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# Class Note for MATH 290 at KU

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Chapter 2 Matrices 21 Operations with Matrices Homework 21 page 56 7 9 13 15 17 25 27 35 37 41 46 49 67 Main points in this section 1 We de ne a few concept regarding matrices This would include addition of matrices scalar multiplication and multiplication of matrices 2 We also represent a system of linear equation as equation with matrices 41 42 CHAPTER 2 MATRICES In this section7 we will do some algebra of matrices That means7 we will add7 subtract7 multiply matrices Matrices will usually be denoted by upper case letters7 A7 B707 Such a matrix 111 112 113 39 39 39 am 121 122 123 39 39 39 a2n A 131 132 133 39 39 39 Ian L aml amZ am 39 39 39 04mm J is also denoted by 127 Remark I used parentheses in the last chapter7 not a square brackets The textbook uses square brackets De nition 211 Two matrices A 127 and B by are equal if both A and B have same size m X n and the entries 127bij forall1 i m and1 j n Read Textbook7 Example 17 p 47 for examples of unequal matrices De nition 212 Following are some standard terminologies 1 A matrix with only one column is called a column matrix or column vector For example7 13 a 19 23 is a column matrix 2 A matrix with only one row is called a row matrix or row vec tor For example b 411 13 19 23 is a row matrix 21 OPERATIONS WITH MATRICES 43 3 Bold face lower case letters7 as above7 will often be used to denote TOW or column matrices 211 Matrix Addition De nition 213 We di ne addition of two matrices of same size Suppose A 127 and B by be two matrices of same size m X n Then the sum A B is de ned to be the matrix of size m X n given by 14 B 04339 1 For example7 with 4 75 9 15 13 10 A 3 78 7 B 75 18 we have AB 72 10 10 14 11 1 21 15 2 Also7 for example7 with 735 72 3 722 05 27 73 75 0 7 we have CD 3 07 0 472 39 3 While7 the sum A C is not de ned because A and 0 do not have same size 4 Read Textbook7 Example 27 p 48 for more such examples 212 Scalar Multiplication Recall7 in some contexts7 real numbers are referred to as scalars in contrast with vectors We de ne7 multiplication of a matrix A by a scalar c 44 CHAPTER 2 MATRICES De nition 214 Let A 127 be an m X 71 matrix and C be a scalar We de ne Scalar multiple CA of A by C as the matrix of same size given by CA caml 1 For a matrix A7 the negative of 7A denotes 71A Also AiB A 71B 2 Let C 11 and 4 45 9 15 A 3 is B 7518 10 14 11 1 Then7 With C 11 we have 4 75 44 755 CA 11 3 78 33 788 10 14 110 154 Likewise7 A 7 B A 71B 4 45 9 15 4 45 49 415 3 is 71 7518 3 is 5 718 10 14 11 1 10 14 411 41 449 45415 is 420 35 78718 8 726 l10411 1441 41 13 3 Read Textbook7 Example 37 p 49 for more such computations 21 OPERATIONS WITH MATRICES 45 213 Matrix Multiplication The textbook gives a helpful motivation for de ning matrix multipli cation in page 49 Read it7 if you are not already motivated De nition 215 Suppose A 127 is a matrix of size m X n and B by is a matrix of size n X p Then the product AB 027 is a matrix size m X p where TL 02739 3 121517 12 2sz 12353 39 ambm Zaikbkj k1 1 Note that number of columns of A must be equal to number of rows of B7 for the product AB to be de ned 2 Note the number of rows of AB is equal to is same as that of A and number of columns of AB is equal to is same as that of B 3 Read Textbook7 Example 4 57 p51 52 I will workout a few below Exercise 216 Ex 12 p 57 Let 1 71 7 1 1 2 A 2 71 8 7 B 2 1 1 3 1 71 1 73 2 Compute AB abd BA We have 1171271117117731271172 AB 2171281217118732271182 3112711311171733211712 46 CHAPTER 2 MATRICES 6 21 15 8 23 19 4 1 5 Now7 we compute BA We have 111223 171 121 17182 BA 211213 271 7111 271817 73 712117382 Also note that AB a BA Exercise 217 Ex 16 p 57 Let 0 1 3 2 A 4 0 2 7 and B 3 8 1 7 1 Compute AB and