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Matrices and Determinants question

2023 · Shift 1 · Q31
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  5. /2023 · Shift 1 · Q31

Matrices and Determinants question

2023 · Shift 1 · Q31

JEE AdvancedMathematicsMatrices and DeterminantsMCQ+3 / −1
Let α,β\alpha, \betaα,β and γ\gammaγ be real numbers. Consider the following system of linear equations

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

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)β=21​(7α−3) and γ=28\gamma=28γ=28, then the system has (1) a unique solution
(Q) If β=12(7α−3)\beta=\frac{1}{2}(7 \alpha-3)β=21​(7α−3) and γeq28\gamma eq 28γeq28, then the system has (2) no solution
(R) If βeq12(7α−3)\beta eq \frac{1}{2}(7 \alpha-3)βeq21​(7α−3) where α=1\alpha=1α=1 and γeq28\gamma eq 28γeq28, then the system has (3) infinitely many solutions
(S) If βeq12(7α−3)\beta eq \frac{1}{2}(7 \alpha-3)βeq21​(7α−3) where α=1\alpha=1α=1 and γ=28\gamma=28γ=28, then the system has (4) x=11,y=−2x=11, y=-2x=11,y=−2 and z=0z=0z=0 as a solution
(5) x=−15,y=4x=-15, y=4x=−15,y=4 and z=0z=0z=0 as a solution

The correct option is:
  1. A
    (P)→(3)  (Q)→(2)  (R)→(1)  (S)→(4)(P) \rightarrow(3) ~~ (Q) \rightarrow(2) ~~(R) \rightarrow(1)~~ (S) \rightarrow(4)(P)→(3)  (Q)→(2)  (R)→(1)  (S)→(4)
  2. B
    (P)→(3)  (Q)→(2)  (R)→(5)  (S)→(4)(P) \rightarrow(3) ~~(Q) \rightarrow(2) ~~(R) \rightarrow(5)~~ (S) \rightarrow(4)(P)→(3)  (Q)→(2)  (R)→(5)  (S)→(4)
  3. C
    (P)→(2)  (Q)→(1)  (R)→(4)  (S)→(5)(P) \rightarrow(2)~~ (Q) \rightarrow(1) ~~ (R) \rightarrow(4) ~~ (S) \rightarrow(5)(P)→(2)  (Q)→(1)  (R)→(4)  (S)→(5)
  4. D
    (P)→(2)  (Q)→(1)  (R)→(1)  (S)→(3)(P) \rightarrow(2) ~~ (Q) \rightarrow(1) ~~ (R) \rightarrow(1) ~~ (S) \rightarrow(3)(P)→(2)  (Q)→(1)  (R)→(1)  (S)→(3)
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