Homework: 2.1 (page 56): 7, 9, 13, 15, 17, 25, 27, 35, 37, 41, 46, 49, 67

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1 Chapter Matrices Operations with Matrices Homework: (page 56):, 9, 3, 5,, 5,, 35, 3, 4, 46, 49, 6 Main points in this section: We define a few concept regarding matrices This would include addition of matrices, scalar multiplication and multiplication of matrices We also represent a system of linear equation as equation with matrices 4

2 4 CHAPTER MATRICES In this section, we will do some algebra of matrices That means, we will add, subtract, multiply matrices Matrices will usually be denoted by upper case letters, A, B, C, Such a matrix A is also denoted by a ij a a a 3 a n a a a 3 a n a 3 a 3 a 33 a 3n a m a m a m3 a mn Remark I used parentheses in the last chapter, not a square brackets The textbook uses square brackets Definition Two matrices A a ij and B b ij are equal if both A and B have same size m n and the entries a ij b ij for all i m and j n Read Textbook, Example, p 4 for examples of unequal matrices Definition Following are some standard terminologies: A matrix with only one column is called a column matrix or column vector For example, 3 a 9 is a column matrix A matrix with only one row is called a row matrix or row vector For example b is a row matrix 3

3 OPERATIONS WITH MATRICES 43 3 Bold face lower case letters, as above, will often be used to denote row or column matrices Matrix Addition Definition 3 We difine addition of two matrices of same size Suppose A a ij and B b ij be two matrices of same size m n Then the sum A + B is defined to be the matrix of size m n given by A + B a ij + b ij For example, with A , B we have A+B Also, for example, with 35 C, D we have C+D While, the sum A + C is not defined because A and C do not have same size 4 Read Textbook, Example, p 48 for more such examples Scalar Multiplication Recall, in some contexts, real numbers are referred to as scalars (in contrast with vectors) We define, multiplication of a matrix A by a scalar c

4 44 CHAPTER MATRICES Definition 4 Let A a ij be an m n matrix and c be a scalar We define Scalar multiple ca of A by c as the matrix of same size given by ca ca ij For a matrix A, the negative of A denotes ( )A Also A B : A + ( )B Let c and A Then, with c we have ca , B Likewise, A B A + ( )B ( ) Read Textbook, Example 3, p 49 for more such computations

5 OPERATIONS WITH MATRICES 45 3 Matrix Multiplication The textbook gives a helpful motivation for defining matrix multiplication in page 49 Read it, if you are not already motivated Definition 5 Suppose A a ij is a matrix of size m n and B b ij is a matrix of size n p Then the product AB c ij is a matrix size m p where c ij : a i b j + a i b j + a i3 b 3j + + a in b nj n a ik b kj k Note that number of columns (n) of A must be equal to number of rows (n) of B, for the product AB to be defined Note the number of rows of AB is equal to is same as that (m) of A and number of columns of AB is equal to is same as that (n) of B 3 Read Textbook, Example 4-5, p5-5 I will workout a few below Exercise 6 (Ex, p 5) Let A 8 3 Compute AB abd BA We have AB, B 3 + ( ) + + ( ) + ( 3) + ( ) + + ( ) ( ) + 8 ( 3) + ( ) ( ) ( ) ( 3) ( )

6 46 CHAPTER MATRICES Now, we compute BA We have ( ) + ( ) ( BA ( ) + ( ) ( + ( 3) + 3 ( ) + ( 3) ( ) + + ( 3) 8 + ( Also note that AB BA Exercise (Ex 6, p 5) Let 0 3 A Compute AB and BA, if definded, and B Solution: Note BA is not defined, because size of B is 3 and size of A is 3 3 But AB is defined and is a 3 matrix (or a column matrix) We have 0 + ( ) ( 3) + 3 AB ( 3) ( ) ( 3) Exercise 8 (Ex 8 p 5) Let A and B 6 4