BA7 if de nded Solution Note BA is not de ned7 because size of B is 3 X 1 and size of A is 3 X 3 But AB is de ned and is a 3 X 1 matrix or a column matrix We have 02717331 6 AB 4207321 10 821371 26 Exercise 218 Ex 18 p 57 Let 103 24 A and B 6 13 8 717 20 16 42 21 OPERATIONS WITH MATRICES 47 Compute AB and BA7 if de nded Solution Note AB is not de ned7 because size of A is 2 X 5 and size of B is 2 X 2 But BA is de ned and we have BA 1166106131368172671714620 4126 40213 4328 4722717 44220 i 37 78 51 7104 124 T 16 26 28 742 5639 214 Systems of linnear equations Systems of linear equations can be represented in a matrix form Given a system of liner equations 111 112 113 air n b1 121 122 123 a2nn 52 131 132 133 a3nn 53 am i am z am33 amr n bm we can write this system in the matrix form as Ax b where 111 112 113 quot 39 am 1 b1 121 122 123 39 39 39 a2n 2 52 A 131 132 133 39 39 39 Ian 7 X 3 7 b 53 Laml amZ am 04an Clearly7 A is the coe icient matrix7 x would be called the unknown or variable matrix or vector and b would be called the constant vector or matrix 48 CHAPTER 2 MATRICES 1 With 111 112 117 am 121 122 127 a2n 31 131 732 132 7quot39aj 137 7quot39an Ian aml amZ amj 04mm we can write or think A a1 a2 aj an as a matrix of matrices 2 The above system of linear equations can be reWriteen as 1a1 2a2jajnanb We say7 b is a linear combination of the column metrices or vectors a17 a27 7an With coe icents 17 2 7m 3 So7 the solutions of the above system are precisely those 017 70 such that b is a linear combination of the vectors a17a27 7an With coe icents 017 027 7on Exercise 219 Ex 34 changed p 58 Consider the system of equation 1 2 733 1 71 22 1 1 722 3 2 Write this system in the matrix form Ax b and solve matrix equation for x Solution The equation reduces to 1 1 73 x1 71 71 2 0 x2 1 72 1 x3 2 21 OPERATIONS WITH MATRICES 49 This answers the rst part To solve7 the augmented matrix of the 117371 712 01 l17212 system is Using Tl 847 we can reduce it to Gauss Jordan form 1 0 0 5 0 1 0 3 0 0 1 3 The corresponding linear system is 1 0 0 1 5 0 1 0 2 0 0 1 3 3 Multiplying7 this give the solution 1 5 2 3 3 3 Exercise 2110 Ex 40 58 m 1H 1 Solution Here we have four unknowns 1 b7 07d In any case7 multiply m ing 2a3b ab 203d cd 50 CHAPTER 2 MATRICES Equating respective entries 2a 3b 3 a b 17 20 3d 4 c d 71 The augmented matrix of the system 2 3 0 0 3 1 1 0 0 17 0 0 2 3 4 0 0 1 1 71 Use T1 84 to reduce it to the Gauss Jordan form 1 0 0 48 0 1 0 731 0 0 1 0 77 L0 0 0 1 6 J Looking at the corresponding to this matrix7 we get 0 d a48b7317c777d6 and a y 22 PROPERTIES OF MATRIX OPERATIONS 51 22 Properties of Matrix operations Homework 22 page 70 1 3 79 15 17 21 23 29 31 35 37 39 43 57 59 Main points in this section 1 We write down some of the properties of matrix addition and multiplication 2 We de ne transpose of a matrix 5 We start writing some proofs 52 CHAPTER 2 MATRICES We start developing some algebra of matrices Theorem 221 Properties Let A7B7C be three m X n matrices and c7 d be scalars Then 1 A B B A Commutatiuity of matrid addition 2 A B C A B C Associatiuity of matrid addition 3 cdA cdA Associatiuity of scalar multiplication 4 1A A identity of scalar multiplication 5 cA B cA cB Distributiuity of scalar multiplication 6 c dA cA dA Distributiuity of scalar multiplication Proof One needs to prove all these statements using de nitions of ad dition and scalar multiplication To prove the commutativity of ma trix addition 17 Let A awl7 B lbwl Both A and B have same size m X n7 so A B7 B A are de ned From de nition 43 laijllb jl la jb jl and BA lbijl laijl lb ja jl From commutative property of addition of real numbers7 we have aij bij b ja j Therefore7 from de nition of equality of matrices7 AB B A So7 1 is proved Other properties are proved similarly I Remark For matrices A7B7C as in