Matrices & Determinants
System of linear equations
Grade Class 12
Question:
<p>System of linear equations in x, y, z have infinite solutions which</p><p>2x + y + z = 1</p><p>x - 2y + z = 2</p><p>3x - y + 2z = 3</p><p>(A) can be written as (-3λ -1, λ, 5λ + 3) ∀ λ ∈ R</p><p>(B) can be written as (3λ -1, -λ, -5λ + 3) ∀ λ ∈ R</p><p>(C) are such that every solution satisfy x - 3y + 1 = 0</p><p>(D) are such that none of them satisfy 5x + 3z = 1</p>
(A) can be written as (-3λ -1, λ, 5λ + 3) ∀ λ ∈ R
(B) can be written as (3λ -1, -λ, -5λ + 3) ∀ λ ∈ R
(C) are such that every solution satisfy x - 3y + 1 = 0
(D) are such that none of them satisfy 5x + 3z = 1
Step-by-Step Solution
Key Concept: The system of equations has infinite solutions. By solving the system, we find the parametric form of the solution set and check which options satisfy the conditions.
<p>Given equations: 2x + y + z = 1, x - 2y + z = 2, 3x - y + 2z = 3. Adding the first two equations gives 3x - y + 2z = 3, which is the third equation. Thus, the system is dependent and has infinite solutions. Let y = \lambda. Then 2x + z = 1 - \lambda and x + z = 2 + 2\lambda. Subtracting gives x = -1 - 3\lambda. Then z = 2 + 2\lambda - x = 2 + 2\lambda - (-1 - 3\lambda) = 3 + 5\lambda. So the solution is (-3\lambda - 1, \lambda, 5\lambda + 3). Option (A) is correct. Option (B) is incorrect. For (C), x - 3y + 1 = (-3\lambda - 1) - 3(\lambda) + 1 = -6\lambda \neq 0. So (C) is incorrect. For (D), 5x + 3z = 5(-3\lambda - 1) + 3(5\lambda + 3) = -15\lambda - 5 + 15\lambda + 9 = 4 \neq 1. So (D) is correct.</p>
Correct Answer: 1, 2, 4