Question 119Asked in BPSC TRE 2.0 (2023) Mathematics Class 9-10 (Re-exam)

Consider the system of equations [[5, 2, 1], [−2, 5, 2], [−1, 2, 8]] [x₁, x₂, x₃]ᵀ = [13, −22, 14]ᵀसमीकरणों की प्रणाली पर विचार करें:

Consider the system of equations [[5, 2, 1], [−2, 5, 2], [−1, 2, 8]] [x₁, x₂, x₃]ᵀ = [13, −22, 14]ᵀ With the initial guess of the solution [x₁⁽⁰⁾, x₂⁽⁰⁾, x₃⁽⁰⁾]ᵀ = [1, 1, 1]ᵀ the approximate value of the solution [x₁⁽¹⁾, x₂⁽¹⁾, x₃⁽¹⁾]ᵀ after one iteration by the Gauss-Seidel method is
समीकरणों की प्रणाली पर विचार करें: [[5, 2, 1], [−2, 5, 2], [−1, 2, 8]] [x₁, x₂, x₃]ᵀ = [13, −22, 14]ᵀ समाधान [x₁⁽⁰⁾, x₂⁽⁰⁾, x₃⁽⁰⁾]ᵀ = [1, 1, 1]ᵀ के प्रारंभिक अनुमान के साथ गॉस-सीडेल विधि द्वारा एक पुनरावृत्ति के बाद समाधान [x₁⁽¹⁾, x₂⁽¹⁾, x₃⁽¹⁾]ᵀ का अनुमानित मूल्य है
  1. A[2, −4·4, 1·625]ᵀ[2, −4·4, 1·625]ᵀ
  2. B[2, −4, −3]ᵀ[2, −4, −3]ᵀ
  3. C[2, −4, 3]ᵀ[2, −4, 3]ᵀ
  4. DMore than one of the aboveउपर्युक्त में से एक से अधिक
  5. ENone of the aboveउपर्युक्त में से कोई नहीं
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Correct answer: (C) [2, −4, 3]ᵀ[2, −4, 3]ᵀ

Explanation

x₁ = (13 − 2·1 − 1)/5 = 2. Using the new x₁: x₂ = (−22 + 2·2 − 2·1)/5 = −4. Then x₃ = (14 + 2 − 2(−4))/8 = 3. The result is [2, −4, 3]ᵀ.x₁ = (13 − 2·1 − 1)/5 = 2। नए x₁ से x₂ = (−22 + 2·2 − 2·1)/5 = −4। फिर x₃ = (14 + 2 − 2(−4))/8 = 3। परिणाम [2, −4, 3]ᵀ है।

From BPSC TRE 2.0 (2023) Mathematics Class 9-10 (Re-exam) — Previous Year Paper

This is question 119 of BPSC TRE 2.0 (2023) Mathematics Class 9-10 (Re-exam) — Previous Year Paper. See all questions of this paper with answers, or attempt the paper as a timed online test to get your score and all-India rank.