1. Chapter 8 Matrices and Determinants Matrix Solutions to Linear Systems

2. A matrix gives us a shortened way of writing a system of equations. The first step in solving a system of linear equations using matrices is to write the augmented matrix. An augmented matrix has a vertical bar separating the columns of the matrix into two groups. The coefficients of each variable are placed to the left of the vertical line, and the constants are placed to the right. If any variable is missing, its coefficient is 0. Here are two examples. 25231 19211 31213 25=2z2z+3y3y+x 19=2z2z+y+x 31=2z2z+y+3x3x 4100 9310 -19-521 25=z 19=3z3z+y 31=5z5z–2y2y+x Solving Linear Systems by Using Matrices

3. Example Write the augmented matrix for the following system: 2x-3y+z = -2 3x-5z = 6 4x-3y+z = 2 Solution:

4. Write the solution set for a system of equations represented by the matrix 4100 9310 -19-521 Solution The system represented by the matrix is This system can be simplified as follows. 4100 9310 -19-521 1x + 2y – 5z = -19 0x + 1y + 3z = 9 0x + 0y + 1z = 4 x + 2y – 5z = -19 y + 3z = 9 z = 4 Equation 1 Equation 2 Equation 3 Text Example

5. Solution y + 3z = 9 Equation 2 y + 3(4) = 9 Substitute 4 for z. y + 12 = 9 Multiply. y = -3 Subtract 12 from both sides. The value of z is known. We can find y by back-substitution. With values for y and z, we can now use back-substitution to find x. We see that x = 7, y = -3, and z = 4. The solution set for the system is {(7, -3, 4)}. x + 2y – 5z = -19 Equation 1 x + 2(-3) – 5(4) = -19 Substitute -3 for y and 4 for z. x – 6 – 20 = -19 Multiply. x – 26 = -19 Add. x = 7 Add 26 to both sides. Text Example cont.

6. Matrix Row Operations These row operations produce matrices that lead to systems with the same solution set as the original system. 1.Two rows of a matrix may be interchanged. This is the same as interchanging two equations in the linear system. 2.The elements in any row may be multiplied by a nonzero number. This is the same as multiplying both sides of an equation by a nonzero number. 3.The elements in any row may be multiplied by a nonzero number, and these products may be added to the corresponding elements in any other row. This is the same as multiplying both sides of an equation by a nonzero number and then adding equations to eliminate a variable. Two matrices are row equivalent if one can be obtained from the other by a sequence of row operations.

7. Use the matrix And perform each indicated row operation: a. R 1 R 2 b. 3R 1 c. 2R 2 + R 3 -64-3-2 5-321 21-12183 Solution a.The notation R 1 R 2 means to interchange the elements in row 1 and row 2. This results in the row-equivalent matrix. -64-3-2 21-12183 5-321 This was row 2; now it’s row 1.This was row 1; now it’s row 2. Text Example

8. Solution b.The notation 3R 1 means to multiply each element in row 1 by 3. This results in the row-equivalent matrix. -64-3-2 21-12183 353(-3)3231 -64-3-2 21-12183 5-321 =. c.The notation 2R 2 + R 3 means to add 2 times the elements in row 2 to the corresponding elements in row 3. Replace the elements in row 3 by these sums. First, we find 2 times the elements in row 2: 2(1) or 2, 2(2) or 4, 2(-3) or –6, 2(5) or 10. Now we add these products to the corresponding elements in row 3. Although we use row 2 to find the products, row 2 does not change. It is the elements in row 3 that change, resulting in the row-equivalent matrix 4-210 5-321 21-12183 -6+104-6-3+4-2+2 21-12183 5-321 = Text Example cont.

9. Solving Linear Systems Using Gaussian Elimination Write the augmented matrix for the system Use matrix row operations to simplify the matrix to one with 1s down the diagonal from upper left to lower right, and 0s below the 1s Write the system of linear equations corresponding to the matrix in step 2, and use back-substitution to find the system’s solutions

10. Use matrices to solve the system 25=2z2z+3y3y+x 19=2z2z+y+x 31=2z2z+y+3x3x Solution Step 1Write the augmented matrix for the system. Linear SystemAugmented Matrix 25=2z2z+3y3y+x 19=2z2z+y+x 31=2z2z+y+3x3x 25231 19211 31213 Text Example

11. Solution Step 2Use matrix row operations to simplify the matrix to one with 1s down the diagonal from upper left to lower right, and 0s below the 1s. Our goal is to obtain a matrix of the form f100 ed10 cba1. Our first step in achieving this goal is to get 1 in the top position of the first column. 25231 19211 31213 To get 1 in this position, we interchange rows 1 and 2. (We could also interchange rows 1 and 3 to attain our goal.) 25231 31213 19211 This was row 2; now it’s row 1.This was row 1; now it’s row 2.We want 1 in this position. Text Example cont.

