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Linear Algebra Chapter 2 Matrices 2.1 Addition, Scalar Multiplication, and Multiplication of Matrices • aij: the element of matrix A in row i and column j. • For a square nn matrix A, the main diagonal is: a11 a12 a a22 21 A an1 an 2 a1n a2 n ann Definition Two matrices are equal if they are of the same size and if their corresponding elements are equal. Thus A = B if aij = bij i, j. ( for every, for all) Ch2_2 Addition of Matrices Definition Let A and B be matrices of the same size. Their sum A + B is the matrix obtained by adding together the corresponding elements of A and B. The matrix A + B will be of the same size as A and B. If A and B are not of the same size, they cannot be added, and we say that the sum does not exist. Thus if C A B, then cij aij bij i,j. Ch2_3 Example 1 4 7, B 2 5 6, and C 5 4. Let A 1 8 0 2 3 3 1 2 7 Determine A + B and A + C, if the sum exist. Solution 4 7 2 5 6 (1) A B 1 8 0 2 3 3 1 4 5 7 6 1 2 0 3 2 1 3 8 3 9 1. 3 1 11 (2) Because A is 2 3 matrix and C is a 2 2 matrix, there are not of the same size, A + C does not exist. Ch2_4 Scalar Multiplication of matrices Definition Let A be a matrix and c be a scalar. The scalar multiple of A by c, denoted cA, is the matrix obtained by multiplying every element of A by c. The matrix cA will be the same size as A. Thus if B cA, then bij caij i, j. Example 2 Let A 1 2 4. 7 2 0 3 1 3 (2) 3 4 3 6 12 3A . 3 7 3 (3) 3 0 21 9 0 Observe that A and 3A are both 2 3 matrices. Ch2_5 Negation and Subtraction Definition We now define subtraction of matrices in such a way that makes it compatible with addition, scalar multiplication, and negative. Let A – B = A + (–1)B Example 3 Suppose A 5 0 2 and B 2 8 1. 3 6 5 0 4 6 5 2 0 8 2 (1) 3 8 1 A B . 2 11 3 0 6 4 5 6 3 Ch2_6 Multiplication of Matrices Definition Let the number of columns in a matrix A be the same as the number of rows in a matrix B. The product AB then exists. Let A: mn matrix, B: nk matrix, The product matrix C=AB has elements cij ai1 ai 2 b1 j b 2j ain ai1b1 j ai 2b2 j ainbnj bnj C is a mk matrix. If the number of columns in A does not equal the number of row B, we say that the product does not exist. Ch2_7 Example 4 0 1 1 3 5 Let A ,B , and C 6 2 5. 2 0 3 2 6 Determine AB, BA, and AC , if the products exist. Solution. AB 1 2 5 1 3 3 5 2 0 3 3 5 0 3 0 2 0 1 3 2 0 2 0 2 1 6 1 1 3 6 1 2 0 6 (1 5) (3 3) (1 0) (3 (2)) (11) (3 6) (2 5) (0 3) (2 0) (0 (2)) (2 1) (0 6) 14 6 19 . 10 0 2 BA and AC do not exist. Note. In general, ABBA. Ch2_8 Example 5 1 2 Let A 7 0 and B 1 0. Determine AB. 3 5 3 2 2 1 1 2 10 3 5 1 2 1 0 1 0 7 05 AB 7 0 7 0 3 5 3 3 2 1 0 3 2 3 2 3 5 5 2 3 0 5 1 7 0 0 0 7 0 3 6 0 10 3 10 Example 6 Let C = AB, A 2 1 and B 7 3 2 Determine c23. 3 4 5 0 1 2 (3 2) (4 1) 2 3 4 c23 1 Ch2_9 Size of a Product Matrix If A is an m r matrix and B is an r n matrix, then AB will be an m n matrix. A B = AB mr rn mn Example 7 If A is a 5 6 matrix and B is an 6 7 matrix. Because A has six columns and B has six rows. Thus AB exits. And AB will be a 5 7 matrix. Ch2_10 Special Matrices Definition A zero matrix is a matrix in which all the elements are zeros. A diagonal matrix is a square matrix in which all the elements not on the main diagonal are zeros. An identity matrix is a diagonal matrix in which every diagonal element is 1. 