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Brief Introduction to Vectors and Matrices

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Brief Introduction to Vectors and Matrices CHAPTER 1 Brief Introduction to Vectors and Matrices In this chapter, we will discuss some needed concepts found in in- troductory course in linear algebra. We will introduce matrix, vector, vector-valued function, and linear independency of a group of vectors and vector-valued functions. 1. Vectors and Matrices A matrix is a group of numbers(elements) that are arranged in rows and columns. In general, an m £ n matrix is a rectangular array of mn numbers (or elements) arranged in m rows and n columns. If m = n the matrix is called a square matrix. For example a 2£2 matrix is · ¸ a11 a12 a21 a22 and an 3 £ 3 matrix is 2 3 a11 a12 a13 4 a21 a22 a23 5 a31 a32 a33 Generally, we use bold phase letter, like A, to denote a matrix, and lower case letters with subscripts, like aij; to denote element of th th a matrix. Here aij would be the element at i row and j column. st So a11 is an element at 1 row and column. Sometime we use the abbreviation A = (aij) for a matrix with elements aij: 1.1. Special matrices. 0 denotes the zero matrix whose elements are all zeroes. So 2 £ 2 and 3 £ 3 zero matrices are 2 3 · ¸ 0 0 0 0 0 and 4 0 0 0 5 0 0 0 0 0 Another special matrix is the identity matrix, denoted by I, a iden- tity matrix is an matrix whose main diagonal elements are 1, and all 1 2 1. BRIEF INTRODUCTION TO VECTORS AND MATRICES other elements are 0. So 2 £ 2 and 3 £ 3 zero matrices are 2 3 · ¸ 1 0 0 1 0 and 4 0 1 0 5 0 1 0 0 1 A vector is a matrix with one row or one column. In this chapter, a vector is always a matrix with one column as · ¸ x1 x2 for a two-dimensional vector and 2 3 x1 4 x2 5 x3 for a three dimensional vector. Here the element has only one index that denotes the row position (Sometimes we use different variable to denote number in different position such as using · ¸ x y for a 2-dimensional vector). We use bold lower case, such as v, to denote a vector. 1.2. Operations on Matrices. Arrange number in rectangular fashion, as a matrix, itself is not something terribly interesting. The most important advantage from that kind arrangement is that we can define matrix addition, multiplication, and scalar multiplication. Definition 1.1. (i) Equality: Two matrix A = (aij) and B = (bij) are equal if corresponding elements are equal, i.e. aij = bij: (ii) Addition: If A = (aij) and B = (bij) and the sum of Aand B is A + B = (cij) = aij + bij: (iii) Scalar Product: If A = (aij) is matrix and k is num- ber(scalar), the kA = (kaij) is product of k and A. From the above definition, we see that, to multiply a matrix by a number k, we simply multiply each of its entries by k; to add two matrices we just add their corresponding entries; A¡B = A+(¡1)B. Example 1.1. Let · ¸ 2 3 A = ¡1 4 1. VECTORS AND MATRICES 3 and · ¸ 0 5 B = ; 3 ¡4 find (a) A + B, (b) 3A, (c) 4A ¡ B: Solution (a) · ¸ · ¸ 2 3 0 5 A + B = + ¡1 4 3 ¡4 · ¸ · ¸ 2 + 0 3 + 5 2 8 = = ¡1 + 3 4 + (¡4) 2 0 (b) · ¸ · ¸ 2 3 6 9 3A = 3 = ¡1 4 ¡3 12 (c) · ¸ · ¸ · ¸ 8 12 0 5 8 7 4A ¡ B = ¡ = ¡4 16 3 ¡4 ¡7 20 a The following fact lists all properties of matrix addition and scalar product. Theorem 1.1. Let A; bB; and C be matrices. Let a; b be scalars (numbers). We have (1) A + 0 = 0 + A = A; A ¡ A = 0; (2) A + B = B + A (commutativity); (3) A + (B + C) = (A + B) + C; (ab)A = a(bA) (associa- tivity); (4) a(A+B) = aA+aB; (a+b)A = aA+bA (distributivity) When we have a row vector and a column vector with the same number of elements, we can define the dot product as Definition 1.2. Dot Product: · ¸ £ ¤ y1 ² in 2-dimension: Let x = x1 x2 and y = , the y2 dot product of x and y is, x ¢ y = x1y1 + x2y2 4 1. BRIEF INTRODUCTION TO VECTORS AND MATRICES 2 3 £ ¤ y1 ² in 3-dimension: Let x = x1 x2 x3 and y = 4 y2 5, y3 the dot product of x and y is, x ¢ y = x1y1 + x2y2 + x3y3 Definition 