12. Solution Now we want to get 0s below the 1 in the first column. 25231 31213 19211 Now add these products to the corresponding numbers in row 2. Notice that although we use row 1 to find the products, row 1 does not change. We want 0 in these positions. Let’s first get a 0 where there is now a 3. If we multiply the top row of numbers by –3 and add these products to the second row of numbers, we will get 0 in this position. The top row of numbers multiplied by –3 gives -3(1) or –3,-3(1) or –3, -3(2) or –6, -3(19) or –57. 25231 31 + (-3)2 + (-3)1 + (-3)3 + (-3) 19211 25231 -26-4-20 19211 = We want 0 in this position. Text Example cont.

13. Solution Now add these products to the corresponding numbers in row 3. We are not yet done with the first column. If we multiply the top row of numbers by –1 and add these products to the third row of numbers, we will get 0 in this position. The top row of numbers multiplied by –1 gives -1(1) or –1,-1(1) or –1, -1(2) or –2, -1(19) or –19. 25+(-19)2 + (-2)3 + (-1)1 + (-1) -26-4-20 19211 6020 -26-4-20 19211 = We want 1 in this position. 6020 -26-4-20 19211 We move on to the second column. We want 1 in the second row, second column. Text Example cont.

14. Solution To get 1 in the desired position, we multiply –2 by its reciprocal, -1/2. Therefore, we multiply all the numbers in the second row by –1/2 to get 6020 › (-26)› (-4)› (-2)› (0) 19211 6020 13210 19211 = We want 0 in this position. Now add these products to the corresponding numbers in row 3. We are not yet done with the second column. If we multiply the top row of numbers by –2 and add these products to the third row of numbers, we will get 0 in this position. The second row of numbers multiplied by –2 gives -2(0) or 0,-2(1) or –2, -2(2) or –4, -2(13) or –26. 6+(-26)0 + (-4)2 + (-2)0 + 0 13210 19211 -20-400 13210 19211 = Text Example cont.

15. Solution We want 1 in this position. -20-400 13210 19211 We move on to the third column. We want 1 in the third row, third column. To get 1 in the desired position, we multiply –4 by its reciprocal, -1/4. Therefore, we multiply all the numbers in the third row by –1/4 to get -1/4(-20)-1/4(-4)-1/4(0) 13210 19211 5100 13210 19211 = We now have the desired matrix with 1s down the diagonal and 0s below the 1s. Step 3Write the system of linear equations corresponding to the matrix in step 2, and use back-substitution to find the system’s solution. The system represented by the matrix in step 2 is Text Example cont.

16. Solution We immediately see that the value for z is 5. To find y, we back-substitute 5 for z in the second equation. 5=z 13=2z2z+y 19=2z2z+y+x 25231 19211 31213 y + 2z = 13 Equation 2 y + 2(5) = 13 Substitute 5 for x. y = 3 Solve for y. Finally, back-substitute 3 for y and 5 for z in the first equation: x + y + 2z = 19 Equation 1 x + 3 + 2(5) = 19 Substitute 3 for y and 5 for x. x + 13 = 19 Multiply and add. x = 6 Subtract 13 from both sides. The solution set for the original system is {(6, 3, 5)}. Text Example cont.

17. Solving Linear Systems Using Gauss-Jordan Elimination 1.Write the augmented matrix for the system. 2.Use matrix row operations to simplify the matrix to one with 1s down the diagonal from upper left to lower right, and 0s above and below the 1s. 3.Use the reduced row-echelon form of the matrix in step 2 to write the system’s solutions set. (Back-substitution is not necessary.)

18. Inconsistent and Dependent Systems and Their Applications

19. Use Gaussian elimination to solve the system 12=7z7z+4y4y–3x3x 5=6z6z+3y3y–2x2x 2=2z2z–y–x 4-43 56-32 2-21 12=4z4z+4y4y–3x3x 5=6z6z+3y3y–2x2x 2=2z2z–y–x Solution Step 1 Write the augmented matrix for the system. Linear System Augmented Matrix Text Example

20. Solution Step 2 Attempt to simplify the matrix to one with 1s down the diagonal and 0s below the 1s. Notice that the augmented matrix already has a 1in the top position of the first column. Now we want 0s below the 1. To get the first 0, multiply row 1 by  2 and add these products to row 2. To get the second 0, multiply row 1 by  3 and add these products to row 3. Performing these operations, we obtain the following matrix. 6100 1100 2-21 We want 1 in this position. Use the Previous matrix and: Replace row 2 by -2R 1 + R 2. Replace row 3 by -3R 1 + R 3. Text Example cont.

21. Solution Step 2 Moving on to the second column, we obtain 1 in the desired position by multiplying row 2 by  1. Text Example cont.

