--- title: "Power method: 15,-4,-3 matrix" subject: "Numerical Method" chapter: 3 key: "semester_4/numerical_method:power-method-15-4-3" source: https://www.bctengineer.com/q/semester_4/numerical_method/power-method-15-4-3 site: BCT Engineering --- # Power method: 15,-4,-3 matrix Numerical Method · Chapter 3: Solution of system of linear algebraic equations - Asked in: 2073 Bhadra, 8 marks - Video: [By Using the Power method find the Eigenvalues And EigenVector of square matrix||Numerical Analysis](https://www.youtube.com/watch?v=hfAIhoc67ZE) — Alfaizan academy - 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-15-4-3 - Whole chapter: https://www.bctengineer.com/md/semester_4/numerical_method/chapter-3 ## Question Find the dominant Eigen value and the corresponding Eigen vector of the given matrix using power method. [15 -4 -3; -10 12 -6; -20 4 -2] ## Short answer Power method with initial [1,1,1], scaling by third component after sign stabilization. Dominant eigenvalue converges to 20.0000, eigenvector to [-1.0000, 0.5000, 1.0000]. ## 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 one of its components. The scale factor tends to λ and the scaled vector tends to the eigenvector. #### Step 1: Matrix and starting vector ```text A = 15 -4 -3 -10 12 -6 -20 4 -2 ``` Take x(0) = [1, 1, 1]ᵀ. Iterate: A·x(k) = λ(k)·x(k+1), where λ(k) is the scale factor. For k = 1 to 3, λ(k) is the component of largest magnitude in A·x(k). From k = 4 on, the first and third components of A·x(k) are almost equal in size and opposite in sign, so we always scale by the third component. This keeps the vector from flipping sign at every step. (Scaling by whichever of the two happens to be slightly larger would make the vector change sign each time and give the misleading estimate −20.) #### Step 2: Iterations | k | x(k) | A·x(k) | λ(k) (scale factor) | x(k+1) = A·x(k)/λ(k) | |---|---|---|---|---| | 1 | [1.0000, 1.0000, 1.0000] | [8.0000, -4.0000, -18.0000] | -18.0000 | [-0.4444, 0.2222, 1.0000] | | 2 | [-0.4444, 0.2222, 1.0000] | [-10.5556, 1.1111, 7.7778] | -10.5556 | [1.0000, -0.1053, -0.7368] | | 3 | [1.0000, -0.1053, -0.7368] | [17.6316, -6.8421, -18.9474] | -18.9474 | [-0.9306, 0.3611, 1.0000] | | 4 | [-0.9306, 0.3611, 1.0000] | [-18.4028, 7.6389, 18.0556] | 18.0556 | [-1.0192, 0.4231, 1.0000] | | 5 | [-1.0192, 0.4231, 1.0000] | [-19.9808, 9.2692, 20.0769] | 20.0769 | [-0.9952, 0.4617, 1.0000] | | 6 | [-0.9952, 0.4617, 1.0000] | [-19.7749, 9.4923, 19.7510] | 19.7510 | [-1.0012, 0.4806, 1.0000] | | 7 | [-1.0012, 0.4806, 1.0000] | [-19.9406, 9.7793, 19.9467] | 19.9467 | [-0.9997, 0.4903, 1.0000] | | 8 | [-0.9997, 0.4903, 1.0000] | [-19.9565, 9.8803, 19.9550] | 19.9550 | [-1.0001, 0.4951, 1.0000] | | 9 | [-1.0001, 0.4951, 1.0000] | [-19.9816, 9.9423, 19.9820] | 19.9820 | [-1.0000, 0.4976, 1.0000] | | 10 | [-1.0000, 0.4976, 1.0000] | [-19.9900, 9.9705, 19.9899] | 19.9899 | [-1.0000, 0.4988, 1.0000] | | 11 | [-1.0000, 0.4988, 1.0000] | [-19.9952, 9.9854, 19.9952] | 19.9952 | [-1.0000, 0.4994, 1.0000] | | 12 | [-1.0000, 0.4994, 1.0000] | [-19.9975, 9.9927, 19.9975] | 19.9975 | [-1.0000, 0.4997, 1.0000] | | 13 | [-1.0000, 0.4997, 1.0000] | [-19.9988, 9.9963, 19.9988] | 19.9988 | [-1.0000, 0.4998, 1.0000] | | 14 | [-1.0000, 0.4998, 1.0000] | [-19.9994, 9.9982, 19.9994] | 19.9994 | [-1.0000, 0.4999, 1.0000] | | 15 | [-1.0000, 0.4999, 1.0000] | [-19.9997, 9.9991, 19.9997] | 19.9997 | [-1.0000, 0.5000, 1.0000] | | 16 | [-1.0000, 0.5000, 1.0000] | [-19.9998, 9.9995, 19.9998] | 19.9998 | [-1.0000, 0.5000, 1.0000] | | 17 | [-1.0000, 0.5000, 1.0000] | [-19.9999, 9.9998, 19.9999] | 19.9999 | [-1.0000, 0.5000, 1.0000] | | 18 | [-1.0000, 0.5000, 1.0000] | [-20.0000, 9.9999, 20.0000] | 20.0000 | [-1.0000, 0.5000, 1.0000] | From the last two rows, λ and the vector are unchanged to 4 decimal places. Check: A·[−1, 0.5, 1]ᵀ = [−20, 10, 20]ᵀ = 20·[−1, 0.5, 1]ᵀ. **Answer:** dominant eigenvalue λ ≈ 20.0000; corresponding eigenvector ≈ [-1.0000, 0.5000, 1.0000]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. In each power-method step, what is done to A·x(k) to get x(k+1)? A. It is divided by its component of largest magnitude, which is also the estimate of the eigenvalue B. It is divided by its smallest component C. It is multiplied by the determinant of A D. It is subtracted from x(k) Answer: A. The scale factor λ(k) is the largest-magnitude component; it tends to the dominant eigenvalue and the scaled vector tends to the eigenvector. (Key point: Scaling step) 2. For x(0) = [1,1,1], A·x(0) = [8, -4, -18]. What is λ(1) and the new vector? A. λ(1) = 8, vector [1, -0.5, -2.25] B. λ(1) = -18, vector [-0.4444, 0.2222, 1.0000] C. λ(1) = 18, vector [0.4444, -0.2222, -1] D. λ(1) = -4, vector [-2, 1, 4.5] Answer: B. The largest magnitude component is -18 (sign kept). Dividing [8,-4,-18] by -18 gives [-0.4444, 0.2222, 1]. (Key point: First iteration) 3. What is the final dominant eigenvalue and eigenvector? A. λ ≈ 15, vector ≈ [1, 1, 1] B. λ ≈ -18, vector ≈ [-0.4444, 0.2222, 1] C. λ ≈ 20, vector ≈ [-1, 0.5, 1] D. λ ≈ 12, vector ≈ [-1, 0.5, 1] Answer: C. The iterations settle at λ = 20 with [-1, 0.5, 1]; check: A·[-1,0.5,1] = [-20, 10, 20] = 20·[-1,0.5,1]. (Key point: Result) 4. Why does the answer stop at iteration 18? A. λ and the vector are unchanged to 4 decimal places between the last two rows B. The vector became [1,1,1] again C. The determinant became zero D. Exactly 10 iterations are always required Answer: A. Iteration stops when successive values agree to the required accuracy. 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