Dr. Jie Zou PHY 33201 Chapter 3 Solution of Simultaneous Linear Algebraic Equations: Lecture (II) Note: Besides the main textbook, also see Ref: Applied.

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Presentation transcript:

Dr. Jie Zou PHY Chapter 3 Solution of Simultaneous Linear Algebraic Equations: Lecture (II) Note: Besides the main textbook, also see Ref: Applied Numerical Methods with MATLAB for Engineers and Scientists, S. Chapra, Ch. 9.

Dr. Jie Zou PHY Outline How to solve small (n  3) sets of linear algebraic equations (does not require a computer)? The graphical method Cramer’s rule How to solve large (n  3) sets of linear algebraic equations (solution algorithms that can be implemented on a computer)? Gauss elimination method-an introduction

Dr. Jie Zou PHY The graphical method Graphical solution of a set of two linear algebraic equations: The solution is represented by the intersection point of the two lines. Notes: Can work if the system is small (n  3); impractical for larger systems. The resulting solution is not very accurate. Useful for visualizing the properties of the solutions.

Dr. Jie Zou PHY Some examples Singular systems: (a) no solution and (b) infinite solutions Ill-conditioned systems: (c) the slopes of the two lines are so close that the point of intersection is difficult to detect visually.

Dr. Jie Zou PHY Cramer’s rule The determinant |A| of a matrix [A]: For small matrices: Notes: For large matrices, the determinants can be very complicated; The determinant |A| is a single number. Minors

Dr. Jie Zou PHY Cramer’s rule (cont.) Solving a small set of linear equations using Cramer’s rule An example: Solving three linear equations given by [A]x = b, where Solution by Cramer’s rule: Notes: Large computational effort; accuracy limited by round-off errors; impractical for large systems. What should be x 2 and x 3 ?

Dr. Jie Zou PHY An example Example 3.9: Find the solution of the following system of equations using Cramer’s rule: x 1 – x 2 + x 3 = 3, 2x 1 + x 2 – x 3 = 0, and3x 1 + 2x 2 + 2x 3 = 15.

Dr. Jie Zou PHY Naïve Gauss Elimination Naïve: The algorithm does not avoid the problem of division by zero. Gauss elimination: Inspiration: Elimination of unknowns Two major steps: (1) Forward elimination (2) Back substitution.

Dr. Jie Zou PHY An example Solving the following set of equations using “elimination of unknowns”; a 11 x 1 + a 12 x 2 + a 13 x 3 = b 1 a 21 x 1 + a 22 x 2 + a 23 x 3 = b 2 a 31 x 1 + a 32 x 2 + a 33 x 3 = b 3 Can we design a systematic solution algorithm so that it can be implemented on a computer? Yes!-Gauss elimination! Continue in Lecture (III)