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Mathematics Question on linear inequalities

Let α,β, and γ be real numbers. Consider the following system of linear equations:
x + 2y + z = 7
x + αz = 11
2x - 3y + βz = γ
Match each entry in List I to the correct entries in List II

List IList II
(P)If β=12\frac{1}{2}(7α - 3) and γ\gamma=28, then the system has
(Q)If β=12\frac{1}{2}(7α - 3) and \gamma$$\neq28, then the system has
(R)If β\neq$$\frac{1}{2}(7α - 3) where α\alpha=1 and \gamma$$\neq28, then the system has
(S)If β\neq$$\frac{1}{2}(7α - 3) where α\alpha=1 and γ\gamma=28, then the system has
(5)
A

(P) →(3), (Q)→ (2), (R) →(1), (S) →(4)

B

(P) →(3), (Q)→ (2), (R) →(5), (S) →(4)

C

(P) →(2), (Q)→ (1), (R) →(4), (S) →(5)

D

(P) →(2), (Q)→ (1), (R) →(1), (S) →(3)

Answer

(P) →(3), (Q)→ (2), (R) →(1), (S) →(4)

Explanation

Solution

x+2y+z=7x + 2y + z = 7
x+αz=11x + αz = 11
2x−3y+βz=γ2x - 3y + βz = γ

△=\triangle = ∣121 10α 2−3β∣\begin{vmatrix}1&2&1 \\\ 1 & 0 & \alpha \\\ 2 &-3 & \beta \end{vmatrix}= 0
3α−2(β−2α)−3=03\alpha - 2(\beta - 2\alpha) -3 = 0
7α−2β=37\alpha - 2 \beta = 3
⇒ β=12(7α−3)\beta = \frac {1}{2} (7 \alpha -3)

∣721 110α γ−3β∣,△2=∣171 111α 2γβ∣,△3=∣172 1011 2−3γ∣\begin{vmatrix}7&2&1\\\ 11&0 &\alpha \\\ \gamma &-3 &\beta\end{vmatrix} , \triangle_2 = \begin{vmatrix}1&7&1\\\ 1&11 &\alpha \\\ 2&\gamma &\beta\end{vmatrix} , \triangle_3 = \begin{vmatrix}1&7&2\\\ 1&0 &11 \\\ 2& -3 &\gamma\end{vmatrix}

△3=0\triangle_3 =0
⇒ 33−2(γ−22)+7(−3)=033 -2(\gamma -22) + 7(-3) = 0
γ=28\gamma = 28
△1=21α−2(11β−αβ)−33\triangle_1 = 21 \alpha - 2(11\beta - \alpha \beta) - 33
  =21α−22β+22β+2αβ−33\,\,= 21\alpha -22 \beta + 22 \beta + 2 \alpha \beta -33
△2=11β−αβ−7(β−2α)+γ−22\triangle_2 = 11 \beta - \alpha\beta - 7(\beta - 2 \alpha) + \gamma -22
  =14α+4β+γ−αγ−22\,\, = 14 \alpha + 4 \beta + \gamma - \alpha\gamma -22

(P) If\text{(P) If} β=12(7α−3)  and  γ=28\beta = \frac{1}{2}(7\alpha -3) \,\,and \,\,\gamma = 28
△=0, △1=0, △2=0, △3=0\triangle = 0, \,\triangle_1 = 0, \,\triangle_2 = 0, \,\triangle_3 = 0
Infinitely many solutions
x = 11, y = – 2 and z = 0 will satisfy all the three given equations, so it is a solution.

(Q) If\text{(Q) If} β=12(7α−3)  and  γ≠28  then\beta = \frac {1}{2} (7 \alpha -3) \,\, and \,\, \gamma ≠ 28 \, \,then
△=0, but  △3≠0 so no solution\triangle =0 , \,but \,\,\triangle_3\neq 0 \text{ so no solution}

 (R) If β≠12(7α−3),α=1 and  γ≠28\text{ (R) If }\beta \neq \frac{1}{2}(7\alpha - 3), \alpha = 1 \, and \,\, \gamma\neq 28
  △≠0,△3≠0 so a unique solution\,\,\triangle \neq 0, \triangle_3 \neq 0 \text{ so a unique solution}

 (S)   If  β≠12(7α−3), α=1,  γ=28\text{ (S) \, If }\, \beta \neq \frac{1}{2}(7\alpha - 3), \, \alpha = 1, \,\, \gamma = 28
   △≠0, △3=0 ,△1≠0, △2≠0,so a unique solution\,\,\, \triangle \neq 0, \,\triangle_3 = 0 \,, \triangle_1 \neq 0,\, \triangle_2 \neq 0, \text{so a unique solution}

x = 11, y = – 2 and z = 0 will satisfy all the three equations.
Hence, The correct option is (A) (P) →(3), (Q)→ (2), (R) →(1), (S) →(4).