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Course: ENGINEERIN MAT188, Fall 2007
School: University of Toronto
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1 MAT188H1F Chapter Lec03 Burbulla Chapter 1 Lecture Notes Fall 2007 Chapter 1 Lecture Notes MAT188H1F Lec03 Burbulla Chapter 1 Chapter 1 Matrix Multiplicaton Chapter 1 Lecture Notes MAT188H1F Lec03 Burbulla Chapter 1 Matrix Multiplicaton Some Introductory Examples The product ST of the m p matrix S and the q n matrix T is only defined if p = q; that is, the number of columns of S must match the...

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1 MAT188H1F Chapter Lec03 Burbulla Chapter 1 Lecture Notes Fall 2007 Chapter 1 Lecture Notes MAT188H1F Lec03 Burbulla Chapter 1 Chapter 1 Matrix Multiplicaton Chapter 1 Lecture Notes MAT188H1F Lec03 Burbulla Chapter 1 Matrix Multiplicaton Some Introductory Examples The product ST of the m p matrix S and the q n matrix T is only defined if p = q; that is, the number of columns of S must match the number of rows of T . Consider the four matrices 4 1 -1 3 3 7 -2 5 A= ,B = , C = -1 , D = 2 4 -2 2 5 9 4 5 A is 2 3; B is 2 2; C is 3 1; and D is 2 2. Only the following products will be defined: BA, AC , DA, BD, and DB. Why? Because the definition of matrix multiplication requires you to match the elements in the rows of one matrix with the elements in the columns of the other. Matrix multiplication is a complicated matter! but it turns out to be extremely useful. Chapter 1 Lecture Notes MAT188H1F Lec03 Burbulla . Chapter 1 Matrix Multiplicaton Example 1 Consider the product BA. Here is how we multiply them: BA = = = Similarly, DA = -2 5 9 4 1 -1 3 2 4 -2 = 8 22 -16 17 7 19 , 3 7 2 5 1 -1 3 2 4 -2 3(1) + 7(2) 3(-1) + 7(4) 3(3) + 7(-2) 2(1) + 5(2) 2(-1) + 5(4) 2(3) + 5(-2) 17 25 -5 12 18 -4 where I did the calculations in my head. As should you! Chapter 1 Lecture Notes MAT188H1F Lec03 Burbulla Chapter 1 Matrix Multiplicaton Example 2 Both products BD and DB are possible. Let's calculate them: BD = = 3 7 2 5 57 43 41 30 -2 5 9 4 . DB = = -2 5 9 4 4 11 35 83 . 3 7 2 5 Note that BD = DB. Be careful! matrix multiplication has properties different than number multiplication. Chapter 1 Lecture Notes MAT188H1F Lec03 Burbulla Chapter 1 Matrix Multiplicaton Example 3 1 -1 3 2 4 -2 20 -6 4 -1 5 AC = = x Why is such multiplication useful? Consider AX , if X = y : z x 1 -1 3 y AX = 2 4 -2 z = x - y + 3z 2x + 4y - 2z MAT188H1F Lec03 Burbulla Chapter 1 Lecture Notes Chapter 1 Matrix Multiplicaton Thus the system of equations x 2x - y + 4y + 3z - 2z = 20 = -6 can be written in matrix form AX = B, with x 1 -1 3 A= ,X = y ,B = 2 4 -2 z 20 -6 . In general, any system of m linear equations in n variables can be written as a single matrix equation AX = B, with m n coefficient matrix A = [aij ] , x1 b1 x2 b2 X = x3 and B = b3 . ... ... xn bm Chapter 1 Lecture Notes MAT188H1F Lec03 Burbulla Chapter 1 Matrix Multiplicaton The General Definition for the Product AB If A is an m p matrix and B is a p n matrix, then the m n matrix C = [cij ] , defined by cij = ai1 b1j + ai2 b2j + + aip bpj , for 1 i m, 1 j n, is called the product of A and B, and we write C = AB. If you have seen the dot product of vectors in high school, then cij = Ri Cj , where b1j b and Cj = 2j ... bpj Ri = ai1 ai2 . . . aip are the i-th row of A and the j-th column of B, respectively. Chapter 1 Lecture Notes MAT188H1F Lec03 Burbulla Chapter 1 Matrix Multiplicaton Properties of Matrix Multiplication Let c be a scalar and let A, B, C be matrices of appropriate sizes so that the following products can be formed. Then: IA = A and BI = B, where I is an identity matrix. A(BC ) = (AB)C A(B C ) = AB AC (B C )A = BA CA c(AB) = (cA)B = A(cB) (AB)T = B T AT , note the change of order! See the end of Section 1.4 in the textbook for the proofs. Note that AB = BA is not one of the properties listed above. Chapter 1 Lecture Notes MAT188H1F Lec03 Burbulla Chapter 1 Matrix Multiplicaton Some Computational Examples Consider the matrices A= 1 -1 3 2 4 -2 ,B = 3 7 2 5 ,D = -2 5 9 4 . From Example 1, we know BA + DA = = Compare with (B + D)A = 1 12 11 9 1 -1 3 2 4 -2 = 25 47 -21 29 25 15 17 25 -5 12 18 -4 25 47 -21 29 25 15 + 8 22 -16 17 7 19 Chapter 1 Lecture Notes MAT188H1F Lec03 Burbulla Chapter 1 Matrix Multiplicaton From Example 2, we have BD = so (BD)T = Compare with DT BT = Note that BT DT = is totally different. Chapter 1 Lecture Notes MAT188H1F Lec03 Burbulla 57 43 41 30 57 41 43 30 , . -2 9 5 4 3 2 7 5 = 57 41 43 30 . 