the theorem7 by the expression ABC we mean AB 0 or A BC It is well de ned7 because A B C A B C by associatieve property2 of matrix addition Theorem 222 Let Omn denote the m X n matrix whose entries are all zero Let A be a m X n matrix Then 1 Then A Omn A 22 PROPERTIES OF MATRIX OPERATIONS 53 2 We have A 7A Omn 3 If cA 0mm then either c 0 or A 0 We say on the set of allm gtltn matrices Omn is an additive identity property 1 and 7A is the additive inverse ofA property 221 Properties of matrix multiplication Theorem 223 MultProperties Let A7B7C39 be three matrices of varying sizes so that all the products below are de ned and c be a scalars Then ABC ABC Associativity of multiplication A BC A0 BC Left 7 Distributivity of multiplication AB C AB AC Right 7 Distributivity of multiplication CAB cAB an Associativity WHgtWNE ln 17 by AB a BA7 we mean AB is not always equal to BA Proof We Will only prove In this case7 let A be a matrix of size m X n and then B70 would have to be of same size n X p Write 111 112 113 39 39 39 am 121 122 123 39 39 39 a2n A 131 132 133 39 39 39 Ian 7 Laml am2 am 1an AB a BA Failure of Commutativity of multiplication 54 CHAPTER 2 MATRICES and 511 512 513 39 quot blp C11 C12 C13 39 39 39 Clp 521 522 523 39 39 39 pr C21 022 023 39 39 39 02p B 531 532 533 39 39 39 b3p 7 C C31 032 033 39 39 39 03p L bnl an an 39 39 39 bnp L L Cnl CnZ Cn3 39 39 39 Cnp L Therefore7 AB C 111 112 113 39 39 39 117 511 C11 512 012 513 013 quot 39 blp 01p 121 122 123 39 39 39 a2n 521 C21 522 022 523 023 quot 39 pr 02p 131 132 133 39 39 39 Ian 531 C31 532 032 533 033 39 39 39 53p 03p am1 amZ ams 39 39 39 amn bm Cm bnz an an CH 39 39 39 bnp Cnp Which is 221 Imam Cm 221 mam Ck2 39 quot 221 Imam Ckp 221 a2kbk1 Cm 221 a2kbk2 Ck2 39 quot 221 1219ka Ckp 221 a3kbk1 Cm 221 a3kbk2 Ck2 39 quot 221 a3kbkp Ckp L 221 amkbk1 Cm 221 amkbk2 Ck2 quot39 221 amkbkp Ckp L 22 PROPERTIES OF MATRIX OPERATIONS 55 Which is n n n 2191 arkbkr 2191 041kka quot 39 21ml alk bkp n n n 2191 azkbki 2191 azkbkz 39 39 39 2191 121cka n n n 2191 agkbki 2191 043kka 39 39 39 2191 Zak bkp n n n 2191 amkbki 2191 amkbkz 39 39 39 2191 amkbkp AB AC 71 n n 2191 aikcki 2191 11ka2 39 39 39 2191 alkckp n n n 2191 azkckr 2191 a2kck2 39 39 2191 aZkap n n n 2191 agkckr 2191 13ka2 39 39 2191 lamp n n n 2191 amkcm 2191 amkckz 39 39 39 2191 akakp So7 AB C AB AC and the proof of complete I Alternate way to write the same proof Let A be a matrix of size m X n and then B70 would have to be of same size n X p Write A aikl7 B lbkjl7 C lckjl Then B C bkj ckj So7 AB C laikllbkj ijl Owl Say Then the ijW entry 0427 of AB C is given by n n n 062 Z Mk kj an 2 12795197 2 12790197 k1 k1 k1 But AB 7 and AC n n E 12795197 E 12790197 k1 k1 So7 the ijW entry of AB AC is also equal to 0427 Therefore AB C AB AC and the proof is complete I 56 CHAPTER 2 MATRICES Remark For matrices A BC as in the theorem so that all the mul tiplications in the associative law 2 are de ned Then the expres sion ABC would mean ABC or ABC It is well de ned because ABC ABC by associatieve property2 of matrix multiplication Reading assignment 1 Textbook Example 3 p 64 to 7experience7 that associatiVity property 2 works 2 Textbook Example 4 p 64 to see examples that commutativity for multiplication fails property 1 3 Textbook Example 5 p 65 to see examples that cancellation fails That means examples AB A0 for nonzero ABC with B a C This makes solving matrix equation like AX AC dif ferent from solving equation in real numbers like aa 0 Theorem 224 Let In denote the square matrix of order n whose main diagonal entries are 1 and all the entries are zero So 10 0 1000 0100 100 0100 In 0 0 1 0 13 010 14 0010 01 0001 0001 Let A be an m X n matrix Then ImA A and AIn A Because of these two properties In is called the identity matrix of order n Proof Easy 22 PROPERTIES OF MATRIX