0mn 0 0 0 0 0 0 zero matrix 0 0 0 a11 0 0 a 22 A 0 0 0 0 ann 1 0 0 I n 0 1 0 0 0 1 identity matrix diaginal matrix A Ch2_11 Theorem 2.1 Let A be m n matrix and Omn be the zero m n matrix. Let B be an n n square matrix. On and In be the zero and identity n n matrices. Then A + Omn = Omn + A = A BOn = OnB = On BIn = InB = B Example 8 Let A 2 1 3 and B 2 1. 8 4 5 3 4 2 A O23 4 2 BO2 3 2 BI 2 3 1 3 0 0 0 2 1 3 A 5 8 0 0 0 4 5 8 1 0 3 0 1 1 4 0 0 0 0 0 0 2 1 3 0 O2 0 1 B 4 Ch2_12 Homework Exercises: 5, 8, 11, 17 from page 71 to 73. Exercise 17 p.72 Let A be a matrix whose third row is all zeros. Let B be any matrix such that the product AB exists. Prove that the third row of AB is all zeros. Solution b1i b1i b b ( AB)3i [a31 a32 a3n ] 2i [0 0 0] 2i 0, i. bni bni Ch2_13 2.2 Algebraic Properties of Matrix Operations Theorem 2.2 -1 Let A, B, and C be matrices and a, b, and c be scalars. Assume that the size of the matrices are such that the operations can be performed. Properties of Matrix Addition and scalar Multiplication 1. A + B = B + A Commutative property of addition 2. A + (B + C) = (A + B) + C Associative property of addition 3. A + O = O + A = A (where O is the appropriate zero matrix) 4. c(A + B) = cA + cB Distributive property of addition 5. (a + b)C = aC + bC Distributive property of addition 6. (ab)C = a(bC) Ch2_14 Theorem 2.2 -2 Let A, B, and C be matrices and a, b, and c be scalars. Assume that the size of the matrices are such that the operations can be performed. Properties of Matrix Multiplication 1. A(BC) = (AB)C Associative property of multiplication 2. A(B + C) = AB + AC Distributive property of multiplication 3. (A + B)C = AC + BC Distributive property of multiplication 4. AIn = InA = A (where In is the appropriate zero matrix) 5. c(AB) = (cA)B = A(cB) Note: AB BA in general. Multiplication of matrices is not commutative. Ch2_15 Proof of Thm 2.2 (A+B=B+A) Consider the (i,j)th elements of matrices A+B and B+A: ( A B)ij aij bij bij aij ( B A)ij . A+B=B+A Example 9 Let A 1 3, B 3 7, and C 0 2. 1 4 5 8 5 1 A B C 1 3 3 7 0 2 1 5 1 4 5 8 1 3 0 3 7 2 4 6. 5 4 8 5 5 1 1 9 Ch2_16 Arithmetic Operations If A is an m r matrix and B is r n matrix, the number of scalar multiplications involved in computing the product AB is mrn. Consider three matrices A, B and C such that the product ABC exists. Compare the number of multiplications involved in the two ways (AB)C and A(BC) of computing the product ABC Ch2_17 Example 10 4 3, and C 1 . Let A 1 2, B 0 1 Compute ABC. 3 1 1 0 2 0 Solution. Which method is better? Count the number of multiplications. (1) (AB)C 3 2 1 1. AB 1 2 0 1 3 1 1 0 2 1 3 11 26+32 4 9 =12+6=18 2 1 1 1 . ( AB)C 1 3 11 0 1 (2) A(BC) 4 0 1 3 1 1 BC 1 0 2 0 4 A( BC ) 1 2 1 9. 3 1 4 1 32+22 =6+4=10 A(BC) is better. Ch2_18 Caution In algebra we know that the following cancellation laws apply. If ab = ac and a 0 then b = c. If pq = 0 then p = 0 or q = 0. However the corresponding results are not true for matrices. AB = AC does not imply that B = C. PQ = O does not imply that P = O or Q = O. Example 11 1 2 1 2 3 8 (1) Consider t he matrices A ,B , and C . 2 4 2 1 3 2 3 4 Observe that AB AC , but B C. 6 8 1 2 2 6 (2) Consider t he matrices P , and Q . 4 2 1 3 Observe that PQ O, but P O and Q O. Ch2_19 Powers of Matrices Definition If A is a square matrix, then Ak AA A k times Theorem 2.3 If A is an n n square matrix and r and s are nonnegative integers, then 1. ArAs = Ar+s. 2. (Ar)s = Ars. 3. A0 = In (by definition) Ch2_20 Example 12 1 2 4 If A , compute A . 0 1 Solution 1 2 1 2 3 2 A 1 0 1 0 1 2 2 3 2 3 2 11 10 A . 