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Unformatted text preview: MATH 152 Fall 200607 Applied Linear Algebra & Differential Equations Worked Examples Dr. Tony Yee Department of Mathematics The Hong Kong University of Science and Technology September 1, 2006 ii Contents Table of Contents iii 1 Introduction 1 2 FirstOrder Differential Equations 5 3 SecondOrder Linear Equations 39 4 Laplace Transform 83 5 Matrix 127 6 Systems of Linear Equations 135 7 Euclidean Vector 155 8 Eigenvalue and Eigenvector 177 9 Systems of Differential Equations 209 10 Orthogonality 231 iii Chapter 6 Systems of Linear Equations ¥ Example 6.1 (Augmented matrix) Write down the augmented matrix for the system of linear equations. 8 < : x 1 x 3 = 5 , x 1 + x 2 + x 4 = 1 , x 1 3 x 4 = 2 . Solution The augmented matrix is 2 4 1 1 5 1 1 1 1 1 3 2 3 5 . 2 ¥ Example 6.2 (Augmented matrix) Suppose ax + by = 1 represents a straight line which passes through the points (1 , 2) and (3 , 4) . Write down the augmented matrix for the corresponding system of linear equations for a , b . Solution For ax + by = 1 to pass through (1 , 2) and (3 , 4), we have a · 1 + b · 2 = 1 and a · 3 + b · 4 = 1. Then we have two linear equations with two variables a , b . The augmented matrix is » 1 2 1 3 4 1 – . 2 ¥ Example 6.3 (Row echelon form) Which of the following matrices are in row echelon form? A = » 3 4 – , B = 2 4 1 1 1 1 3 5 , C = 2 4 3 1 3 3 5 , P = 2 4 1 1 1 3 5 , Q = 2 4 3 5 , R = 2 4 1 1 1 1 3 5 , X = » 1 1 1 – , Y = » 1 1 1 1 – , Z = 2 4 1 1 1 1 3 5 . Solution Matrices A , C , P , Q , X , Y , Z are in row echelon form. Matrix B is not in row echelon form because the second row which consists of all zero entries is not at the bottom of the matrix. Matrix R is not in row echelon form because the (2 , 1)entry is not zero. 2 135 6. Systems of Linear Equations ¥ Example 6.4 (Gaussian elimination – unique solution) Solve the system of linear equations 8 > < > : x 1 + x 2 2 x 3 = 1 , 2 x 1 +3 x 2 2 x 3 = 2 , 3 x 1 11 x 3 = 4 . Solution We simplify the augmented matrix as follows 2 4 1 1 2 1 2 3 2 2 3 11 4 3 5 2 R 1 + R 2→ 3 R 1 + R 3 2 4 1 1 2 1 1 2 4 3 5 1 3 5 3 R 2 + R 3→ 2 4 1 1 2 1 1 2 4 1 11 3 5 . Interpreted the last matrix as a system of linear equations, we have 8 < : x 1 + x 2 2 x 3 = 1 , x 2 +2 x 3 = 4 , x 3 = 11 . By backward substitution , starting from the third equation, we have x 3 = 11. Substituting this value into the second equation, we have x 2 = 18. Substituting the values of x 2 and x 3 into the first equation, we have x 1 = 39. We conclude that the system has a unique solution ( x 1 ,x 2 ,x 3 ) = ( 39 , 18 , 11). 2 ¥ Example 6.5 (Gaussian elimination – unique solution) Solve the system of linear equations 8 > < > : x 1 +2 x 2 = 7 , 2 x 1 3 x 2 = 7 , 3 x 1 5 x 2 = 12 ....
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