OPERATIONS 57 Reading assignment 1 Textbook7 Example 67 p 66 2 Textbook7 Example 77 p 667 just to get used to computing AT Following the de nition 112 in item 4 we gave a classi cation of system of equation7 which we state as a theorem and prove as follows Theorem 225 Let Ax b be a linear system of m equations in n unknowns Then exactly one of the follwoing is true 1 The system has no solution 2 The system has exactly one solution 3 The system has in nitely many solutions Proof If the system has no solution or have exactly one solution7 then there is nothing to prove So7 assume7 it has at least two distinct solutions x17x2 So7 Axl b and AX2 b Subtracting the second from the rst7 we have Ax1 7 x2 0 Write y x1 7 x2 Then y a 0 and Ay 0 So7 for any scalar 07 we have Ax1 cy Axl Acy b Acy b CAy b CO b So7 x1 cy is a solution7 for each scalar c and they are all distinct So7 the system has in nitely many solutuions This completes the proof I 222 I ranspose of a matrix De nition 226 The transpose of a matrix A is obtained by writ ing the rows as columns andor writing the columns as rows The transpose of A is denoted bt AT So7 for the matrix 58 CHAPTER 2 MATRICES 111 112 113 39 quot am 111 121 131 39 39 39 aml 121 122 123 39 39 39 a2n 112 122 132 39 39 39 amZ T 7 A i 131 132 133 39 39 39 Ian A i 113 123 133 39 39 39 ams aml amZ am 39 39 39 04mm 0417i 0427i 0437i 39 39 39 04mm De nition 227 A square matrix A is said to be symmetric if A AT Theorem 228 Let A7 B be matrices of varying sizes so that the sum or product is de ned and c be a scalar Then7 1 ATT A That means transpose of transpose is the same 2 A BT AT ET 3 CAT CAT 4 ABT BTAT This is the most important one in this list In fact properties 2 4 extends to sum of higher number of matrices Reading assignment 1 Textbook7 Example 87 p 68 to compute transpose of a matrix 2 Textbook7 Example 97 p 667 about transpose and product of matrices 22 PROPERTIES OF MATRIX OPERATIONS 59 Exercise 229 Ex 8 p 70 Let 72 71 0 3 A 1 0 B 2 0 3 i4 i4 71 Then 1 Solve X 3A 7 2B Solution In this case 72 71 0 3 X 3A 7 2B 3 1 0 7 2 2 0 3 74 74 71 Which is 76 73 0 6 76 79 3 0 7 4 0 71 0 9 712 78 72 17 710 2 Solve 2X 2A 7 B Solution We have 72 71 0 3 74 72 0 3 2X 2A7B 2 1 0 7 2 0 2 0 7 2 0 3 74 74 71 6 78 74 71 OR 74 75 2X 0 0 10 77 Divide by 2 1e multiply by 7 we have 1 74 75 72 725 X 5 0 0 0 0 10 77 5 735 60 CHAPTER 2 MATRICES 3 Solve 2X 3A B Solution We have 2X 3A B Subtracting 3A from both sides7 2X B 7 3A So7 0 3 72 71 6 6 2X B7314 2 0 3 1 0 71 0 74 71 3 74 713 11 Divide by 2 1e multiply by 7 we have him ELM El 2 l 4713 11 2765 55 J Exercise 2210 Ex 10changed p 70 Let 0 1 7C fl 0 Solution Note ABC is not de ned because A has 3 column and BC has two rows We compute CBA We have CBA CBA and 13 i A 2 42 7 42 mowmmmA 1 3H3 iHi 2 Exercise 2211 Ex 18 p 70 Let 1 72 705 l A2 7 B 24 123 0171 13 712 A 7 Compute CBA and ABC7 if de ned 01 710 GB 22 PROPERTIES OF MATRIX OPERATIONS ThenAB0butA3 0norB7 0 61 Note such a phenomenon will not occur with real numbers That is why with real numbers aw 07a a 0 i w 0 We cannot do similar algebra with matrices With A a 0 the solution to the equation AX 0 is not necessarily X 0 Solution Obvious Exercise 2212 Ex 22 p 70 Let 1 0 Solution We have Compute A IA 1 AIAAA2A20 Exercise 2213 Ex 24 p 70 Let 1 1 A 3 4 0 2 Compute AT7ATA and AAT Solution We have AT 1 3 0 I 1 4 2 SoATA 1 3 0 1 71 190 1 4 2 3 4 1120 0 2 2 17 and 112 i 10 01 1 l 2 0 1120 1164 4 2 l 10 11 11 21 62 CHAPTER 2 MATRICES AlsoAAT 171 l l 374 0 2 l l 13 0 l l 3 4 374 916 078 l0 72 l02 078 04 which is Exercise and comment Notice in this exercise 24 above7 both AAT and ATA are symmetric matrices This is not a surprise ln fact7 for any matrix A7 both AAT and ATA are symmetric ma trices Proof AATT ATTAT AAT So7 AAT is symmetric Similarly7 ATA is symmetric l Exercise 2214 