2 1 2 5 6 1 4 Example 13 Simplify the following matrix expression. A( A 2B) 3B(2 A B) A2 7 B 2 5 AB Solution A( A 2 B) 3B(2 A B) A2 7 B 2 5 AB A2 2 AB 6 BA 3B 2 A2 7 B 2 5 AB 3 AB 6 BA 4 B 2 We can’t add the two matrices Ch2_21 Systems of Linear Equations A system of m linear equations in n variables as follows a11 x1 a1n xn b1 am1 x1 amn xn bm Let a11 a1n x1 b1 A , X , and B am1 amn xn bm We can write the system of equations in the matrix form AX = B Ch2_22 Idempotent and Nilpotent Matrices (Exercises 24-30 p.83 and 14 p.93) Definition (1) A square matrix A is said to be idempotent if A2=A. (2) A square matrix A is said to nilpotent if there is a p positive integer p such that A =0. The least integer p such that Ap=0 is called the degree of nilpotency of the matrix. Example 14 3 6 2 3 6 (1) A ,A A. 1 2 1 2 3 9 2 0 0 (2) B ,B . The degree of nilpotency : 2 1 3 0 0 Ch2_23 Homework Exercises 3, 6, 12, 18, 27, 32, p.82 to p.83 Ch2_24 2.3 Symmetric Matrices Definition The transpose of a matrix A, denoted At, is the matrix whose columns are the rows of the given matrix A. i.e., A : m n At : n m, ( At )ij Aji i, j. Example 15 A 2 7, B 1 2 7, and C 1 3 4. 6 8 0 4 5 1 1 4 t t C 3. At 2 8 B 2 5 0 7 4 7 6 Ch2_25 Theorem 2.4: Properties of Transpose Let A and B be matrices and c be a scalar. Assume that the sizes of the matrices are such that the operations can be performed. 1. (A + B)t = At + Bt Transpose of a sum 2. (cA)t = cAt Transpose of a scalar multiple 3. (AB)t = BtAt Transpose of a product 4. (At)t = A Ch2_26 Symmetric Matrix Definition A symmetric matrix is a matrix that is equal to its transpose. A At , i.e., aij a ji i, j Example 16 5 2 5 4 match 1 0 1 4 0 8 1 7 2 4 8 3 4 0 2 4 7 3 9 3 2 3 9 3 6 match Ch2_27 Remark: If and only if Let p and q be statements. Suppose that p implies q (if p then q), written p q, and that also q p, we say that “p if and only if q” (in short iff ) Ch2_28 Example 17 Let A and B be symmetric matrices of the same size. Prove that the product AB is symmetric if and only if AB = BA. Proof *We have to show (a) AB is symmetric AB = BA, and the converse, (b) AB is symmetric AB = BA. () Let AB be symmetric, then AB= (AB)t by definition of symmetric matrix = BtAt by Thm 2.4 (3) = BA since A and B are symmetric () Let AB = BA, then (AB)t = (BA)t = AtBt by Thm 2.4 (3) = AB since A and B are symmetric Ch2_29 Example 18 Let A be a symmetric matrix. Prove that A2 is symmetric. Proof ( A2 )t ( AA) t ( At At ) AA A2 Ch2_30 Exercises 1, 2, 6, 7, 14, p.93 to p.94 2.4 The Inverse of a Matrix Definition Let A be an n n matrix. If a matrix B can be found such that AB = BA = In, then A is said to be invertible and B is called the inverse of A. If such a matrix B does not exist, then A has no inverse. (denote B = A1, and Ak=(A1)k ) Example 19 2 1 Prove that the matrix A 1 2 has inverse B 3 1 . 3 4 2 2 Proof 1 2 AB 1 2 3 1 1 0 I 2 3 4 2 0 1 2 1 1 2 1 0 2 BA 3 1 I2 3 4 0 1 2 2 Thus AB = BA = I2, proving that the matrix A has inverse B. Ch2_32 Theorem 2.5 The inverse of an invertible matrix is unique. Proof Let B and C be inverses of A. Thus AB = BA = In, and AC = CA = In. Multiply both sides of the equation AB = In by C. C(AB) = CIn Thm2.2 (CA)B = C InB = C B=C Thus an invertible matrix has only one inverse. Ch2_33 Gauss-Jordan Elimination for finding the Inverse of a Matrix Let A be an n n matrix. 1. Adjoin the identity n n matrix In to A to form the matrix [A : In]. 