3 2 7 5 -2 9 5 4 = 4 35 11 83 Chapter 1 Matrix Multiplicaton Powers of a Matrix If A is a square matrix we define A2 = AA, A3 = AAA, etc. So for B= you can check that B2 = and B 3 = B 2B = 23 56 16 3 39 7 2 5 = 181 441 126 307 . 3 7 2 5 3 7 2 5 = 23 56 16 39 3 7 2 5 In general, it is very cumbersome to calculate higher powers of a matrix! Chapter 1 Lecture Notes MAT188H1F Lec03 Burbulla Chapter 1 Matrix Multiplicaton Example 4 Suppose A and B are square matrices. Then (A + B)2 = (A + B)(A + B) = A2 + BA + AB + B 2 . So (A + B)2 = A2 + 2AB + B 2 AB = BA. Similarly, check that (A + B)(A - B) = A2 - B 2 AB = BA. Chapter 1 Lecture Notes MAT188H1F Lec03 Burbulla Chapter 1 Matrix Multiplicaton Example 5 For numbers, the only solutions to the equation x 2 = 1 are x = 1. Things are very different for matrices. Consider the matrix equation A2 = I . Two obvious solutions are A = I , but there are infinitely many more. For example, if A is 2 2: A= Can you find any others? 0 1/c c 0 , for c = 0. Chapter 1 Lecture Notes MAT188H1F Lec03 Burbulla Chapter 1 Matrix Multiplicaton Connections Between Homogeneous and Nonhomogeneous Linear Systems As we have seen, the general system of m linear equations in n variables can be represented by a single matrix equation: AX = B. The corresponding homogeneous system is AX = O. Theorem 2, page 31: Suppose X0 is any particular solution to the system AX = B. 1. If Y is any solution to the system AX = O, then X = X0 + Y is a solution to the system AX = B. 2. Every solution X to AX = B has the form X = X0 + Y , for some solution Y to the system AX = O. Chapter 1 Lecture Notes MAT188H1F Lec03 Burbulla Chapter 1 Matrix Multiplicaton Proof of Theorem 2, page 31 We are given that AX0 = B. Part 1: Assume AY = O, and let X = X0 + Y . Then AX = A(X0 + Y ) = AX0 + AY = B + O = B. Part 2: Let AX = B. Then AX - AX0 = B - B = O. On the other hand, AX - AX0 = A(X - X0 ). Thus A(X - X0 ) = O Y = X - X0 is a solution to AX = O; so X = X0 + Y . Chapter 1 Lecture Notes MAT188H1F Lec03 Burbulla Chapter 1 Matrix Multiplicaton Example 5: From Example 9 of Section 1.2 Consider the system x1 -x1 2x1 The solution was of equations - 2x2 + x3 + x4 = 1 + 2x2 + x4 = 3 - 4x2 + x3 = -2 x1 -3 + 2s + t x2 s = X = x3 4 - 2t x4 t , for s, t R Chapter 1 Lecture Notes MAT188H1F Lec03 Burbulla Chapter 1 Matrix Multiplicaton Write the solution as -3 2s + t 0 s + X = 4 -2t 0 t you can take -3 2s + t 0 s and Y = X0 = 4 -2t 0 t You can check that 1 AX0 = 3 and that AY = O. -2 Chapter 1 Lecture Notes MAT188H1F Lec03 Burbulla ; . Chapter 1 Matrix Multiplicaton Block Multiplication Sometimes it is convenient to partition the entries of a matrix into blocks. Especially if there are lots of zeros in the matrix. Consider the matrix 2 3 -1 5 6 P Q , A = 1 4 5 -2 3 = O R 0 0 2 7 3 where the 2 2 matrix P, the 2 3 matrix Q, the 1 2 matrix O, and the 1 3 matrix R are all called blocks. If 2 3 -2 5 U B= 1 , 3 = V 5 -1 4 2 where U and V are also blocks, then AB can be calculated as Chapter 1 Lecture Notes MAT188H1F Lec03 Burbulla Chapter 1 Matrix Multiplicaton AB = = = P Q O R PU + QV OU + RV PU + QV RV U V -2 + 48 21 + 4 = -6 + 7 23 + 23 49 5 46 25 = 1 46 , 49 5 as you may check. Chapter 1 Lecture Notes MAT188H1F Lec03 Burbulla Chapter 1 Matrix Multiplicaton Rows and Columns as Blocks If R1 R A = 2 and B = ... Rm C1 C2 C3 . . . Cn , then AB = [Ri Cj ] , for 1 i m, 1 j n. More interestingly, if A= C1 C2 C3 . . . Cn and X = x1 x2 x3 ... xn , Chapter 1 Lecture Notes MAT188H1F Lec03 Burbulla Chapter 1 Matrix Multiplicaton then x1 x2 C1 C2 C3 . . . Cn x3 = ... xn = x1 C1 + x2 C2 + x3 C3 + + xn Cn , AX as you may check. Thus the system AX = B has a solution x1 x2 X = x3 x1 C1 + x2 C2 + x3 C3 + + xn Cn = B, ... xn which is a new way of looking at a solution to a system of linear equations. Chapter 1 Lecture Notes MAT188H1F Lec03 Burbulla Chapter 1 Matrix Multiplicaton Back to Example 4 of Section 1.2 The system of equations 2x + y x + y -3x + 2y - z + z + 4z = 1 = 6 = 13 can be rewritten as 2 1 -1 1 x 1 +y 1 +z 1 = 6 xC1 +yC2 +zC3 = B. -3 2 4 13 The solution was x 1 y = 2 C1 + 2C2 + 3C3 = B. z 3 This is a preview of results to come in Chapter 4. Chapter 1 Lecture Notes MAT188H1F Lec03 Burbulla
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