Ex 38 p 71 Let 1 Find scalars a7b such that Z 1X bY Solution This is7 in fact7 a problem of solving a system of linear equations Suppose Z 1X bY Then7 X Z OR WMH NOH 22 PROPERTIES OF MATRIX OPERATIONS 63 Here a7b are unknown to be solved for The augmented matrix of this system is 1 1 1 2 0 4 3 2 4 By Tl 847 the matrix reduced to the Gauss Jordan form 1 0 2 0 1 71 0 0 0 This gives a 27b 71 We can check 2X 7 Y Z 2 Show that there do not exist scalar a b such that W 1X bY Solution This is7 in fact7 a problem of solving a system of linear equations Suppose W 1X bY Then7 1 1 Xil 1 W OR 20 a b b 3 2 Here a7b are unknown to be solved for The augmented matrix of this system is Here a7b are unknown to be solved for The augmented matrix of this system is WNH 1 0 2 H00 By Tl 847 the matrix reduced to the Gauss Jordan form OOH OHO HOD 64 CHAPTER 2 MATRICES The system corresponding to this matrix is a 0b 00 1 which is inconsistent ShowthataXbYcWOthenabc0 Solution This is again a problem of solving a system of linear equations Suppose O 1X bY CW Then a 110 a 0 XYWbOOR 200 b0 c 321 c 0 The augmented matrix of this system is given by NOH 1 0 0 2 0 0 3 1 0 From Tl 8384 the matrix reduces to the following Gauss Jordan form 1 0 0 0 1 0 0 0 1 COO So a 0 b 00 0 as was required to show Find the scalars a b c not all zero nontrivial solution such that 777 aXbYCZO Solution This is again a problem of solVing a system of linear equations Suppose O 1X bY CZ Then 1 111a 0 XYZbOOR 204 b0 c 324 c 0 22 PROPERTIES OF MATRIX OPERATIONS The augmented matrix of this system is given by 1 2 3 NOH HgtgtJgtH 0 0 0 From Tl 83847 the matrix reduces to the following Gauss Jordan form 1 0 2 0 0 1 71 0 0 0 0 0 The linear system corresponding to this matrix is 1 20 0 b 70 0 0 Using c t as parameter7 we have a 72t7b 750 t So7 for any scalar if7 we have 72tX 751 tZ 0 Exercise 2215 Ex 44 p 71 Let 5 4 f2776 and A 1 2 Compute fA Solution We have 5 4 5 4 29 28 A2 AA 1 2 1 2 7 8 So fA A2 7 7A 612 1 0 0 1 2928 754 78 12 66 CHAPTER 2 MATRICES 00 0039 So7 fA 0 One can say that A is a matrix root of 60 06 29 28 35 28 7 8 7 14 23 THE INVERSE OF A MATRIX 67 23 The Inverse of a matrix Homework Textbook7 EX 97 137 257 277 337 357 377 417 45 page 84 Main point in the section is to de ne and compute inverse of matrices Z We give a formula to compute inverse of a 2 X 2 matrices 2 We describe the method of row reduction to compute inverse of a matrix of any size 5 We use inverse of matrices to solve system of linear equations 68 CHAPTER 2 MATRICES De nition 231 For a square matrix A of size n X n we say that A is invertible or nonsingular if there is an n X 71 matrix B such that AB BA In where In is the identity matrix of order n The matrix B is called the multiplicative inverse of A If a matrix does not have an inverse7 it is called a noninvertible or singular matrix A hon square matrix does have an inverse Because ifA is an m X 72 matrix with m a n then for AB and BA to be de ned the size of B has to be n X m The size of AB would be m X m and size of BA would be n X 72 S0 AB a BA Theorem 232 If A is invertible7 then the inverse is unique The inverse is denoted by A l Proof Suppose B and C are two inverses of A We nned to prove B C We have ABBAI and ACCAI So 7 B BI BAC BAC 10 0 So7 the proof is complete I Remark 1 Note that the proof works if C was only a right inverse and B was a left inverse of A 2 Also note that the proof did not use any property of matrices7 except associativity 231 Finding Inverse of a matrix There are many methods of nding inverse of a matrix We will discuss at least two of them The easy one7 for a 2 X 2 matrices is given by the theorem 23 THE INVERSE OF A MATRIX 69 Theorem 233 Suppose is a non zero 2 X 2 matrix Then 1 If ad7 bc 0 then A has no invierse ie A is a singular matrix 2 If ad 7 be a 0 then