--- title: "Power method: 25,1,2 matrix" subject: "Numerical Method" chapter: 3 key: "semester_4/numerical_method:power-method-25-1-2" source: https://www.bctengineer.com/q/semester_4/numerical_method/power-method-25-1-2 site: BCT Engineering --- # Power method: 25,1,2 matrix Numerical Method · Chapter 3: Solution of system of linear algebraic equations - Asked in: 2071 Bhadra, 8 marks - Video: [Rayleigh's Power Method | Largest eigen value and eigen vector | Engineering mathematics|Mathspedia|](https://www.youtube.com/watch?v=rwa_o0FbAhc) — Mathspedia - 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-25-1-2 - 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. [25 1 2; 1 3 0; 2 0 -4] ## Short answer Starting from [1,1,1] and iterating with the power method, scaling by the largest component gives the dominant eigenvalue 25.1821 and eigenvector [1.0000, 0.0451, 0.0685] (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 = 25 1 2 1 3 0 2 0 -4 ``` 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] | [28.0000, 4.0000, -2.0000] | 28.0000 | [1.0000, 0.1429, -0.0714] | | 2 | [1.0000, 0.1429, -0.0714] | [25.0000, 1.4286, 2.2857] | 25.0000 | [1.0000, 0.0571, 0.0914] | | 3 | [1.0000, 0.0571, 0.0914] | [25.2400, 1.1714, 1.6343] | 25.2400 | [1.0000, 0.0464, 0.0647] | | 4 | [1.0000, 0.0464, 0.0647] | [25.1759, 1.1392, 1.7410] | 25.1759 | [1.0000, 0.0453, 0.0692] | | 5 | [1.0000, 0.0453, 0.0692] | [25.1836, 1.1358, 1.7234] | 25.1836 | [1.0000, 0.0451, 0.0684] | | 6 | [1.0000, 0.0451, 0.0684] | [25.1820, 1.1353, 1.7263] | 25.1820 | [1.0000, 0.0451, 0.0686] | | 7 | [1.0000, 0.0451, 0.0686] | [25.1822, 1.1353, 1.7258] | 25.1822 | [1.0000, 0.0451, 0.0685] | | 8 | [1.0000, 0.0451, 0.0685] | [25.1821, 1.1352, 1.7259] | 25.1821 | [1.0000, 0.0451, 0.0685] | From the last two rows, λ and the vector are unchanged to 4 decimal places. **Answer:** dominant eigenvalue λ ≈ 25.1821; corresponding eigenvector ≈ [1.0000, 0.0451, 0.0685]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 = [25 1 2; 1 3 0; 2 0 -4], x(0) = [1,1,1], what is x(1)? A. [1, 0.25, -0.5] B. [1, 0.1429, -0.0714] C. [0.8929, 0.1429, -0.0714] D. [28, 4, -2] Answer: B. A·x(0) = [28, 4, -2]; dividing by 28 gives [1, 0.1429, -0.0714]. (Key point: Starting step) 2. For x(2) = [1, 0.1429, -0.0714], A·x(2) = [25, 1.4286, 2.2857]. What is λ(2)? A. 2.2857 B. 1.4286 C. 25 D. 28 Answer: C. λ(k) is the largest-magnitude component of A·x(k), which is 25. (Key point: Scale factor in iteration 2) 3. What does the method give as the dominant eigenvalue and eigenvector? A. λ ≈ 25.1821, vector ≈ [1, 0.0451, 0.0685] B. λ ≈ 28, vector ≈ [1, 0.1429, -0.0714] C. λ ≈ -4, vector ≈ [0, 0, 1] D. λ ≈ 3, vector ≈ [0, 1, 0] Answer: A. λ and the vector are unchanged to 4 decimals at iterations 7 and 8. 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