7 OPERATIONS WITH MATRICES 4 Compute AB and BA, if definded Solution Note AB is not defined, because size of A is 5 and size of B is But BA is defined and we have BA ( ) + 6 ( ) ( ) + ( ) Systems of linnear equations Systems of linear equations can be represented in a matrix form Given a system of liner equations a x + a x + a 3 x a n x n b a x + a x + a 3 x a n x n b a 3 x + a 3 x + a 33 x a 3n x n b 3 a m x + a m x + a m3 x a mn x n b m we can write this system in the matrix form as where A a a a 3 a n a a a 3 a n a 3 a 3 a 33 a 3n a m a m a m3 a mn Ax b, x x x x 3 x n, b Clearly, A is the coefficient matrix, x would be called the unknown (or variable) matrix (or vector) and b would be called the constant vector (or matrix) b b b 3 b n

8 48 CHAPTER MATRICES With a a a a 3 a m, a we can write (or think) as a matrix of matrices a a a 3 a m, a j a j a j a 3j a mj A a a a j a n, a n The above system of linear equations can be (re)writeen as x a + x a + + x j a j + + x n a n b a n a n a 3n a mn We say, b is a linear combination of the column metrices (or vectors) a, a,, a n with coefficents x, x,, x n 3 So, the solutions of the above system are precisely those c,, c n such that b is a linear combination of the vectors a, a,, a n with coefficents c, c,, c n Exercise 9 (Ex 34 (changed), p 58) Consider the system of equation: x +x 3x 3 x +x x x +x 3 Write this system in the matrix-form Ax b and solve matrix equation for x Solution: The equation reduces to 3 0 x x x 3

9 OPERATIONS WITH MATRICES 49 This answers the first part system is To solve, the augmented matrix of the 3 0 Using TI-84, we can reduce it to Gauss-Jordan form: The corresponding linear system is: Multiplying, this give the solution: x x x 3 x x x Exercise 0 (Ex 40, 58) a c b d Solution: Here we have four unknowns a, b, c, d In any case, multiplying: a + 3b a + b 3 c + 3d c + d 4

10 50 CHAPTER MATRICES Equating respective entries: a +3b 3 a +b c +3d 4 c +d The augmented matrix of the system: Use TI-84 to reduce it to the Gauss-Jordan form: Looking at the corresponding to this matrix, we get a b a 48, b 3, c, d 6 and c d

11 PROPERTIES OF MATRIX OPERATIONS 5 Properties of Matrix operations Homework: (page 0):, 3,, 9, 5,,, 3, 9, 3, 35, 3, 39, 43, 5, 59 Main points in this section: We write down some of the properties of matrix addition and multiplication We define transpose of a matrix 3 We start writing some proofs

12 5 CHAPTER MATRICES We start developing some algebra of matrices Theorem (Properties) Let A, B, C be three m n matrices and c, d be scalars Then A + B B + A Commutativity of matrix addition (A + B) + C A + (B + C) Associativity of matrix addition 3 (cd)a c(da) Associativity of scalar multiplication 4 A A identity of scalar multiplication 5 c(a + B) ca + cb Distributivity of scalar multiplication 6 (c + d)a ca + da Distributivity of scalar multiplication Proof One needs to prove all these statements using definitions of addition and scalar multiplication To prove the commutativity of matrix addition (), Let A a ij, B b ij Both A and B have same size m n, so A + B, B + A are defined From definition A + B a ij + b ij a ij + b ij and B + A b ij + a ij b ij + a ij From commutative property of addition of real numbers, we have a ij + b ij b ij +a ij Therefore, from definition of equality of matrices, A+B B + A So, () is proved Other properties are proved similarly Remark For matrices A, B, C as in the theorem, by the expression A + B + C we mean (A + B) + C or A + (B + C) It is well defined, because (A + B) + C A + (B + C) by associatieve property() of matrix addition Theorem Let O mn denote the m n matrix whose entries are all zero Let A be a m n matrix Then Then A + O mn A

13 PROPERTIES OF MATRIX OPERATIONS 53 We have A + ( A) O mn 3 If ca O mn, then either c 0 or A 0 We say, on the set of all m n matrices, O mn is an additive identity (property ()), and ( A) is the additive inverse of A (property ()) Properties of matrix multiplication Theorem 3 (Mult-Properties) Let A, B, C be three matrices (of varying sizes) so that all the products below are defined and c be a scalars Then AB BA Failure of Commutativity of multiplication (AB)C A(BC) Associativity of multiplication 3 (A + B)C AC + BC Left Distributivity of multiplication 4 A(B + C) AB + AC Right Distributivity of multiplication 5 c(ab) (ca)b an Associativity In (), by AB BA, we mean AB is not always equal to BA Proof We will only prove (4) In this case, let A be a matrix of size m n and then B, C would have to be of same size n p Write A a a a 3 a n a a a 3 a n a 3 a 3 a 33 a 3n a m a m a m3 a mn,