22. It is impossible to convert this last matrix to the desired form of 1s down the main diagonal. If we translate the last row back into equation form, we get 0x  0y  0z  5, which is false. Regardless of which values we select for x, y, and z, the last equation can never be a true statement. Consequently, the system has no solution. The solution set is , the empty set. Solution Now we want a 0 below the 1 in column 2. To get the 0, multiply row 2 by 1 and add these products to row 3. (Equivalently, add row 2 to row 3.) We obtain the following matrix. Text Example cont.

23. Up to this point, we have encountered only square systems in which the number of equations is equal to the number of variables. In a nonsquare system, the number of variables differs from the number of equations. Nonsquare Systems

24. Example Use Gaussian elimination to solve the following system: 3x-y+z = 2 2x+3y-z = 7 z = t Solution: First, write the system in augmented matrix form:

25. Example cont. (1/3)R 1

26. Example cont. -2R 1 + R 2

27. Example cont. 3/11 R 2

28. Example cont. x - (1/3)y + (1/3)z = 2/3 y - (5/11)z = 17/11 z = t y - (5/11)t = 17/11 y = x - (1/3)(17+5t)/11 + (1/3)t = 2/3 33x - 17 - 5t + 11t = 22 33x + 6t = 39 33x = 39 - 6t

29. Use Gaussian elimination to solve the system 8=z+2y2y+x 26=6z6z+7y7y+3x3x Text Example SolutionWe begin with the augmented matrix. Because we now have 1s down the diagonal that begins with the upper  left entry and a 0 below this 1, we translate the matrix back into equation form. x  2y  z  8 Equation 1 y  3z  2 Equation 2

30. We can let z equal any real number and use back  substitution to express x and y in terms of z. Equation 2 Equation 1 y  3z  2 x  2y  z  8 y   3z  2x  2(  3z  2)  z  8 x  6z  4  z  8 x  5z  4  8 x  5z  4 Solution Text Example cont.

31. With z = t, the ordered solution (x, y, z) enables us to express the system's solution set as {(5t  4,  3t  2, t)} where t is any real number. Solution Text Example cont.

32. Matrix Operations and Their Applications

33. We have seen that an array of numbers, arranged in rows and columns and placed in brackets, is called a matrix. We can represent the matrix in two different ways. A capital letter, such as A, B, or C, can denote a matrix. A lowercase letter enclosed in brackets, such as that shown below, can denote a matrix. A  [a ij ] A general element in matrix A is denoted by a ij. This refers to the element in the ith row and jth column. For example, a 32 is the element of A located in the third row, second column. A matrix of order m X n has m rows and n columns. If m  n, a matrix has the same number of rows as columns and is called a square matrix. Matrix A with elements a i j Notations for Matrices

34. Let a. What is the order of A? b. If A  [a ij ], identify a 23 and a 12. Solution a. The matrix has 2 rows and 3 columns, so it is of order 2 X 3. b. The element a 23 is in the second row and third column. Thus, a 23  1 / 5. The element a 12 is in the first row and second column, and consequently a 12  2. Text Example

35. Definition of Equality of Matrices Two matrices A and B are equal if and only if they have the same order m X n and a ij = b ij for i = 1, 2, …, m and j = 1, 2, …, n

36. Properties of Matrix Addition If A, B, and C are m X n matrices and 0 is an m X n zero matrix, then the following properties are true. 1.A+B = B+A Commutative Property 2.(A+B)+C = A+(B+C) Associative Property 3.A + 0 = 0 + A Additive Identity 4.A+(-A) = (-A) + A = 0 Additive Inverse

37. Example Add the following matrices Solution:

38. Definition of Scalar Multiplication If A  [a ij ] is a matrix of order m X n and c is a scalar, then the matrix cA is the m X n matrix given by cA  [ca ij ]. This matrix is obtained by multiplying each element of A by the real number c. We call cA a scalar multiple of A.

39. Example

40. Example cont. Solution:

41. Properties of Scalar Multiplication If A and B are m X n matrices, and c and d scalars, then the following properties are true. 1.(cd)A = c(dA) Associative Property 2.1A = A Scalar Identity Property 3.c(A+B) = cA + cB Distributive Property 4.(c+d)A = cA + dA Distributive Property

42. Definition of Matrix Multiplication: 2X2 Matrices

43. Example Find the product:

44. Example cont. Solution:

45. Definition of Matrix Multiplication The product of an m X n matrix, A, and an n X p matrix, B, is an m X p matrix, AB, whose elements are found as follows. The element in the ith row and jth column of AB is found by multiplying the each element in the ith row of A by the corresponding element in the jth column of B and adding the products.

