--- title: "Power method: 1,2,1 matrix" subject: "Numerical Method" chapter: 3 key: "semester_4/numerical_method:power-method-1-2-1" source: https://www.bctengineer.com/q/semester_4/numerical_method/power-method-1-2-1 site: BCT Engineering --- # Power method: 1,2,1 matrix Numerical Method · Chapter 3: Solution of system of linear algebraic equations - Asked in: 2072 Asoj, 8 marks - Video: [Eigen Values and Eigen Vectors by Power Method | Power Method](https://www.youtube.com/watch?v=VmW_eIj5BWc) — The Random Club - 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-1-2-1 - Whole chapter: https://www.bctengineer.com/md/semester_4/numerical_method/chapter-3 ## Question Find the largest Eigen value and the corresponding Eigen vector of the following matrix using the power method with an accuracy of 2 decimal points. [1 2 1; 2 1 2; 1 2 -1] ## Short answer Start with initial vector [1,1,1] and iterate, scaling by the largest component. After 13 iterations, the dominant eigenvalue converges to 3.93 and the corresponding eigenvector is [0.88, 1.00, 0.58] (any scalar 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 = 1 2 1 2 1 2 1 2 -1 ``` 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, 5.0000, 2.0000] | 5.0000 | [0.8000, 1.0000, 0.4000] | | 2 | [0.8000, 1.0000, 0.4000] | [3.2000, 3.4000, 2.4000] | 3.4000 | [0.9412, 1.0000, 0.7059] | | 3 | [0.9412, 1.0000, 0.7059] | [3.6471, 4.2941, 2.2353] | 4.2941 | [0.8493, 1.0000, 0.5205] | | 4 | [0.8493, 1.0000, 0.5205] | [3.3699, 3.7397, 2.3288] | 3.7397 | [0.9011, 1.0000, 0.6227] | | 5 | [0.9011, 1.0000, 0.6227] | [3.5238, 4.0476, 2.2784] | 4.0476 | [0.8706, 1.0000, 0.5629] | | 6 | [0.8706, 1.0000, 0.5629] | [3.4335, 3.8670, 2.3077] | 3.8670 | [0.8879, 1.0000, 0.5968] | | 7 | [0.8879, 1.0000, 0.5968] | [3.4847, 3.9693, 2.2911] | 3.9693 | [0.8779, 1.0000, 0.5772] | | 8 | [0.8779, 1.0000, 0.5772] | [3.4551, 3.9102, 2.3007] | 3.9102 | [0.8836, 1.0000, 0.5884] | | 9 | [0.8836, 1.0000, 0.5884] | [3.4720, 3.9440, 2.2952] | 3.9440 | [0.8803, 1.0000, 0.5820] | | 10 | [0.8803, 1.0000, 0.5820] | [3.4623, 3.9246, 2.2984] | 3.9246 | [0.8822, 1.0000, 0.5856] | | 11 | [0.8822, 1.0000, 0.5856] | [3.4678, 3.9357, 2.2966] | 3.9357 | [0.8811, 1.0000, 0.5835] | | 12 | [0.8811, 1.0000, 0.5835] | [3.4647, 3.9293, 2.2976] | 3.9293 | [0.8817, 1.0000, 0.5847] | | 13 | [0.8817, 1.0000, 0.5847] | [3.4665, 3.9330, 2.2970] | 3.9330 | [0.8814, 1.0000, 0.5840] | From the last two rows, λ and the vector are unchanged to 2 decimal places. **Answer:** dominant eigenvalue λ ≈ 3.93; corresponding eigenvector ≈ [0.88, 1.00, 0.58]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 2 1; 2 1 2; 1 2 -1] and x(0) = [1,1,1], what are A·x(0) and λ(1)? A. [4, 5, 2] and λ(1) = 5 B. [4, 5, 2] and λ(1) = 2 C. [3, 4, 1] and λ(1) = 4 D. [1, 2, 1] and λ(1) = 2 Answer: A. Row sums give [4,5,2]; the largest component is 5, so x(1) = [0.8, 1, 0.4]. (Key point: First iteration) 2. What does the scale factor λ(k) approach as iterations continue? A. The smallest eigenvalue B. The determinant of A C. The dominant (largest magnitude) eigenvalue D. The trace of A Answer: C. The power method converges to the dominant eigenvalue, here about 3.93. (Key point: Meaning of λ(k)) 3. What is the final answer to 2 decimal places? A. λ ≈ 5.00, vector ≈ [0.80, 1.00, 0.40] B. λ ≈ 3.93, vector ≈ [0.88, 1.00, 0.58] C. λ ≈ 2.30, vector ≈ [0.88, 1.00, 0.58] D. λ ≈ 3.93, vector ≈ [1.00, 0.88, 0.58] Answer: B. Iterations 12 and 13 agree to 2 decimals, giving 3.93 and [0.88, 1.00, 0.58]. (Key point: Accuracy and answer) ## Close variants set in other exams - [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: 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: 1,4,-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: 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 - [Vector between two points in cylindrical](https://www.bctengineer.com/question.md?key=semester_3%2Felectromagnetism%3Avector-a-to-b-cylindrical) — Electromagnetism, chapter 1 - [Cylindrical points to cartesian vector](https://www.bctengineer.com/question.md?key=semester_3%2Felectromagnetism%3Acylindrical-vector-to-cartesian-components) — Electromagnetism, chapter 1 - [Half power points, bandwidth and effect of Q](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_circuit_theory%3Aseries-resonance-bandwidth-q-effect) — Electrical Circuit Theory, chapter 1 - [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