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Slide 9.2- 1 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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OBJECTIVES Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Matrix Algebra Learn the definition of equality of matrices. Learn the definition of matrix addition and scalar multiplication. Learn the definition of matrix multiplication. Learn an application of matrix multiplication to computer graphics. SECTION 9.2 1 2 3 4
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Slide 9.2- 3 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EQUALITY OF MATRICES Two matrices A = [a ij ] and B = [b ij ] are said to be equal, written A = B, if 1.A and B have the same order m 5 n (that is, A and B have the same number m of rows and the same number n of columns.) 2. a ij = b ij for all i and j. (The (i, j)th entry of A is equal to the corresponding (i, j)th entry of B.)
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Slide 9.2- 4 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley MATRIX ADDITION If A =[a ij ] and B = [b ij ] are two m 5 n matrices, their sum A + B is the matrix defined by the m 5 n matrix defined by for all i and j.
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Slide 9.2- 5 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley SCALAR MULTIPLICATION If A = [a ij ] be an m 5 n matrices, and let c be a real number. Then the scalar product of A and c is denoted by cA and is defined by
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Slide 9.2- 6 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley MATRIX SUBTRACTION If A and B are two m 5 n matrices, their difference is defined by Subtraction A – B is performed by subtracting the corresponding entries of B from A.
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Slide 9.2- 7 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley MATRIX ADDITION AND SCALAR MULTIPLICATION PROPERTIES Let A, B, and C be m 5 n matrices and c and d be scalars. 1. A + B = B + A 2. A + (B + C) = (A + B) + C 3. A + 0 = 0 + A = A 4. A + (–A) = (–A) + A = 0 5. (cd)A = c(dA) 6. 1A = A 7. c(A + B) = cA + cB 8. (c + d)A = cA + dA
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Slide 9.2- 8 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 4 Solving a Matrix Equation Solve the matrix equation 3A + 2X = 4B for X, Solution where Substitute for A and B.
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Slide 9.2- 9 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 4 Solving a Matrix Equation Solution continued You should check that the matrix X satisfies the given matrix equation.
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Slide 9.2- 10 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley RULE FOR DEFINING THE PRODUCT AB In order to define the product AB of two matrices A and B, the number of columns of A must be equal to the number of rows of B. If A is an m 5 p matrix and B is a p 5 n matrix, then the product AB is an m 5 n matrix.
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Slide 9.2- 11 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley PRODUCT OF 1 x n AND n x 1 MATRICES Suppose is a 1 5 n row matrix and is a n 5 1 column matrix. We define the product
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Slide 9.2- 12 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley MATRIX MULTIPLICATION Let A = [a ij ] be an m 5 p matrix and B = [b ij ] be a p 5 n matrix. Then the product AB is the m 5 n matrix C = [c ij ], where the entry c ij of C is obtained by multiplying the ith row (matrix) of A by the jth column (matrix) of B. The definition of the product AB says that
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Slide 9.2- 13 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 8 Finding the Product of Two Matrices Find the products AB and BA, where Solution Since A is of order 2 5 2 and the order of B is 2 5 3, the product AB is defined and has order 2 5 3. If AB = C = [c ij ], each entry c ij of C is obtained by multiplying the ith row of A by the jth column of B.
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Slide 9.2- 14 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 8 Finding the Product of Two Matrices Solution continued C 11 is obtained by multiplying the first row of A by the first column of B. Thus
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Slide 9.2- 15 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 8 Finding the Product of Two Matrices Solution continued The product BA is not defined, because B is of order 2 5 3 and A is of order 2 5 2; that is, the number of columns of B is not the same as the number of rows of A.
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Slide 9.2- 16 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley PROPERTIES OF MATRIX MULTIPLICATION Let A, B, and C be matrices and let c be a scalar. Assume that each product and sum is defined. Then 1.(AB)C = A(BC) 2.(i) A(B + C) = AB + AC (ii) (A + B)C = AC + BC 3. c(AB) = (cA)B = A(cB)
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Slide 9.2- 17 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 10 Transforming a Letter The capital letter L in the figure is determined by six points (or vertices) P 1 – P 6. The coordinates of the six points can be stored in a data matrix D, together with instructions stating that these vertices are connected by lines.
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Slide 9.2- 18 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 10 Transforming a Letter Given x-coordinate y-coordinate Vertex compute AD. Graph the figure corresponding to the matrix AD by connecting the images of the vertices with the
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Slide 9.2- 19 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 10 Transforming a Letter Solution The columns of the product matrix AD represent the images of the vertices of the letter L. appropriate lines. This produces the complete transformed image of the original figure. …
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Slide 9.2- 20 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 10 Transforming a Letter Solution continued The transformed vertices are plotted on the next slide, along with connecting line segments that correspond to those in the original figure. …
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Slide 9.2- 21 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 10 Transforming a Letter Solution continued The figure is a result of the transformation represented by the product matrix AD.
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