1 Linear Equations in Linear Algebra 1.4 THE MATRIX EQUATION A = b
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 1 1 2 2 n n A is defined only if the number of columns of A equals the number of entries in.. Slide 1.4-2
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 1 2 3 1 2 3 7 m. Slide 1.4-3
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 = 4 1 2 3 5 + 3 = 1 2 3 1 2 1 4 1 2 3 0 + + = 5 3 1 ----(1). ----(2) Slide 1.4-4
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 1 1 2 1 4 2 0 5 3 = 1 3. ----(3) Equation (3) has the form is called a matri equation. A = b. Such an equation Slide 1.4-5
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 1 1 2 2 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
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
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
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
COMPUTATION OF A Eample 2: Compute A, where 1 = 2 and. 3 Solution: From the definition, 2 3 4 A = 1 5 3 6 2 8 2 3 4 2 3 4 1 1 5 3 = 1 + 5 + 3 2 1 2 3 6 2 8 6 2 8 3 Slide 1.4-10
COMPUTATION OF A 2 3 4 1 2 3 = + 5 + 3 1 2 3 6 2 8 1 2 3 2 + 3 + 4 = + 5 3 6 2 + 8 1 2 3 1 2 3 1 2 3. ---(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 1.4-11
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. 2 3 4 2 3 4 1 + + 1 2 3 = 2 3 1 1 5 3 = + 5 3 2 1 2 3 3 Slide 1.4-12
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, 1 0 0 0 1 0 0 0 1 is an identity matri. Slide 1.4-13
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 1 2 3 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 1.4-14
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 1 1 1 2 3 2 2 3 3 = ( u + v )a + ( u + v )a + ( u + v )a 1 1 1 2 2 2 3 3 3 Entries in u + v 1 1 2 2 3 3 1 1 2 2 3 3 Columns of A = ( u a + u a + u a ) + ( v a + v a + v a ) = Au + Av Slide 1.4-15
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 ) 1 2 3 2 1 1 2 2 3 3 1 1 2 2 3 3 = c( u a + u a + u a ) = c( Au) 1 1 2 2 3 3 Slide 1.4-16