46. Properties of Matrix Multiplication (AB)C = A(BC) Associative Property A(B+C) = AB+AC (A+B)C = AC + BC Distributive Properties c(AB) = (cA)B Associative Property of Scalar Multiplication

47. Multiplicative Inverses of Matrices and Matrix Equations

48. Definition of the Multiplicative Inverse of a Square Matrix Let A be an n X n matrix. If there exists an n X n matrix A -1 such that AA -1 = I n and -1 A = I n then A -1 is the multiplicative inverse of A.

49. Show that B is the multiplicative inverse of A, where A = and B = -52 3 12 35 Text Example SolutionTo show that B is the multiplicative inverse of A, we must find the products AB and BA. If B is the multiplicative inverse of A, then AB will be the multiplicative identity matrix and BA will be the multiplicative identity matrix. Because A and B are 2 X 2 matrices, n  2. Thus, we denote the multiplicative identity matrix as I 2 ; it is also a 2 X 2 matrix. We must show that

50. Show that B is the multiplicative inverse of A, where A = and B = -52 3 12 35 Text Example cont. Solution Let’s first show AB=I 2 Both products give the multiplicative identity matrix. Thus, B is the multiplicative inverse of A and we can designate B as A -1  Let’s now show BA=I 2

51. Example Find the multiplicative inverse of A. Solution

52. Example cont. Find the multiplicative inverse of A. Solution

53. Example cont. Find the multiplicative inverse of A. Solution

54. Example cont. Find the multiplicative inverse of A. Solution

55. Example cont. Find the multiplicative inverse of A. Solution

56. Multiplicative Inverse of a 2x2 Matrix The matrix A is invertible if and only if ad-bc  0. If ad-bc=0, then A does not have a multiplicative inverse.

57. Procedure for Finding the Multiplicative Inverse of an Invertible Matrix To find A -1 for any n X n matrix A for which A -1 exists: 1.Form the augmented matrix [A | I ], where I is the multiplicative identity matrix of the same order as the given matrix A. 2.Perform row transformations on [A | I ] to obtain a matrix of the form [I | B ]. This is equivalent to using Gauss-Jordan elimination to change A into the identity matrix. 3.Matrix B is A -1. 4.Verify the result by showing that AA -1  I and A -1 A  I.

58. Solving a System Using A -1 If AX  B has a unique solution, X  A -1 B. To solve a linear system of equations, multiply A -1 and B to find X.

59. Solve the system by using A -1, the inverse of the coefficient matrix Solution The linear system can be written as 1/2=3y3y–-2x 2=z+2y2y  2=z+ y–x 0-3-2 1 0 11 z y x 1/2 2 2 = The solution is given by X  A -1 B. Consequently, we must find A -1. Using the inverse of matrix A that we found previously, Thus, x  1/2, y  -1/2, and z  1. The solution set is {(1/2, -1/2, 1)}. Text Example

60. Encoding a Word or Message 1.Express the word or message numerically. 2.List the numbers in step 1 by columns and form a square matrix. If you do not have enough numbers to form a square matrix, put zeros in any remaining spaces in the last column. 3.Select any square invertible matrix, called the coding matrix, the same size as the matrix in step 2. Multiply the coding matrix by the square matrix that expresses the message numerically. The resulting matrix is the coded matrix. 4.Use the numbers, by columns, from the coded matrix in step 3 to write the encoded message.

61. Decoding a Word or Message That Was Encoded 1.Find the multiplicative inverse of the coding matrix. 2.Multiply the multiplicative inverse of the coding matrix and the coded matrix. 3.Express the numbers, by columns, from the matrix in step 2 as letters.

62. Determinants and Cramer’s Rule

63. Definition of the Determinant of a 2 x 2 Matrix We also say that the value of the second-order determinant is a 1 b 2 – a 2 b 1. The determinant of the matrix is denoted by And is defined by

64. Example Evaluate the determinant of: Solution:

65. Solving a Linear System in Two Variables Using Determinants Cramer’s Rule

66. Example Use Cramer’s rule to solve the system: Solution:

67. Example cont. Use Cramer’s rule to solve the system: Solution:

68. Definition of a Third-Order Determinant

69. Definition of a 3 X 3 Matrix A third-order determinant is defined by

70. Evaluating the Determinant of a 3 X 3 Matrix 1.Each of the three terms in the definition contains two factors - a numerical factor and a second- order determinant. 2.The numerical factor in each term is an element from the first column of the third-order determinant. 3.The minus sign precedes the second term. 4.The second-order determinant that appears in each term is obtained by crossing out the row and the column containing the numerical factor.

71. Evaluate: 241 0-3-2 059 Solution Note that the last column has two 0s. We will expand the determinant about the elements in that column. Text Example

72. Determinants: Inconsistent and Dependent-Systems 1.If D  0 and at least one of the determinants in the numerator is not 0, then the system is inconsistent. The solution set is Ø. 2.If D  0 and all the determinants in the numerators are 0, then the equations in the system are dependent.