--- title: "Power method: 2,5,1 matrix" subject: "Numerical Method" chapter: 3 key: "semester_4/numerical_method:power-method-2-5-1" source: https://www.bctengineer.com/q/semester_4/numerical_method/power-method-2-5-1 site: BCT Engineering --- # Power method: 2,5,1 matrix Numerical Method · Chapter 3: Solution of system of linear algebraic equations - Asked in: 2070 Magh, 8 marks - Video: [Eigenvalue Power Method | Lecture 30 | Numerical Methods for Engineers](https://www.youtube.com/watch?v=_PDyi5BVY-E) — Jeffrey Chasnov - 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-2-5-1 - Whole chapter: https://www.bctengineer.com/md/semester_4/numerical_method/chapter-3 ## Question Find the largest Eigen value and corresponding vector of the following matrix using power method. [2 5 1; 5 -2 3; 1 3 10] ## Short answer Using the power method from initial vector [1,1,1] and scaling by the largest component, after 12 iterations the dominant eigenvalue is 11.289 and the corresponding eigenvector is [0.287, 0.334, 1.000] (any multiple). ## 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 = 2 5 1 5 -2 3 1 3 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] | [8.0000, 6.0000, 14.0000] | 14.0000 | [0.5714, 0.4286, 1.0000] | | 2 | [0.5714, 0.4286, 1.0000] | [4.2857, 5.0000, 11.8571] | 11.8571 | [0.3614, 0.4217, 1.0000] | | 3 | [0.3614, 0.4217, 1.0000] | [3.8313, 3.9639, 11.6265] | 11.6265 | [0.3295, 0.3409, 1.0000] | | 4 | [0.3295, 0.3409, 1.0000] | [3.3637, 3.9658, 11.3523] | 11.3523 | [0.2963, 0.3493, 1.0000] | | 5 | [0.2963, 0.3493, 1.0000] | [3.3393, 3.7828, 11.3443] | 11.3443 | [0.2944, 0.3335, 1.0000] | | 6 | [0.2944, 0.3335, 1.0000] | [3.2560, 3.8049, 11.2947] | 11.2947 | [0.2883, 0.3369, 1.0000] | | 7 | [0.2883, 0.3369, 1.0000] | [3.2609, 3.7676, 11.2989] | 11.2989 | [0.2886, 0.3335, 1.0000] | | 8 | [0.2886, 0.3335, 1.0000] | [3.2445, 3.7761, 11.2890] | 11.2890 | [0.2874, 0.3345, 1.0000] | | 9 | [0.2874, 0.3345, 1.0000] | [3.2473, 3.7680, 11.2909] | 11.2909 | [0.2876, 0.3337, 1.0000] | | 10 | [0.2876, 0.3337, 1.0000] | [3.2438, 3.7706, 11.2888] | 11.2888 | [0.2873, 0.3340, 1.0000] | | 11 | [0.2873, 0.3340, 1.0000] | [3.2448, 3.7687, 11.2894] | 11.2894 | [0.2874, 0.3338, 1.0000] | | 12 | [0.2874, 0.3338, 1.0000] | [3.2440, 3.7694, 11.2889] | 11.2889 | [0.2874, 0.3339, 1.0000] | From the last two rows, λ and the vector are unchanged to 3 decimal places. **Answer:** dominant eigenvalue λ ≈ 11.289; corresponding eigenvector ≈ [0.287, 0.334, 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. The power method on A = [2 5 1; 5 -2 3; 1 3 10] finds which eigenvalue? A. The one with the largest magnitude B. The smallest eigenvalue C. The one closest to zero D. All eigenvalues at once Answer: A. It converges to the dominant eigenvalue and its eigenvector. (Key point: What power method finds) 2. For x(0) = [1,1,1], what is x(1)? A. [1, 0.75, 1.75] B. [0.5714, 0.4286, 1.0000] C. [8, 6, 14] D. [0.2874, 0.3339, 1] Answer: B. A·x(0) = [8, 6, 14]; dividing by 14 gives [0.5714, 0.4286, 1]. (Key point: First iteration) 3. What final result is obtained, to 3 decimals? A. λ ≈ 14.000, vector ≈ [0.571, 0.429, 1.000] B. λ ≈ 11.289, vector ≈ [1.000, 0.334, 0.287] C. λ ≈ 11.289, vector ≈ [0.287, 0.334, 1.000] D. λ ≈ 10.000, vector ≈ [0.287, 0.334, 1.000] Answer: C. Iterations 11 and 12 agree to 3 decimals at 11.289 and [0.287, 0.334, 1]. (Key point: Result and accuracy) ## Close variants set in other exams - [Power method: 1,4,-1 matrix](https://www.bctengineer.com/question.md?key=semester_4%2Fnumerical_method%3Apower-method-1-4-minus1) — Numerical Method, chapter 3, asked 3× - [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: 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× ## Related in Numerical Method - [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 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 - [Sallen-Key LPF at 10 kHz, largest C 0.01 μF](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Asallen-key-10khz-largest-capacitance) — Filter Design, chapter 6 - [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