--- title: "Power method: 1,4,-1 matrix" subject: "Numerical Method" chapter: 3 key: "semester_4/numerical_method:power-method-1-4-minus1" source: https://www.bctengineer.com/q/semester_4/numerical_method/power-method-1-4-minus1 site: BCT Engineering --- # Power method: 1,4,-1 matrix Numerical Method · Chapter 3: Solution of system of linear algebraic equations - Asked in: 2069 Poush, 8 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/numerical_method/power-method-1-4-minus1 - Whole chapter: https://www.bctengineer.com/md/semester_4/numerical_method/chapter-3 ## Question Find the dominant Eigen value and the corresponding vector of the following matrix using power method. [1 4 -1; 4 2 5; -1 5 10] ## Short answer Starting from [1,1,1]ᵀ, repeatedly multiply by the matrix and normalize by the largest component. After 10 iterations the dominant eigenvalue converges to approximately 12.471, with corresponding eigenvector approximately [0.092, 0.512, 1.000]ᵀ. ## Answer **Power method:** to find the dominant (largest in magnitude) eigenvalue λ and its eigenvector x of A, start with a trial vector x(0), compute A·x(k) repeatedly and scale it by its largest component. The scale factor tends to λ and the scaled vector tends to the eigenvector. #### Step 1: Matrix and starting vector ```text A = 1 4 -1 4 2 5 -1 5 10 ``` Take x(0) = [1, 1, 1]ᵀ. Iterate: A·x(k) = λ(k)·x(k+1), where λ(k) is the component of largest magnitude in A·x(k). #### Step 2: Iterations | k | x(k) | A·x(k) | λ(k) (largest component) | x(k+1) = A·x(k)/λ(k) | |---|---|---|---|---| | 1 | [1.0000, 1.0000, 1.0000] | [4.0000, 11.0000, 14.0000] | 14.0000 | [0.2857, 0.7857, 1.0000] | | 2 | [0.2857, 0.7857, 1.0000] | [2.4286, 7.7143, 13.6429] | 13.6429 | [0.1780, 0.5654, 1.0000] | | 3 | [0.1780, 0.5654, 1.0000] | [1.4398, 6.8429, 12.6492] | 12.6492 | [0.1138, 0.5410, 1.0000] | | 4 | [0.1138, 0.5410, 1.0000] | [1.2777, 6.5373, 12.5911] | 12.5911 | [0.1015, 0.5192, 1.0000] | | 5 | [0.1015, 0.5192, 1.0000] | [1.1783, 6.4443, 12.4945] | 12.4945 | [0.0943, 0.5158, 1.0000] | | 6 | [0.0943, 0.5158, 1.0000] | [1.1574, 6.4088, 12.4846] | 12.4846 | [0.0927, 0.5133, 1.0000] | | 7 | [0.0927, 0.5133, 1.0000] | [1.1460, 6.3975, 12.4740] | 12.4740 | [0.0919, 0.5129, 1.0000] | | 8 | [0.0919, 0.5129, 1.0000] | [1.1433, 6.3932, 12.4725] | 12.4725 | [0.0917, 0.5126, 1.0000] | | 9 | [0.0917, 0.5126, 1.0000] | [1.1420, 6.3919, 12.4713] | 12.4713 | [0.0916, 0.5125, 1.0000] | | 10 | [0.0916, 0.5125, 1.0000] | [1.1417, 6.3913, 12.4711] | 12.4711 | [0.0915, 0.5125, 1.0000] | From the last two rows, λ and the vector are unchanged to 3 decimal places. **Answer:** dominant eigenvalue λ ≈ 12.471; corresponding eigenvector ≈ [0.092, 0.512, 1.000]ᵀ (any multiple of it is also an eigenvector). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For A = [1 4 -1; 4 2 5; -1 5 10] and x(0) = [1,1,1], what is A·x(0)? A. [4, 11, 14] B. [4, 2, 10] C. [1, 4, -1] D. [14, 11, 4] Answer: A. Row sums: 1+4-1 = 4; 4+2+5 = 11; -1+5+10 = 14. (Key point: First iteration) 2. For x(2) = [0.2857, 0.7857, 1], A·x(2) = [2.4286, 7.7143, 13.6429]. What is λ(2)? A. 7.7143 B. 13.6429 C. 2.4286 D. 14 Answer: B. The largest component is 13.6429. (Key point: Second iteration scale factor) 3. What is the result of the power method? A. λ ≈ 14.000, vector ≈ [0.286, 0.786, 1.000] B. λ ≈ 12.471, vector ≈ [1.000, 0.512, 0.092] C. λ ≈ 12.471, vector ≈ [0.092, 0.512, 1.000] D. λ ≈ 6.391, vector ≈ [0.092, 0.512, 1.000] Answer: C. Iterations 9 and 10 agree to 3 decimals: λ = 12.471 and vector [0.092, 0.512, 1]. (Key point: Final result) ## Close variants set in other exams - [Power method: 15,-4,-3 matrix](https://www.bctengineer.com/question.md?key=semester_4%2Fnumerical_method%3Apower-method-15-4-3) — Numerical Method, chapter 3, asked 2× - [Power method: 2,5,1 matrix](https://www.bctengineer.com/question.md?key=semester_4%2Fnumerical_method%3Apower-method-2-5-1) — Numerical Method, chapter 3, asked 4× ## Related in Numerical Method - [Power method: 1,2,1 matrix](https://www.bctengineer.com/question.md?key=semester_4%2Fnumerical_method%3Apower-method-1-2-1) — Numerical Method, chapter 3, asked 2× - [Power method: 25,1,2 matrix](https://www.bctengineer.com/question.md?key=semester_4%2Fnumerical_method%3Apower-method-25-1-2) — Numerical Method, chapter 3, asked 2× - [Power method pseudo-code](https://www.bctengineer.com/question.md?key=semester_4%2Fnumerical_method%3Apower-method-algorithm) — Numerical Method, chapter 3, asked 2× ## Related in other subjects - [Value of a 10-year-old building](https://www.bctengineer.com/question.md?key=semester_8%2Fconstruction_management%3Avalue-building-10-years-2071) — Construction Management, chapter 13, asked 2× - [EPS and book value; Rs 10,00,000, 3500 shares](https://www.bctengineer.com/question.md?key=semester_7%2Fproject_engineering%3Aeps-bv-capital-1000000-3500-shares) — Project Engineering, chapter 6 - [Stiffness matrix: frame with 20 kN/m, 10 kN](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Astiffness-frame-20knm-10kn-2i) — Theory of Structures II, chapter 3 - [AM transmitter 10 kW/11.8 kW: modulation index and power](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aam-10kw-11-8kw) — Communication Systems, chapter 2