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1 Linear Equations in Linear Algebra 1.4 THE MATRIX EQUATION Ax b © 2012 Pearson Education, Inc. MATRIX EQUATION Ax b Definition: If A is an m n matrix, with columns a1, n …, an, and if x is in , then the product of A and x, denoted by Ax, is the linear combination of the columns of A using the corresponding entries in x as weights; that is, Ax a1 a 2 x1 x 2 a n x1a1 x2a 2 ... xna n . x n Ax is defined only if the number of columns of A equals the number of entries in x. © 2012 Pearson Education, Inc. Slide 1.4- 2 MATRIX EQUATION Ax b m Example 1: For v1, v2, v3 in , write the linear combination 3v1 5v 2 7v3 as a matrix times a vector. Solution: Place v1, v2, v3 into the columns of a matrix A and place the weights 3, 5 , and 7 into a vector x. That is, 3v1 5v 2 7v3 v1 © 2012 Pearson Education, Inc. v2 3 v3 5 Ax . 7 Slide 1.4- 3 MATRIX EQUATION Ax b Now, write the system of linear equations as a vector equation involving a linear combination of vectors. For example, the following system x1 2 x2 x3 4 ----(1) 5 x2 3x3 1 is equivalent to 1 2 1 4 x1 x2 x3 . 0 5 3 1 © 2012 Pearson Education, Inc. ----(2) Slide 1.4- 4 MATRIX EQUATION Ax b As in the given example (1), the linear combination on the left side is a matrix times a vector, so that (2) becomes x1 1 2 1 4 0 5 3 x2 1. x 3 ----(3) Equation (3) has the form Ax b. Such an equation is called a matrix equation. © 2012 Pearson Education, Inc. Slide 1.4- 5 MATRIX EQUATION Ax b Theorem 3: If A is an m n matrix, with columns m a1, …, an, and if b is in , then the matrix equation Ax b has the same solution set as the vector equation x1a1 x2a 2 ... xn an b , which, in turn, has the same solution set as the system of linear equations whose augmented matrix is a n b. a1 a 2 © 2012 Pearson Education, Inc. Slide 1.4- 6 EXISTENCE OF SOLUTIONS The equation Ax b has a solution if and only if b is a linear combination of the columns of A. Theorem 4: Let A be an m n matrix. Then the following statements are logically equivalent. That is, for a particular A, either they are all true statements or they are all false. m a. For each b in , the equation Ax b has a solution. b. Each b in m is a linear combination of the columns of A. m c. The columns of A span . d. A has a pivot position in every row. © 2012 Pearson Education, Inc. Slide 1.4- 7 PROOF OF THEOREM 4 Statements (a), (b), and (c) are logically equivalent. So, it suffices to show (for an arbitrary matrix A) that (a) and (d) are either both true or false. Let U be an echelon form of A. m Given b in , we can row reduce the augmented matrix A b to an augmented matrix U d for m some d in : A b ... U d If statement (d) is true, then each row of U contains a pivot position, and there can be no pivot in the augmented column. © 2012 Pearson Education, Inc. Slide 1.4- 8 PROOF OF THEOREM 4 So Ax b has a solution for any b, and (a) is true. If (d) is false, then the last row of U is all zeros. Let d be any vector with a 1 in its last entry. Then U d represents an inconsistent system. Since row operations are reversible, U d can be transformed into the form A b. The new system Ax b is also inconsistent, and (a) is false. © 2012 Pearson Education, Inc. Slide 1.4- 9 COMPUTATION OF Ax 3 4 2 Example 2: Compute Ax, where A 1 5 3 x1 6 2 8 and x x2 . x3 Solution: From the definition, 3 4 x1 2 2 3 4 1 5 3 x x 1 x 5 x 3 2 1 2 3 6 2 8 x3 6 2 8 © 2012 Pearson Education, Inc. Slide 1.4- 10 COMPUTATION OF Ax 2 x1 3x2 4 x3 x1 5 x2 3x3 ---(1) 6 x1 2 x2 8 x3 2 x1 3x2 4 x3 x1 5 x2 3x3 6 x1 2 x2 8 x3 . The first entry in the product Ax is a sum of products (a dot product), using the first row of A and the entries in x. © 2012 Pearson Education, Inc. Slide 1.4- 11 COMPUTATION OF Ax 2 3 4 x1 2 x1 3x2 4 x3 x . That is, 2 x3 Similarly, the second entry in Ax can be calculated by multiplying the entries in the second row of A by the corresponding entries in x and then summing the resulting products. x1 1 5 3 x x 5 x 3x 2 3 2 1 x3 © 2012 Pearson Education, Inc. Slide 1.4- 12 ROW-VECTOR RULE FOR COMPUTING Ax Likewise, the third entry in Ax can be calculated from the third row of A and the entries in x. If the product Ax is defined, then the ith entry in Ax is the sum of the products of corresponding entries from row i of A and from the vertex x. The matrix with 1s on the diagonal and 0s elsewhere is called an identity matrix and is denoted by I. 1 0 0 For example, 0 1 0 is an identity matrix. 0 0 1 © 2012 Pearson Education, Inc. Slide 1.4- 13 PROPERTIES OF THE MATRIX-VECTOR PRODUCT Ax Theorem 5: If A is an m n matrix, u and v are n vectors in , and c is a scalar, then a. A(u v) Au Av; b. A(cu) c( Au). Proof: For simplicity, take n 3, A a1 a 2 a 3 , 3 and u, v in . For i 1,2,3, let ui and vi be the ith entries in u and v, respectively. © 2012 Pearson Education, Inc. Slide 1.4- 14 PROPERTIES OF THE MATRIX-VECTOR PRODUCT Ax To prove statement (a), compute A(u v) as a linear combination of the columns of A using the entries in u v as weights. A(u v) a1 a 2 u1 v1 a 3 u2 v2 u3 v3 Entries in (u1 v1 )a1 (u2 v2 )a 2 (u3 v3 )a 3 uv Columns of A (u1a1 u2a 2 u3a 3 ) (v1a1 v2a 2 v3a 3 ) Au Av © 2012 Pearson Education, Inc. Slide 1.4- 15 PROPERTIES OF THE MATRIX-VECTOR PRODUCT Ax To prove statement (b), compute A(cu) as a linear combination of the columns of A using the entries in cu as weights. cu1 A(cu) a1 a 2 a 3 cu2 (cu1 )a1 (cu2 )a 2 (cu3 )a 3 cu3 c(u1a1 ) c(u2a 2 ) c(u3a 3 ) c(u1a1 u2a 2 u3a 3 ) c( Au) © 2012 Pearson Education, Inc. Slide 1.4- 16