1.3. Matrix product Let A = (aij) and B = (bij); if the number of columns of A is the same as number of rows of B, then the product of A and B is th th given by AB = (cij) where cij is dot product of i row of A with j column of B: Example 1.2. Let · ¸ 2 3 A = ¡1 4 and · ¸ 0 5 B = ; 3 ¡4 find AB Solution · ¸ · ¸ 2 3 0 5 AB = ¡1 4 3 ¡4 · ¸ · ¸ 2 ¢ (0) + 3 ¢ (3) 2 ¢ 5 + 3 ¢ (¡4) 9 ¡2 = = (¡1) ¢ (0) + 4 ¢ 3 ¡1 ¢ 5 + 4 ¢ (¡4) 12 ¡21 Notice, the first element of AB is 2 ¢ (0) + 3 ¢ (3) which· is the¸ dot £ ¤ 0 product of first row of A; 2 3 and first column of B; a 3 The following fact gives properties of matrix product, Theorem 1.2. Let A; B; C be three matrices and r be a scalar, we have ² A(BC) = (AB)C; r(AB) = A(rB) (associativity) ² A(B + C) = AB + AC (distributivity) Notice, in general AB 6= BA, that is for most of the times, AB is not equal to BA: Using the matrix notation and matrix product, we can write the following system of equations ½ ax1 + bx2 = y1 cx1 + dx2 = y2 1. VECTORS AND MATRICES 5 as Ax = y with · ¸ a b A = ; c d · ¸ · ¸ x y x = 1 ; and y = 1 : x2 y2 Definition 1.4. A square (ex. 2 £ 2 or 3 £ 3) matrix A is invertible if there is a matrix A-1 such that AA-1 = A-1A = I: Theorem 1.3. Let · ¸ a b A = c d be a 2 £ 2 matrix, if A is invertible, we have · ¸ 1 d ¡b A-1 = ad ¡ bc ¡c d So if A is invertible, to solve Ax = y, we need to simply multiply both sides with A-1; that is x = A-1y: Example 1.3. Solve the system of equation ½ 3x1 ¡ 4x2 = 2 ¡2x1 + 5x2 = 7 Solution The equation can be rewrite Ax = y with · ¸ 3 ¡4 A = ; ¡2 5 · ¸ · ¸ x 2 x = 1 ; and y = : So in matrix form the system of x2 7 equation is · ¸ · ¸ · ¸ 3 ¡4 x 2 1 = : ¡2 5 x2 7 Now the inverse of A is · ¸ 1 5 4 A-1 = ; 3(5) ¡ (¡2)(¡4) 2 3 so the solution is · ¸ · ¸ · ¸ 1 5 4 2 38 x = A-1y = = 7 2 3 7 25 7 7 a 6 1. BRIEF INTRODUCTION TO VECTORS AND MATRICES Example 1.4. Solve the system of equation 8 < 3x1 ¡ 4x2 + 5x3 = 2 ¡2x + 5x = 7 : 1 2 x1 ¡ 5x2 + 8x3 = ¡1 Solution The equation can be rewrite Ax = y with 2 3 3 ¡4 5 A = 4 ¡2 5 0 5 ; 1 ¡5 8 2 3 2 3 x1 2 x = 4 x2 5 ; and y = 4 7 5 : So in matrix form the system of x3 ¡1 equation, Ax = y; is 2 3 2 3 2 3 3 ¡4 5 x1 2 4 ¡2 5 0 5 4 x2 5 = 4 7 5 : 1 ¡5 8 x3 ¡1 It is a little harder to compute the inverse of a 3 £ 3 matrix, we will use Mathcad to solve the equation. Here is how to do it, ² Type A:[Ctrl][M] at a blank area to bring up the matrix definition screen, put 3 in the both input boxes and click OK, you will get a 3 £ 3 matrix place holder like 2 3 A := 4 5 Fill the entries of A in the corresponding position, using [Tab] key to navigate among the place holders(or just click each one). ² Type b:[Ctrl][M] in another blank area, the matrix defini- tion screen is up again. This time put 3 in the number of row box, and 1 in2 the3 number of column box and click OK. You will get b:=4 5 put the values of y in the corresponding position. ² Type A^-1 *b= you will get the solution, which is, 2 154 3 81 4 175 5 81 80 81 1. VECTORS AND MATRICES 7 ² Notice, by default, Mathcad will display the results as dec- imal, you can double click on the result vector to change it to fraction, after you double click the result you will have a popup menu such as Figure 1. Format Result a The next example shows how we can determine the unknown con- stants typically found in the initial value problems of system of differ- ential equations. t 2t t Example 1.5. Let x1(t) = C1e + C2e and x2 = 2C1e ¡ 2t C2e : If x1(0) = 2; x2(0) = 3; find C1 and C2 t 2t Solution From x1(t) = C1e + C2e , set t = 0 we have 0 2(0) x1(0) = C1e + C2e = C1 + C2: t 2t ( Similarly, x2(t) = 2C1e ¡ C2e ; gives x2(0) = 2C1e 0) ¡ 2(0) C2e = 2C1 ¡ C2: Together with x1(0) = 2; x2(0) = 3 we have the following system of equations, ½ C1 + C2 = 1 2C1 ¡ C2 = 3 Rewrite the equation in matrix form · ¸ · ¸ · ¸ 1 1 C 1 1 = ; 2 ¡1 C2 3 and using Mathcad we find the solution is · ¸ · ¸ C 4 1 = 3 C 1 2 ¡ 3 8 1.
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