14 54 CHAPTER MATRICES and b b b 3 b p c c c 3 c p b b b 3 b p c c c 3 c p B b 3 b 3 b 33 b 3p, C c 3 c 3 c 33 c 3p b n b n b n3 b np c n c n c n3 c np Therefore, A(B + C) a a a 3 a n a a a 3 a n a 3 a 3 a 33 a 3n b + c b + c b 3 + c 3 b p + c p b + c b + c b 3 + c 3 b p + c p b 3 + c 3 b 3 + c 3 b 33 + c 33 b 3p + c 3p a m a m a m3 a mn b n + c n b n + c n b n3 + c n3 b np + c np which is n k a k(b k + c k ) n k a k(b k + c k ) n k a k(b kp + c kp ) n k a k(b k + c k ) n k a k(b k + c k ) n k a k(b kp + c kp ) n k a 3k(b k + c k ) n k a 3k(b k + c k ) n k a 3k(b kp + c kp ) n k a mk(b k + c k ) n k a mk(b k + c k ) n k a mk(b kp + c kp )

15 PROPERTIES OF MATRIX OPERATIONS 55 which is n k a n kb k k a kb k n k a kb kp n k a n kb k k a kb k n k a kb kp n k a n 3kb k k a 3kb k n k a 3kb kp n k a n mkb k k a mkb k n k a mkb kp + n k a kc k n k a kc k n k a kc kp AB + AC n k a n kc k k a kc k n k a kc kp n k a n 3kc k k a 3kc k n k a 3kc kp n k a mkc k ) n k a mkc k n k a mkc kp So, A(B + C) AB + AC and the proof of complete Alternate way to write the same proof: Let A be a matrix of size m n and then B, C would have to be of same size n p Write A a ik, B b kj, C c kj Then B + C b kj + c kj So, A(B + C) a ik b kj + c kj α ij (say) Then the (ij) th entry α ij of A(B + C) is given by n n n α ij a ik (b kj + c kj ) a ik b kj + a ik c kj k k k But n n AB a ik b kj, and AC a ik c kj k k So, the (ij) th entry of AB + AC is also equal to α ij Therefore A(B + C) AB + AC and the proof is complete

16 56 CHAPTER MATRICES Remark For matrices A, B, C as in the theorem so that all the multiplications in the associative law () are defined Then the expression ABC would mean (AB)C or A(BC) It is well defined, because (AB)C A(BC) by associatieve property() of matrix multiplication Reading assignment Textbook, Example 3, p 64 to experience that associativity (property ) works Textbook, Example 4, p 64 to see examples that commutativity for multiplication fails (property ) 3 Textbook, Example 5, p 65 to see examples that cancellation fails That means, examples AB AC for nonzero A, B, C with B C This makes solving matrix equation like AX AC, different from solving equation in real numbers, like ax c Theorem 4 Let I n denote the square matrix of order n whose main diagonal entries are and all the entries are zero So, I n 0 0 0, I 3 0 0, I Let A be an m n matrix Then I m A A and AI n A Because of these two properties, I n is called the identity matrix of order n Proof Easy

17 PROPERTIES OF MATRIX OPERATIONS 5 Reading assignment Textbook, Example 6, p 66 Textbook, Example, p 66, just to get used to computing A r Following the definition in item (4) we gave a classification of system of equation, which we state as a theorem and prove as follows Theorem 5 Let Ax b be a linear system of m equations in n unknowns Then exactly one of the follwoing is true: The system has no solution The system has exactly one solution 3 The system has infinitely many solutions Proof If the system has no solution or have exactly one solution, then there is nothing to prove So, assume, it has at least two distinct solutions x, x So, Ax b and Ax b Subtracting the second from the first, we have A(x x ) 0 Write y x x Then y 0 and Ay 0 So, for any scalar c, we have A(x + cy) Ax + A(cy) b + (Ac)y b + c(ay) b + c0 b So, x +cy is a solution, for each scalar c (and they are all distinct) So, the system has infinitely many solutuions This completes the proof Transpose of a matrix Definition 6 The transpose of a matrix A is obtained by writing the rows as columns (and/or writing the columns as rows) The transpose of A is denoted bt A T So, for the matrix

