Matrices & Determinants
System of Linear Equations
Grade Class 12
Question:
<p>Let α, β and γ be real numbers. consider the following system of linear equations</p><p>x + 2y + z = 7</p><p>x + αz = 11</p><p>2x - 3y + βz = γ</p><p>Match each entry in List-I to the correct entries in List-II</p><table border="1"><tr><td>List-I</td><td>List-II</td></tr><tr><td>(P) If β = 1/2(7α - 3) and γ = 28, then the system has</td><td>(1) a unique solution</td></tr><tr><td>(Q) If β = 1/2(7α - 3) and γ ≠ 28, then the system has</td><td>(2) no solution</td></tr><tr><td>(R) If β ≠ 1/2(7α - 3) where α = 1 and γ ≠ 28, then the system has</td><td>(3) infinitely many solutions</td></tr><tr><td>(S) If β ≠ 1/2(7α - 3) where α = 1 and γ = 28, then the system has</td><td>(4) x = 11, y = -2 and z = 0 as a solution</td></tr><tr><td></td><td>(5) x = -15, y = 4 and z = 0 as a solution</td></tr></table>
(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)
Step-by-Step Solution
Key Concept: The system of linear equations can be analyzed using the determinant of the coefficient matrix. If the determinant is non-zero, there is a unique solution. If the determinant is zero, the system may have no solution or infinitely many solutions depending on the consistency of the equations.
The coefficient matrix is A = [[1, 2, 1], [1, 0, α], [2, -3, β]]. The determinant |A| = 1(0 - (-3α)) - 2(β - 2α) + 1(-3 - 0) = 3α - 2β + 4α - 3 = 7α - 2β - 3. For a unique solution, |A| ≠ 0, i.e., β ≠ 1/2(7α - 3). If β = 1/2(7α - 3), |A| = 0, the system has either no solution or infinitely many solutions. Checking the consistency for β = 1/2(7α - 3) and γ = 28 leads to infinitely many solutions. For γ ≠ 28, it leads to no solution. For α = 1, β ≠ 1/2(7-3) = 2, the system has a unique solution. Checking the specific values for (R) and (S) confirms the mapping.
Correct Answer: A