beta+4 \alpha-3 \\\\& =7 \alpha-2 \beta-3\end{aligned}

Δxamp;=∣7amp;2amp;111amp;0amp;αγamp;−3amp;β∣amp;=7(3α)−2(11β−γα)+1(−33)amp;=21α−22β+22γ−33\begin{aligned} \Delta_x & =\left|\begin{array}{ccc}7 & 2 & 1 \\ 11 & 0 & \alpha \\ \gamma & -3 & \beta\end{array}\right| \\\\ & =7(3 \alpha)-2(11 \beta-\gamma \alpha)+1(-33) \\\\ & =21 \alpha-22 \beta+22 \gamma-33\end{aligned}

Δyamp;=∣1amp;7amp;11amp;11amp;α2amp;γamp;β∣amp;=1(11β−αγ)−7(β−2α)+1(γ−22)amp;=11β−αγ−7β+14α+γ−22amp;=14α+4β+γ−αγ−22\begin{aligned} \Delta_y & =\left|\begin{array}{ccc}1 & 7 & 1 \\ 1 & 11 & \alpha \\ 2 & \gamma & \beta\end{array}\right| \\\\ & =1(11 \beta-\alpha \gamma)-7(\beta-2 \alpha)+1(\gamma-22) \\\\ & =11 \beta-\alpha \gamma-7 \beta+14 \alpha+\gamma-22 \\\\ & =14 \alpha+4 \beta+\gamma-\alpha \gamma-22\end{aligned}

Δzamp;=∣1amp;2amp;71amp;0amp;112amp;−3amp;γ∣amp;=1(33)−2(γ−22)+7(−3)amp;=33−2γ+44−21amp;=−2γ+56\begin{aligned} \Delta_z & =\left|\begin{array}{ccc}1 & 2 & 7 \\ 1 & 0 & 11 \\ 2 & -3 & \gamma\end{array}\right| \\\\ & =1(33)-2(\gamma-22)+7(-3) \\\\ & =33-2 \gamma+44-21 \\\\ & =-2 \gamma+56\end{aligned}

For unique solution Δ≠0\Delta \neq 0

For infinite solution

Δ=Δx=Δy=Δz=0\Delta=\Delta x=\Delta y=\Delta z=0

For no solution Δ=0\Delta=0 and atleast one in Δx,Δy,Δz\Delta x, \Delta y, \Delta z is non zero.

Δ=0\Delta=0

⇒β=12(7α−3)\Rightarrow \beta=\frac{1}{2}(7 \alpha-3)

(P) β=12(7α−3)\beta=\frac{1}{2}(7 \alpha-3) and γ=28\gamma=28

then Δ=0,Δx=Δy=Δz=0\Delta=0, \Delta x=\Delta y=\Delta z=0

∴\therefore Infinite solution

(Q) β=12(7α−3)\beta=\frac{1}{2}(7 \alpha-3) and γ≠28\gamma \neq 28

∴Δ=0\therefore \Delta=0 and Δ2≠0\Delta_2 \neq 0

⇒\Rightarrow No solution.

amp; (R) β≠12(7α−3),α=1,γ≠28amp;⇒Δ≠0= unique solution amp;amp; (S) β≠12(7α−3),α=1,γ=28amp;∴Δ≠0,Δ=4−2βamp;Δx=44−22βamp;Δy=4β−8amp;Δz=0\begin{aligned}& \text { (R) } \beta \neq \frac{1}{2}(7 \alpha-3), \alpha=1, \gamma \neq 28 \\\\& \Rightarrow \Delta \neq 0=\text { unique solution } \\\\& \begin{aligned}& \text { (S) } \beta \neq \frac{1}{2}(7 \alpha-3), \alpha=1, \gamma=28 \\\\& \therefore \Delta \neq 0, \Delta=4-2 \beta \\\\& \Delta x=44-22 \beta \\\\& \Delta y=4 \beta-8 \\\\& \Delta z=0\end{aligned}\end{aligned}

∴x=11,y=−2,z=0\therefore x=11, y=-2, z=0 is the solution.

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