The system of linear equations x + y + z = 6, x + 2y + 3z = 14 and 2x + 5y + pz = q have -
(A) infinitely many solution when p = 8, q = 36
(B) unique solution when p ≠ 8, q ≠ 36
(C) no solution when p = 8, q ≠ 36
(D) atleast one solution for q = 36, p ∈ R
Step-by-Step Solution
Key Concept: The system of linear equations can be represented as AX = B. The determinant of the coefficient matrix A is |A| = 1(2p - 15) - 1(p - 6) + 1(5 - 4) = 2p - 15 - p + 6 + 1 = p - 8. For a unique solution, |A| \neq 0, so p \neq 8. For infinite or no solutions, |A| = 0, so p = 8. When p = 8, the system becomes x + y + z = 6, x + 2y + 3z = 14, 2x + 5y + 8z = q. Adding the first two equations gives 2x + 3y + 4z = 20. Comparing this with the third equation, for consistency, q must be 36. If p = 8 and q = 36, there are infinitely many solutions. If p = 8 and q \neq 36, there is no solution.
The system of equations is: x + y + z = 6, x + 2y + 3z = 14, 2x + 5y + pz = q. The determinant of the coefficient matrix is |A| = |1 1 1; 1 2 3; 2 5 p| = 1(2p - 15) - 1(p - 6) + 1(5 - 4) = 2p - 15 - p + 6 + 1 = p - 8. For a unique solution, |A| \neq 0, i.e., p \neq 8. This is independent of q. Thus, (B) is correct. For p = 8, |A| = 0. The system is: x + y + z = 6, x + 2y + 3z = 14, 2x + 5y + 8z = q. Subtracting the first from the second: y + 2z = 8. Subtracting twice the first from the third: 3y + 6z = q - 12. For infinite solutions, 3(y + 2z) = q - 12, so 3(8) = q - 12, which means 24 = q - 12, so q = 36. Thus, (A) is correct. If p = 8 and q \neq 36, the system is inconsistent, so (C) is incorrect. For q = 36, if p \neq 8, there is a unique solution. If p = 8, there are infinitely many solutions. In both cases, there is at least one solution. Thus, (D) is correct.
Correct Answer: A,B,D