examples152-ch5

examples152-ch5 - MATH 152 Fall 2006-07 Applied Linear...

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MATH 152 Fall 2006-07 Worked Examples Dr. Tony Yee Department of Mathematics The Hong Kong University of Science and Technology September 1, 2006
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Contents Table of Contents iii 1 Introduction 1 2 First-Order Differential Equations 5 3 Second-Order 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
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Chapter 5 Matrix ¥ Example 5.1 (Matrix) The formula a ij = 1 / ( i + j ) for 1 6 i 6 3, 1 6 j 6 4 defines a 3 × 4 matrix A = [ a ij ], namely A = 2 6 6 4 1 2 1 3 1 4 1 5 1 3 1 4 1 5 1 6 1 4 1 5 1 6 1 7 3 7 7 5 . 2 ¥ Example 5.2 (Addition of matrices) Given the following two matrices. Then compute A - 5 B . A = » 3 - 2 - 9 1 , B = » - 4 1 0 - 5 . Solution We first multiply all the entries of B by 5 then subtracted corresponding entries to get the entries in the new matrix. A - 5 B = » 3 - 2 - 9 1 - 5 » - 4 1 0 - 5 = » 3 - 2 - 9 1 - » - 20 5 0 - 25 = » 23 - 7 - 9 26 . 2 ¥ Example 5.3 (Multiplication of matrices) Given the following three matrices. Then compute the products BA and CA , if possible. A = 2 4 1 3 - 1 2 0 4 - 7 1 5 3 5 , B = 2 4 - 4 2 0 1 1 3 3 5 , C = 2 4 1 3 8 - 1 5 3 0 0 7 3 5 . Solution BA is not defined. B is 3 × 2 and A is 3 × 3, therefore the number of columns of B does not match the number of rows of A . It is important to note that just because we cannot compute BA doesn’t mean that we cannot compute AB . Even if we can compute both AB and BA they would probably not be the same matrix. That is, AB 6 = BA in general. CA = 2 4 1 3 8 - 1 5 3 0 0 7 3 5 2 4 1 3 - 1 2 0 4 - 7 1 5 3 5 = 2 4 1 + 6 - 56 3 + 0 + 8 - 1 + 12 + 40 - 1 + 10 - 21 - 3 + 0 + 3 1 + 20 + 15 0 + 0 - 49 0 + 0 + 7 0 + 0 + 35 3 5 = 2 4 - 49 11 51 - 12 0 36 - 49 7 35 3 5 . 2 127
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5. Matrix ¥ Example 5.4 (Multiplication of matrices) Given the matrices A = » 2 - 1 1 1 2 1 , B = 2 4 3 1 - 1 3 5 , C = ˆ 1 - 1 ˜ . Find the products ABC , CAB . Solution ABC = » 2 - 1 1 1 2 1 2 4 3 1 - 1 3 5 ˆ 1 - 1 ˜ = » 4 4 ˆ 1 - 1 ˜ = » 4 - 4 4 - 4 , CAB = ˆ 1 - 1 ˜ » 2 - 1 1 1 2 1 2 4 3 1 - 1 3 5 = ˆ 1 - 3 0 ˜ 2 4 3 1 - 1 3 5 = ˆ 0 ˜ = 0 . 2 ¥ Example 5.5 (Multiplication of matrices) Find the products 2 4 2 0 - 1 3 1 - 2 3 5 » 1 0 3 - 1 , 2 4 1 0 3 3 5 ˆ 1 2 - 1 4 ˜ , 2 4 1 1 0 0 1 0 0 0 1 3 5 100 . Solution 2 4 2 0 - 1 3 1 - 2 3 5 » 1 0 3 - 1 = 2 4 2 0 8 - 3 - 5 2 3 5 , 2 4 1 0 3 3 5 ˆ 1 2 - 1 4 ˜ = 2 4 1 2 - 1 4 0 0 0 0 3 6 - 3 12 3 5 , 2 4 1 1 0 0 1 0 0 0 1 3 5 2 = 2 4 1 2 0 0 1 0 0 0 1 3 5 , 2 4 1 1 0 0 1 0 0 0 1 3 5 3 = 2 4 1 3 0 0 1 0 0 0 1 3 5 , ··· , 2 4 1 1 0 0 1 0 0 0 1 3 5 100 = 2 4 1 100 0 0 1 0 0 0 1 3 5 . 2 ¥ Example 5.6 (Multiplication of matrices) Show that C n = C for all n > 1 . C = 2 4 40 - 80 200 2 - 4 10 - 7 14 - 35 3 5 .
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examples152-ch5 - MATH 152 Fall 2006-07 Applied Linear...

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