Proof Write B d 7b 7c a First a b d 7b ad7bc 0 AB7C d 70 ali 0 ad7bcliadibc1239 1 Case 1 Assume ad 7 be 0 Then we have AB 0 So A cannot have an inverse otherwise we will get B A 1AB O which is not the case 2 Case 2 Assume ad 7 be a 0 We need to prove A B 2 BA Multiply the above equation by we get 1 A ad7 chgt 239 Similarly B A 2 So the proof is complete I Exercise 234 Ex 6 p84 Find the inverse of the matrix 1 72 A if em sts 2 73 70 CHAPTER 2 MATRICES Solution We have adi bc 173 7 722 1 So7 by theorem 233 7 73 2 72 1 Reading assignment Read Textbook7 Example 1 and 27 p 74 75 and Textbook7 Example 57 p 79 1 d 7b A 1 1d 7 b0 70 a 232 2nd method of nding inverse of a matrix The second mothod of nding the inverse evolves out of the spirit of solving equation Suppose A is an n X 72 matrix and it has an inverse X7 where X is an n X 72 matrix To nd X7 we have to solve the equation AX In The following method is suggested to solve this equation 1 Write the augmented matrix of this Equation AX In as A 1 In 2 lfpossible7 row reduce A to the identity matrix I using elementary row operations on the entire augmented matrix A 1 If the result is I l C 7 then A 1 C If this is not possible7 the matrix is not invertible or singular 3 Check your work by multiplying AA 1 and A 1A to see AA l AAA I Justifcation or Proof Suppose B is an m X 72 matrix Then7 each elemetary row operation on a matrix B is same as leftmultiplication of B by a matrix 1 For example7 multiplying the ithirow B by a constant is same as left multiplication of B by the diagonal matrix of order 71 whose 17 7entry is c and other digonal entries are 1 and rest of the entries are zero 23 THE INVERSE OF A MATRIX 71 2 lnterchanging of for example rst and second row is left multiplication of B by the partitioned matrix 01 9 101 0 O m72 We use the notation O to denote the zero matrix of appropriate size So7 we mean SB is the matrix obtained by switching rst and second row of B 3 Adding citimes rst row to the second is same as left multiplication of B by the partitioned matrix 1 0 5 c 1 O O m72 So7 we mean 5B is the matrix obtained from B by adding citimes the rst row to second We can write down a similar fact for adding citimes the 1th row to jth row Our method says7 thee is sequence of matrices E17 E27 7 ET such that the left muitiplications gives ElEszaiAiIi 1101 With E E1E2 ET7 we have EAl1IlC gt EAlEIlC This means7 EA I and E C So7 E C is the left inverse of A This completes the justi cationproof I Reading assignment Read Textbook7 Example 3 and 47 p 76 75 Exercise 235 Ex 10 p 84 Compute the inverse of the matrix 1 2 3 A 3 7 9 71 i4 i7 72 CHAPTER 2 MATRICES Solution Augment this matrix with the idetity matrix 3 and we have 1 2 3 1 0 0 3 7 9 0 1 0 71 74 77 0 0 1 According to the above method7 we will try to reduce rst 3 X 3 part to 37 using row operations Subtract 3 times rst row from second and add rst roe to third 1 2 3 1 0 0 0 1 0 73 1 0 0 72 74 1 0 Now subtract 2 times second row from rst and add 2 times second row to third 1 0 3 7 i2 0 0 1 0 73 1 0 0 0 i4 75 2 1 Divide third row by 747 we get 1 0 3 7 72 0 0 1 0 73 1 0 0 0 1 125 75 725 Subtract 3 times third row from rst 1 0 0 325 75 75 0 1 0 73 1 0 0 0 1 125 75 725 So 325 75 75 125 75 725 23 THE INVERSE OF A MATRIX 73 Exercise 236 Ex 12 p 84 Compute the inverse of the matrix 10 5 77 A 75 1 4 3 2 72 Solution Augment this matrix With the idetity matrix 3 and we have 10577100 7514010 3272001 According to the above method7 we Will try to reduce the left 3 X 3 part to 37 using row operations Divide rst row by 10 1 5 77 1 0 0 75 1 4 0 1 0 3 2 72 0 0 1 Add 5 times rst row to second and substract 3 times rst row from third 1 5 77 1 0 0 0 35 5 5 1 0 i 0 5 1 73 0 1 J Divide second row by 35 1 5 77 1 0 0 1 1 2 0 1 7 7 7 0 0 5 1 73 0 1 Subtract 5 times second row from both the rst