18 58 CHAPTER MATRICES A a a a 3 a n a a a 3 a n a 3 a 3 a 33 a 3n A T a a a 3 a m a a a 3 a m a 3 a 3 a 33 a m3 a m a m a m3 a mn a n a n a 3n a mn Definition A (square) matrix A is said to be symmetric, if A A T Theorem 8 Let A, B be matrices (of varying sizes so that the sum or product is defined) and c be a scalar Then, (A T ) T A (That means transpose of transpose is the same) (A + B) T A T + B T 3 (ca) T ca T 4 (AB) T B T A T (This is the most important one in this list) In fact, properties, 4 extends to sum of higher number of matrices Reading assignment Textbook, Example 8, p 68 to compute transpose of a matrix Textbook, Example 9, p 66, about transpose and product of matrices

19 PROPERTIES OF MATRIX OPERATIONS 59 Exercise 9 (Ex 8, p 0) Let A 0 Then Solve X 3A B 3 4, B Solution: In this case X 3A B 3 0 which is Solve X A B Solution: We have X A B 0 OR 3 4 X Divide by (ie multiply by ), we have X

20 60 CHAPTER MATRICES 3 Solve X + 3A B Solution: We have X + 3A B Subtracting 3A from both sides, X B 3A So, 0 3 X B 3A Divide by (ie multiply by ), we have X Exercise 0 (Ex 0-changed, p 0) Let 3 3 A, B, C 0 Compute C(BA) and A(BC), if defined 0 0 Solution: Note A(BC) is not defined because A has 3 column and BC has two rows We compute C(BA) We have C(BA) (CB)A and 0 3 CB 0 3 So, C(BA) (CB)A Exercise (Ex 8, p 0) Let 4 A, B 4 05

21 PROPERTIES OF MATRIX OPERATIONS 6 Then AB 0 but A 0 nor B 0 (Note, such a phenomenon will not occur with real numbers That is why with real numbers, ax 0, a 0 x 0 We cannot do similar algebra with matrices With A 0 the solution to the equation AX 0 is not necessarily X 0) Solution: Obvious Exercise (Ex, p 0) Let A, and I I 0 Compute A + IA 0 0 Solution: We have A + IA A + A A Exercise 3 (Ex 4 p 0) Let A 3 4 Compute A T, A T A and AA T 0 Solution: We have A T So, A T A

22 6 CHAPTER MATRICES Also AA T which is Exercise and comment: Notice in this exercise 4 above, both AA T and A T A are symmetric matrices This is not a surprise In fact, for any matrix A, both AA T and A T A are symmetric matrices Proof ( AA T ) T ( A T ) T A T AA T So, AA T is symmetric Similarly, A T A is symmetric Exercise 4 (Ex 38 p ) Let X, Y 0, Z 4 3 4, W Find scalars a, b such that Z ax + by 0 0, O Solution: This is, in fact, a problem of solving a system of linear equations Suppose Z ax + by Then, a a X Y Z OR 0 4 b b 3 4

23 PROPERTIES OF MATRIX OPERATIONS 63 Here a, b are unknown to be solved for The augmented matrix of this system is: By TI-84, the matrix reduced to the Gauss-Jordan form: This gives a, b (We can check X Y Z) Show that there do not exist scalar a, b such that W ax + by Solution: This is, in fact, a problem of solving a system of linear equations Suppose W ax + by Then, 0 a a X Y W OR 0 0 b b 3 Here a, b are unknown to be solved for The augmented matrix of this system is: Here a, b are unknown to be solved for The 3 augmented matrix of this system is: By TI-84, the matrix reduced to the Gauss-Jordan form:

24 64 CHAPTER MATRICES The system corresponding to this matrix is a 0, b 0, 0, which is inconsistent 3 Show that ax + by + cw O then a b c 0 Solution: This is, again, a problem of solving a system of linear equations Suppose O ax + by + cw Then, X Y W a b c O OR 0 a 0 0 b 3 The augmented matrix of this system is given by c From TI-83/84, the matrix reduces to the following Gauss-Jordan form: So, a 0, b 0, c 0, as was required to show 4 Find the scalars a, b, c, not all zero (nontrivial solution), such that ax + by + cz O Solution: This is, again, a problem of solving a system of linear equations Suppose O ax + by + cz Then, X Y Z a b c O OR a 0 4 b 3 4 c