2. Compute the reduced echelon form of [A : In]. If the reduced echelon form is of the type [In : B], then B is the inverse of A. If the reduced echelon form is not of the type [In : B], in that the first n n submatrix is not In, then A has no inverse. An n n matrix A is invertible if and only if its reduced echelon form is In. Ch2_34 Example 20 1 1 2 Determine the inverse of the matrix A 2 3 5 3 5 1 Solution 1 1 2 1 0 [ A : I3 ] 2 3 5 0 1 3 5 0 0 1 1 1 2 1 0 (1)R2 0 1 1 2 1 3 1 0 0 2 1 0 0 0 1 1 2 0 R2 (2) R10 1 1 2 1 0 3 1 0 1 1 R3 R1 0 2 0 3 1 0 1 0 1 0 R1 R2 0 1 1 2 1 0 1 R3 (2)R2 0 0 1 3 2 1 0 1 1 1 0 0 R1 R3 0 1 0 5 3 1 R2 (1)R3 0 0 1 3 2 1 1 1 0 Thus, A1 5 3 1. 3 2 1 Ch2_35 Example 21 Determine the inverse of the following matrix, if it exist. 1 1 5 A 1 2 7 2 1 4 Solution 1 5 1 0 0 1 1 1 5 1 0 0 1 2 1 1 0 [ A : I 3 ] 1 2 7 0 1 0 R2 (1)R1 0 2 1 4 0 0 1 R3 (2)R1 0 3 6 2 0 1 2 1 0 1 0 3 R1 (1)R2 0 1 2 1 1 0 R3 3R2 0 0 0 5 3 1 There is no need to proceed further. The reduced echelon form cannot have a one in the (3, 3) location. The reduced echelon form cannot be of the form [In : B]. Thus A–1 does not exist. Ch2_36 Properties of Matrix Inverse Let A and B be invertible matrices and c a nonzero scalar, Then 1. ( A1 ) 1 A 1 1 1 2. (cA) A c 1 1 1 3. ( AB) B A 4. ( An ) 1 ( A1 ) n 5. ( At ) 1 ( A1 )t Proof 1. By definition, AA1=A1A=I. 2. (cA)( 1c A1 ) I ( 1c A1 )(cA) 3. ( AB)( B 1 A1 ) A( BB 1 ) A1 AA1 I ( B 1 A1 )( AB) 1 1 1 n n 4. An ( A1 ) n A A A A I ( A ) A n times n times 5. AA1 I , ( AA1 )t ( A1 )t At I , A1 A I , ( A1 A)t At ( A1 )t I , Ch2_37 Example 22 If A 4 1, then it can be shown that A1 1 1. Use this 3 1 3 4 informatio n to compute ( At ) 1. Solution t 1 1 1 3 (A ) (A ) . 3 4 1 4 t 1 1 t Ch2_38 Theorem 2.6 Let AX = B be a system of n linear equations in n variables. If A–1 exists, the solution is unique and is given by X = A–1B. Proof (X = A–1B is a solution.) Substitute X = A–1B into the matrix equation. AX = A(A–1B) = (AA–1)B = In B = B. (The solution is unique.) Let Y be any solution, thus AY = B. Multiplying both sides of this equation by A–1 gives A–1A Y= A–1B In Y= A–1B Y = A–1B. Then Y=X . Ch2_39 Example 22 x1 x2 2 x3 1 Solve the system of equations 2 x1 3x2 5 x3 3 x1 3 x2 5 x3 2 Solution This system can be written in the following matrix form: 1 1 2 x1 1 2 3 5 x2 3 3 5 x3 2 1 If the matrix of coefficients is invertible, the unique solution is 1 x1 1 1 2 1 x2 2 3 5 3 x 1 3 5 2 3 This inverse has already been found in Example 20. We get x1 0 1 1 1 1 x2 5 3 1 3 2 x 3 2 1 2 1 3 The unique solution is x1 1, x2 2, x3 1. Ch2_40 Elementary Matrices Definition An elementary matrix is one that can be obtained from the identity matrix In through a single elementary row operation. Example 23 1 0 0 I 3 0 1 0 0 0 1 R2 R3 5R2 R2+ 2R1 1 0 0 E1 0 0 1 0 1 0 1 0 0 E2 0 5 0 0 0 1 1 0 0 E3 2 1 0 0 0 1 Ch2_41 Elementary Matrices 。 Elementary row operation 。 Elementary matrix a b R2 R3 g h d e a b A d e g h c f i 5R2 a 5d g c 1 0 0 i 0 0 1 A E1 A f 0 1 0 b c 1 0 0 5e 5 f 0 5 0 A E2 A h i 0 0 1 b a d 2a e 2b R2+ 2R1 g h c 1 0 0 f 2c 2 1 0 A E3 A i 0 0 1 Ch2_42 Notes for elementary matrices Each elementary matrix is invertible. Example 24 I E1 E1 I , i.e., E2 E1 I R1 2 R 2 1 0 0 I 0 1 0 0 0 1 R1 2 R 2 1 2 0 E1 0 1 0 0 0 1 1 2 0 E2 0 1 0 0 0 1 If A and B are row equivalent matrices and A is invertible, then B is invertible. Proof If A … B, then B=En … E2 E1 A for some elementary matrices En, … , E2 and E1. So B1 = (En … E2 E1A)1 =A1E11 E21 … En1. Ch2_43 Homework Exercises 6, 9, 14, 15, 16, 19, 21, 32 from p.105 to 107. Exercise 7 d b 1 a b 1 . If A , show that A (ad bc) c a c d Ch2_44

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