and third 54 1 1 1 1 2 0 1 7 7 7 0 1 13 1 0 0 g 7E 77 1 74 CHAPTER 2 MATRICES Multiply last row by 35 1 1 1 1 2 0 1 7 7 7 0 0 0 1 713 75 35 Subtract 7 times third row from second and add times third row from the rst 1 0 0 710 74 27 0 1 0 2 1 75 0 0 1 713 75 35 So 233 Properties of Inverses Following is a list properties of inverses Theorem 237 Suppose A is an invertible matrix and c a 0 is a scalar Also7 let k be a positive integer Then 23 THE INVERSE OF A MATRIX 75 Proof From de nitions7 B is an inverse of A7 if AB BA I So7 1 is obvious To prove 27 we need to show AkA 1k A 1kAk I For k 27 we need to prove A2A 12 A 12A2 I But A2A 12 AAA 1A 1 AAA 1A 1 AIA 1 AA l 1 Similarly7 A 12A2 I For any integer k gt 27 we prove it similarly or we can prove it by method of induction To do this we assume that A 1k is the inverse of Ak and use it to prove thatAquotlk l is the inverse of A19 7 as follows Ak1A71k1 AkltAA71A71k AkIA71k AkltA71k I So7 2 is proved Also 1 1 CA A 1gt c AA 1 1 I c c Similarly7 iA l CA I So7 3 is proved Finally7 AT A 1T A lAT 1T 1 Similarly7 A lT AT I So7 all the proofs are complete I Reading Assignment Read Textbook7 Example 67 7 p 80 81 Theorem 238 Let A7 B be two invertible matrices of order n Then AB is invertible and AB 1 B lA l Proof We need to prove7 AB B lA l 1 B lA l AB But AB B lA l ABB 1A 1 A IA l AA l 1 Similarly7 we prove B lA l AB I The proof is complete I 76 CHAPTER 2 MATRICES Theorem 239 Cancellation Property Let A7B be two invert ible matrices of order n and C be an invertible matrix of the same order Then ACBCgtAB and CACBgtAB Proof Suppose AC BC Multiply this equation by 0 1 from the right side we can do this because it is given that C has an inverse7 we get 1400 1 BCC 1 So ACC 1 BCC 1 So A B So A B Similaraly7 we prove the other one The proof is complete I Theorem 2310 Let A be an invertible matrices of order n Then the system of linear equations Ax b has a unique solution given by x A lb Proof Proof is exactly same as that of theorem 239 Suppose Ax b Multiply this eqyuation by A l7 and we get A lAX A lb There fore Ix A lb Hence x A lb The proof is complete I Exercise 2311 Ex 26 p 85 Solve the following three linear sys tems 1 Solve the linear system of equations 2a 7y 73 2a y 7 Solution With 23 THE INVERSE OF A MATRIX 77 the system can be written as Ax b 1 1 72 2 39 The solution is Xilb gillillil 2 Solve the linear system of equations By theorem 2337 A l gtJgtH 2a 7y 71 2a y 3 Solution The system can be writeen as Ax b Where A7X are same as in 1 and 1 1 1 1 b We alread com uted A71 The solution is miliillillil 3 Solve the linear system of equations 2a 7y 6 2a y 10 Solution The system can be writeen as Ax b Where A7X are same as in 1 and 6 1 1 1 b We already computed A l 10 4 2 2 78 CHAPTER 2 MATRICES The solution is Nib iilllillil Exercise 2312 Ex 28 p 85 Solve the following two linear sys tems 1 Solve the linear system of equations 1 2 733 1 722 3 1 7 i3 71 Solution With 1 1 73 x1 0 A 1 72 1 X 32 b 0 1 71 71 3 71 the system can be written as Ax b We need to nd the inverse of A To do this augment the identity matrix 37 to A and we get 1 1 73 1 0 0 1 72 1 0 1 0 1 71 71 0 0 1 Subtract rst row from second and third 1 1 73 1 0 0 0 73 4 71 1 0 0 72 2 71 0 1 Divide second row by 737 we get 1173100 411 017 E70 07227101 THE IN VERSE OF A MATRIX 79 Subtract second row from rst and add 2 times second to third row 10 7 10 017 1710 o o 7 71 71 Multiply third row by 7 we get 10 73 o 017171 o 0 0 1 17 Add 3 times third row to rst and add times third row to second 10 0 g 2 i 010 11 72 0 01 1 7 By theorern 2337 3 5 5 2 5 A47 11 72 1 3 517 The solution is 3 5 5 5 2 5 5 xA 1b 1172 2 1 3 3 517 1 5 2 Solve the linear system of equations 1 2 733 1 1 722 3 1 42 i3 80 CHAPTER 2 MATRICES Solution With A and m as in 1 and with the system can be written as Ax