25 PROPERTIES OF MATRIX OPERATIONS 65 The augmented matrix of this system is given by From TI-83/84, the matrix reduces to the following Gauss-Jordan form: The linear system corresponding to this matrix is a +c 0 b c Using c t as parameter, we have a t, b t, c t So, for any scalar t, we have tx + ty + tz O Exercise 5 (Ex 44, p ) Let f(x) x x + 6 and A Compute f(a) 5 4 Solution: We have A AA So, f(a) A A + 6I

26 66 CHAPTER MATRICES So, f(a) O (One can say that A is a (matrix) root of f(x))

27 3 THE INVERSE OF A MATRIX 6 3 The Inverse of a matrix Homework: Textbook, Ex 9, 3, 5,, 33, 35, 3, 4, 45 ; page 84- Main point in the section is to define and compute inverse of matrices: We give a formula to compute inverse of a matrices We describe the method of row reduction to compute inverse of a matrix of any size 3 We use inverse of matrices to solve system of linear equations

28 68 CHAPTER MATRICES Definition 3 For a square matrix A of size n n, we say that A is invertible (or nonsingular) if there is an n n, matrix B such that AB BA I n where I n is the identity matrix of order n The matrix B is called the (multiplicative ) inverse of A If a matrix does not have an inverse, it is called a noninvertible (or singular) matrix A non-square matrix does have an inverse Because if A is an m n matrix with m n, then for AB and BA to be defined, the size of B has to be n m The size of AB would be m m and size of BA would be n n So, AB BA Theorem 3 If A is invertible, then the inverse is unique inverse is denoted by A The Proof Suppose B and C are two inverses of A We nned to prove B C We have So, So, the proof is complete AB BA I and AC CA I B BI B(AC) (BA)C IC C Remark () Note that the proof works if C was only a right-inverse and B was a left-inverse of A () Also note that the proof did not use any property of matrices, except associativity 3 Finding Inverse of a matrix There are many methods of finding inverse of a matrix We will discuss at least two of them The easy one, for a matrices is given by the theorem:

29 3 THE INVERSE OF A MATRIX 69 Theorem 33 Suppose A a c b d is a non-zero matrix Then If ad bc 0, then A has no invierse (ie A is a singular matrix) If ad bc 0, then A ad bc d c b a Proof Write B d c b a First, a b AB c d d c b a ad bc 0 0 ad bc (ad bc)i (Case :) Assume ad bc 0 Then, we have AB O So, A cannot have an inverse (otherwise we will get B A (AB) O, which is not the case) (Case :) Assume ad bc 0 We need to prove A( B) ad bc I ( B)A Multiply the above equation by, we get ad bc ad bc ( ) A ad bc B I Similarly, ( ad bc B) A I So, the proof is complete Exercise 34 (Ex 6, p84) Find the inverse of the matrix A, if exists 3

30 0 CHAPTER MATRICES Solution: We have ad bc ( 3) ( ) So, by theorem 33, A ad bc d c b a 3 Reading assignment: Read Textbook, Example and, p 4-5 and Textbook, Example 5, p 9 3 nd method of finding inverse of a matrix The second mothod of finding the inverse evolves out of the spirit of solving equation Suppose A is an n n matrix and it has an inverse X, where X is an n n matrix To find X, we have to solve the equation AX I n The following method is suggested to solve this equation: Write the augmented matrix of this Equation AX I n as A I n If possible, row reduce A to the identity matrix I using elementary row operations on the entire augmented matrix A I If the result is I C, then A C If this is not possible, the matrix is not invertible (or singular) 3 Check your work by multiplying AA and A A to see AA A A I Justifcation or Proof Suppose B is an m n matrix Then, each elemetary row operation on a matrix B is same as left-multiplication of B by a matrix: For example, multiplying the i th row B by a constant is same as left-multiplication of B by the diagonal matrix of order n, whose (i, i) entry is c and other digonal entries are (and rest of the entries are zero)

