Linear Equations in Linear Algebra

Transcription

1 1 Linear Equations in Linear Algebra 1.4 THE MATRIX EQUATION A = b

2 MATRIX EQUATION A = b m n Definition: If A is an matri, with columns a 1, n, a n, and if is in, then the product of A and, denoted by A, is the linear combination of the columns of A using the corresponding entries in as weights; that is, 1 A [ L ] M n 2 = a a a = a + a a 1 2 n n n A is defined only if the number of columns of A equals the number of entries in.. Slide 1.4-2

3 MATRIX EQUATION A = b Eample 1: For v 1, v 2, v 3 in combination + vector. 3v 5v 7v 1 2 3, write the linear as a matri times a Solution: Place v 1, v 2, v 3 into the columns of a matri A and place the weights 3, 5, and 7 into a vector. That is, 3 3v 5v + 7v = [ v v v ] 5 = A m. Slide 1.4-3

4 MATRIX EQUATION A = b Now, write the system of linear equations as a vector equation involving a linear combination of vectors. For eample, the following system is equivalent to + 2 = = = (1). ----(2) Slide 1.4-4

5 MATRIX EQUATION A = b As in the given eample (1), the linear combination on the left side is a matri times a vector, so that (2) becomes = (3) Equation (3) has the form is called a matri equation. A = b. Such an equation Slide 1.4-5

6 MATRIX EQUATION A = b m n Theorem 3: If A is an matri, with columns m a1,, an, and if b is in, then the matri equation A = has the same solution set as the vector equation which, in turn, has the same solution set as the system of linear equations whose augmented matri is b a + a a = b [ a a L a b] 1 2 n n n., Slide 1.4-6

7 EXISTENCE OF SOLUTIONS A = b The equation 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 matri. Then the following statements are logically equivalent. That is, for a particular A, either they are all true statements or they are all false. A = a. For each b in m, the equation has a solution. b. Each b in m is a linear combination of the columns of A. c. The columns of A span m. d. A has a pivot position in every row. b Slide 1.4-7

8 PROOF OF THEOREM 4 Statements (a), (b), and (c) are logically equivalent. So, it suffices to show (for an arbitrary matri 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 matri [ A b] to an augmented matri [ 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. Slide 1.4-8

9 PROOF OF THEOREM 4 So A = 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, transformed into the form A b. The new system is false. A = b [ ] [ U d] can be is also inconsistent, and (a) Slide 1.4-9

10 COMPUTATION OF A Eample 2: Compute A, where 1 = 2 and. 3 Solution: From the definition, A = = Slide

11 COMPUTATION OF A = = (1) The first entry in the product A is a sum of products (a dot product), using the first row of A and the entries in. Slide

12 COMPUTATION OF A That is,. Similarly, the second entry in A can be calculated by multiplying the entries in the second row of A by the corresponding entries in and then summing the resulting products = = Slide

13 ROW-VECTOR RULE FOR COMPUTING A Likewise, the third entry in A can be calculated from the third row of A and the entries in. If the product A is defined, then the ith entry in A is the sum of the products of corresponding entries from row i of A and from the verte. The matri with 1s on the diagonal and 0s elsewhere is called an identity matri and is denoted by I. For eample, is an identity matri. Slide

14 PROPERTIES OF THE MATRIX-VECTOR PRODUCT A Theorem 5: If A is an matri, u and v are n vectors in, and c is a scalar, then a. b.. m n A(u + v) = Au + Av; A( cu) = c( Au) n = 3 A = a a a Proof: For simplicity, take,, 3 and u, v in. i =1,2,3, For let u i and v i be the ith entries in u and v, respectively. Slide

15 PROPERTIES OF THE MATRIX-VECTOR PRODUCT A A (u + v) To prove statement (a), compute as a linear combination of the columns of A using the entries in u + v as weights. u + v A(u + v) = [ a a a ] u + v u + v = ( u + v )a + ( u + v )a + ( u + v )a Entries in u + v Columns of A = ( u a + u a + u a ) + ( v a + v a + v a ) = Au + Av Slide

16 PROPERTIES OF THE MATRIX-VECTOR PRODUCT A A( cu) To prove statement (b), compute as a linear combination of the columns of A using the entries in cu as weights. cu1 A( cu) = [ a a a ] cu = ( cu )a + ( cu )a + ( cu )a cu3 = c( u a ) + c( u a ) + c( u a ) = c( u a + u a + u a ) = c( Au) Slide