b We already computed A l So7 the solution is 3 5 5 2 75 71 25 xA 1b 1 72 2 1 1 g 0 15 Exercise 2313 Ex 34 p 85 Suppose A7B arer two 2 X 2 ma trices and i l 71 11 11 and B i i ii 11 11 Solution By theorem 2387 we have IIW IIN IIM le 1 Compute AB 1 5 71 71 71 AB B A E H H 1quotquot IIH 1 1 03 U l H N 1 1 1 l 03 N3 N H H 1 1 l H Ieclw 1 NIH 1 1 1 1 39 l w L Imam m Nah t 1 1 1 1 2 Compute AT 1 Solution By theorem 2377 we have AT71 A71T 7i llm IIW 1 1 mm 111 1 24 ELEMENTARY MATRICES 81 3 Compute A In page 80 for a positive integer k it was de ned A k A 1k We have A 2 A lA l 4 Compute 2A 1 IIW IIN lm and l I IIW IIN lm and Solution By theorem 2377 we have 1 1 7 2A 1 A 1 2 Exercise 2314 Ex 38 p 85 Let A 2 5 i1 72 Find 7 so that A is its own inverse Solution If A it its own inverse then A2 2 So7 we have 1 2H 2113 Multiplying7 we have 47 0 10 30 471 so 3 0 41 01 24 Elementary Matrices We skip this section I included some of it in subsection 232 82 CHAPTER 2 MATRICES 25 Application of matrix operations Homework Textbook7 25 EX 157 177 197 217 23 p 113 Main point in this section is to do some applications of matrices Z We discuss stochastic matrices But we will not go deeper into it 2 We discuss application of matrices in Cryptography 25 APPLICATION OF MATRIX OPERATIONS 83 251 Stochastic matrices De nition 251 A square matrix P11 P12 P13 39 39 39 pln P21 P22 P23 39 39 39 P27 P P31 P32 P33 39 39 39 P3n of 8226 R X n pnl p712 p713 39 39 39 prm is called a stochastic matrix if 1 we have 0 S p731 and 2 sum of all the entries in a column is 1 For example7 P11 P21 P31Pn1 1 Remark We Will not emphasize on this topic of stochastic matrices very much in this course We encounter such matrices in probability theory and statistics Reading Assignment Read Textbook7 Example 172 p 99 100 and the discussion preceding this 252 Cryptography A cryptogram is a message written accoring to a secret code We describle a method of using matrix multiplication to encode and de code 84 CHAPTER 2 MATRICES First7 we assign a number to each letter in the letter as follows 03pace 1A 2B 30 4D 5E 6F 8H 9110J11K12L13M14N 15 16P17Q18R19S 20T 21U 22V 23 24X 25Y 26Z Example Write uncoded matrices of size 1 X 3 for the message LINEAR ALGEBRA Solution LIN EAR AL GEB RA 12914 5118 0112 752 1810 253 Encoding Following is the method of encoding and decoing a message 1 Given a message7 rst it is written as sequence of row matrices of a xed size This was done in the above example7 by writting the message in a sequence of uncoded matrices of size 1 X 3 These will be called uncoded matrices 2 The message is encoded by using a square matrix A of appro priate size and simply multiplying the uncoded matrices by A 3 Decoding of the encoded matrices is done by multiplying the encoded matrices by the inverse A 1 of A Exercise 252 Ex 16 p113 Use the endoing matrix 14 2 1 A 73 73 71 3 2 1 25 APPLICATION OF MATRIX OPERATIONS 85 to encode the message PLEASE SEND MONEY Solution We rst write the message in uncoded row matrices of size 1gtlt3asfollows PLE ASE SE N D 16125 1195 0195 1440 MON EY 131514 5 25 0 So7 the encoded matrices are given by uncoded X A encoded 16 12 5 A 203 6 9 1 19 5 A 728 745 713 0 19 5 A 742 747 714 14 4 0 A 184 16 10 13 15 14 A 179 9 12 5 25 0 A 5 765 720 Exercise 253 Ex 22 p 113 Let 3442 A 021 4 i5 3 be the encoding matrix Decode the cryptogram 112 140 8319 25 13 72 76 6195 118 7120 2138 35 23 36 42 48 32 Solution We use T1 to nd 11 2 i8 86 CHAPTER 2 MATRICES So7 the uncoded matrices and corresponding letters are given by encoded X A 1 unencoded word 112 7140 83 A 1 8 1 22 HAV 19 72513 A 1 5 01 E A 72 776 61 A 1 0 7 18 GR 95 7118 71 A 1 5 1 20 EAT 20 21 38 A 1 0 23 5 WE 35723 36 A 1 5 11 5 EKE 42 748 32 A 1 14 4 0 ND So the message is HAVE A GREAT WEEKEND

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