31 3 THE INVERSE OF A MATRIX Interchanging of (for example) first and second row is left-multiplication of B by the (partitioned) matrix: 0 S O 0 O I m We use the notation O to denote the zero matrix of appropriate size So, we mean SB is the matrix obtained by switching first and second row of B 3 Adding c times first row to the second is same as left-multiplication of B by the (partitioned) matrix: 0 E O c O I m So, we mean EB is the matrix obtained from B by adding c times the first row to second We can write down a similar fact for adding c times the i th row to j th row Our method says, thee is sequence of matrices E, E,, E r such that the left-muitiplication(s) gives With E E E E r, we have (E E E r )A I I C EA I I C EA E I C This means, EA I and E C So, E C is the (left) inverse of A This completes the justification/proof Reading assignment: Read Textbook, Example 3 and 4, p 6-5 Exercise 35 (Ex 0, p 84) Compute the inverse of the matrix 3 A 3 9 4

32 CHAPTER MATRICES Solution: Augment this matrix with the idetity matrix I 3 and we have According to the above method, we will try to reduce first 3 3 part to I 3, using row operations Subtract 3 times first row from second and add first roe to third: Now subtract times second row from first and add times second row to third: Divide third row by 4, we get Subtract 3 times third row from first: So, A

33 3 THE INVERSE OF A MATRIX 3 Exercise 36 (Ex, p 84) Compute the inverse of the matrix 0 5 A Solution: Augment this matrix with the idetity matrix I 3 and we have According to the above method, we will try to reduce the left 3 3 part to I 3, using row operations Divide first row by 0: Add 5 times first row to second and substract 3 times first row from third: Divide second row by 35: Subtract 5 times second row from both the first and third:

34 4 CHAPTER MATRICES Multiply last row by 35: Subtract 54 times third row from second and add 0 from the first: times third row So, A Properties of Inverses Following is a list properties of inverses: Theorem 3 Suppose A is an invertible matrix and c 0 is a scalar Also, let k be a positive integer Then (A ) A (A k ) (A ) k 3 (ca) c A 4 (A T ) (A ) T

35 3 THE INVERSE OF A MATRIX 5 Proof From definitions, B is an inverse of A, if AB BA I So, () is obvious To prove (), we need to show (A k )(A ) k ) (A ) k )A k I For k, we need to prove (A )(A ) ) (A ) )A I But (A )(A ) ) (AA)(A A ) A((AA )A ) A(IA ) AA I Similarly, (A ) )A I For any integer k >, we prove it similarly or we can prove it by method of induction: To do this we assume that (A ) k is the inverse of A k and use it to prove that(a ) k+ is the inverse of A k+, as follows: (A k+ )(A ) k+ A k (AA )(A ) k A k I(A ) k A k (A ) k I So, () is proved Also ( ) (ca) c A (c c )(AA ) I I Similarly, ( c A ) (ca) I So, (3) is proved Finally, A T ( A ) T ( A A ) T I T I Similarly, (A ) T A T I So, all the proofs are complete Reading Assignment: Read Textbook, Example 6, p 80-8 Theorem 38 Let A, B be two invertible matrices of order n Then AB is invertible and (AB) B A Proof We need to prove, (AB) ( B A ) I ( B A ) (AB) But (AB) ( B A ) A (( BB ) A ) A ( IA ) AA I Similarly, we prove (B A ) (AB) I The proof is complete

36 6 CHAPTER MATRICES Theorem 39 (Cancellation Property) Let A, B be two invertible matrices of order n and C be an invertible matrix of the same order Then AC BC A B and CA CB A B Proof Suppose AC BC Multiply this equation by C from the right side (we can do this because it is given that C has an inverse), we get (AC)C (BC)C So A(CC ) B(CC ) So AI BI So A B Similaraly, we prove the other one The proof is complete Theorem 30 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 b Proof (Proof is exactly same as that of theorem 39) Suppose Ax b Multiply this eqyuation by A, and we get (A A)x A b Therefore Ix A b Hence x A b The proof is complete Exercise 3 (Ex 6, p 85) Solve the following three linear systems Solve the linear system of equations: x y 3 x +y Solution: With A x x y b 3

37 3 THE INVERSE OF A MATRIX the system can be written as Ax b By theorem 33, A 4 The solution is x A b Solve the linear system of equations: x y x +y 3 Solution: The system can be writeen as Ax b where A, x are same as in () and b W e already computed A 3 4 The solution is x A b Solve the linear system of equations: x y 6 x +y 0 Solution: The system can be writeen as Ax b where A, x are same as in () and 6 b W e already computed A 0 4

38 8 CHAPTER MATRICES The solution is x A b Exercise 3 (Ex 8, p 85) Solve the following two linear systems Solve the linear system of equations: Solution: With 3 A x +x 3x 3 0 x x +x 3 0 x x x 3 x x x x 3 b the system can be written as Ax b We need to find the inverse of A To do this augment the identity matrix I 3, to A and we get Subtract first row from second and third: Divide second row by 3, we get

39 3 THE INVERSE OF A MATRIX 9 Subtract second row from first and add times second to third row: Multiply third row by 3, we get Add 5 3 times third row to first and add 4 3 second: times third row to By theorem 33, A The solution is x A b Solve the linear system of equations: x +x 3x 3 x x +x 3 x x x 3 0

40 80 CHAPTER MATRICES Solution: With A and x as in () and with b the system can be written as Ax b We already computed A So, the solution is x A b Exercise 33 (Ex 34, p 85) Suppose A, B arer two matrices and A 3 Compute (AB) and B Solution: By theorem 38, we have (AB) B A Compute (A T ) Solution: By theorem 3, we have (A T ) (A ) T ( 3 ) T 3

41 4 ELEMENTARY MATRICES 8 3 Compute A (In page 80, for a positive integer k it was defined A k : (A ) k ) We have A A A Compute (A) Solution: By theorem 3, we have (A) A Exercise 34 (Ex 38, p 85) Let x A Find x, so that A is its own inverse Solution: If A it its own inverse then A I So, we have x x 0 0 Multiplying, we have 4 x x 0 0 so 4 x ; so x 3 4 Elementary Matrices We skip this section I included some of it in subsection 3

42 8 CHAPTER MATRICES 5 Application of matrix operations Homework: Textbook, 5 Ex 5,, 9,, 3; p 3 Main point in this section is to do some applications of matrices We discuss stochastic matrices But we will not go deeper into it We discuss application of matrices in Cryptography

43 5 APPLICATION OF MATRIX OPERATIONS 83 5 Stochastic matrices Definition 5 A square matrix p p p 3 p n p p p 3 p n p 3 p 3 p 33 p 3n P of size n n p n p n p n3 p nn is called a stochastic matrix if we have 0 p ij and sum of all the entries in a column is For example, p + p + p p n 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 Textbook, Example, p and the discussion preceding this 5 Cryptography A cryptogram is a message written accoring to a secret code We describle a method of using matrix multiplication to encode and decode

44 84 CHAPTER MATRICES First, we assign a number to each letter in the letter as follows: 0 space A B 3 C 4 D 5 E 6 F G 8 H 9 I 0 J K L 3 M 4 N 5 O 6 P Q 8 R 9 S 0 T U V 3 W 4 X 5 Y 6 Z Example Write uncoded matrices of size 3 for the message LINEAR ALGEBRA Solution: L I N 9 4 E A R 5 8 A L 0 G E B 5 R A Encoding Following is the method of encoding and decoing a message: Given a message, first it is written as sequence of row matrices of a fixed size This was done in the above example, by writting the message in a sequence of uncoded matrices of size 3 These will be called uncoded matrices The message is encoded by using a square matrix A of appropriate 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 of A Exercise 5 (Ex 6, p3) Use the endoing matrix A

45 5 APPLICATION OF MATRIX OPERATIONS 85 to encode the message: PLEASE SEND MONEY Solution: We first write the message in uncoded row matrices of size 3 as follows: P L E 6 5 A S E 9 5 M O N So, the encoded matrices are given by: S E E Y uncoded A encoded 6 5 A A A A A A Exercise 53 (Ex p 3) Let 3 4 A be the encoding matrix Decode the cryptogram: N D Solution: We use TI to find: A

46 86 CHAPTER MATRICES So, the uncoded matrices (and corresponding letters) are given by: encoded A unencoded word A 8 HAV A 5 0 E A 6 6 A 0 8 GR 95 8 A 5 0 EAT 0 38 A W E A 5 5 EKE A ND So the message is HAVE A GREAT WEEKEND

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