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P-BLTZMC08_805-872-hr 21-11-2008 13:26 Page 805 Matrices and Determinants 8 You are being drawn deeper into cyberspace, spending more time online each week. With constantly improving high-resolution images, cyberspace is reshaping your life by nourishing shared enthusiasms. The people who built your computer talk of bandwidth that will give you the visual experience, in high-definition 3-D format, of being in the same room with a person who is actually in another city. Rectangular arrays of numbers, called matrices, play a central role in representing computer images and in the forthcoming technology of tele-immersion. The use of rectangular arrays of numbers in the digital representation of images and the manipulation of images on a computer screen is discussed in Examples 8 and 9 in Section 8.3. 805 P-BLTZMC08_805-872-hr 21-11-2008 13:26 Page 806 806 Chapter 8 Matrices and Determinants Section 8.1 Matrix Solutions to Linear Systems Objectives � Write the augmented matrix � � � for a linear system. Perform matrix row operations. Use matrices and Gaussian elimination to solve systems. Use matrices and GaussJordan elimination to solve systems. he data below show that we spend a lot of time sprucing up. T Average Number of Minutes per Day Americans Spend on Grooming Men Ages 15–19 Ages 20–24 Ages 45–54 Ages 65ⴙ Married Single 37 37 34 28 31 34 59 49 46 46 44 50 B Women R Source: Bureau of Labor Statistics’ American Time-Use Survey The 12 numbers inside the brackets are arranged in two rows and six columns. This rectangular array of 12 numbers, arranged in rows and columns and placed in brackets, is an example of a matrix (plural: matrices). The numbers inside the brackets are called elements of the matrix. Matrices are used to display information and to solve systems of linear equations. Because systems involving two equations in two variables can easily be solved by substitution or addition, we will focus on matrix solutions to systems of linear equations in three or more variables. � Write the augmented matrix for a linear system. Augmented Matrices 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: System of Linear Equations 3x + y + 2z = 31 c x + y + 2z = 19 x + 3y + 2z = 25 c x + 2y - 5z = - 19 y + 3z = 9 z = 4 Augmented Matrix 3 C1 1 1 1 3 2 31 2 3 19 S 2 25 1 C0 0 2 1 0 -5 -19 3 3 9 S. 1 4 P-BLTZMC08_805-872-hr 21-11-2008 13:26 Page 807 Section 8.1 Matrix Solutions to Linear Systems 807 Our goal in solving a system of linear equations in three variables using matrices is to produce a matrix with 1s down the diagonal from upper left to lower right on the left side of the vertical bar, called the main diagonal, and 0s below the 1s. In general, the matrix will be of the form 1 C0 0 a 1 0 b c d 3 e S, 1 f where a through f represent real numbers. The third row of this matrix gives us the value of one variable. The other variables can then be found by back-substitution. � Perform matrix row operations. Matrix Row Operations A matrix with 1s down the main diagonal and 0s below the 1s is said to be in rowechelon form. How do we produce a matrix in this form? We use row operations on the augmented matrix. These row operations are just like what you did when solving a linear system by the addition method. The difference is that we no longer write the variables, usually represented by x, y, and z. Matrix Row Operations The following row operations produce matrices that represent systems with the same solution set: 1. Two rows of a matrix may be interchanged. This is the same as interchanging two equations in a 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. Study Tip When performing the row operation Two matrices are row equivalent if one can be obtained from the other by a sequence of row operations. kRi + Rj you use row i to find the products. However, elements in row i do not change. It is the elements in row j that change: Add k times the elements in row i to the corresponding elements in row j. Replace elements in row j by these sums. Each matrix row operation in the preceding box can be expressed symbolically as follows: 1. Interchange the elements in the ith and jth rows: Ri 4 Rj . 2. Multiply each element in the ith row by k: kRi . 3. Add k times the elements in row i to the corresponding elements in row j: kRi + Rj . EXAMPLE 1 Performing Matrix Row Operations Use the matrix 3 C 1 -2 18 2 -3 -12 21 -3 3 5 S 4 -6 and perform each indicated row operation: a. R1 4 R2 b. 1 3 R1 c. 2R2 + R3 . P-BLTZMC08_805-872-hr 21-11-2008 13:26 Page 808 808 Chapter 8 Matrices and Determinants 3 C 1 -2 18 2 -3 -12 21 -3 3 5 S 4 -6 Solution a. The notation R1 4 R2 means to interchange the elements in row 1 and row 2. This results in the row-equivalent matrix This was row 2; now it’s row 1. 1 2 –3 5 C 3 18 –12 3 21 S . –2 –3 4 –6 The given matrix (repeated) This was row 1; now it’s row 2. b. The notation 13 R1 means to multiply each element in row 1 by 13 . This results in the row-equivalent matrix 1 3 132 C 1 -2 1 3 1182 1 3 1-122 2 -3 -3 4 1 3 1212 1 5 S = C 1 -6 -2 3 6 2 -3 -4 7 - 3 3 5 S. 4 -6 c. The notation 2R2 + R3 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, namely, 1, 2, -3, and 5: 2112 or 2, 2122 or 4, 21-32 or - 6, 2152 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 Replace row 3 by the sum of itself and 2 times row 2. 3 C 1 –2+2=0 18 2 –3+4=1 Check Point 1 –12 21 3 3 S = C1 –3 5 4+(–6)=–2 –6+10=4 0 18 2 1 –12 21 –3 3 5 S. –2 4 Use the matrix 4 12 -20 8 C 1 6 -3 3 7 S - 3 -2 1 -9 and perform each indicated row operation: a. R1 4 R2 � Use matrices and Gaussian elimination to solve systems. b. 1 4 R1 c. 3R2 + R3 . Solving Linear Systems Using Gaussian Elimination The process that we use to solve linear systems using matrix row operations is called Gaussian elimination, after the German mathematician Carl Friedrich Gauss (1777–1855). Here are the steps used in Gaussian elimination: Solving Linear Systems of Three Equations with Three Variables Using Gaussian Elimination 1. Write the augmented matrix for the system. 2. Use matrix row operations to simplify the matrix to a row-equivalent matrix in row-echelon form, with 1s down the main diagonal from upper left to lower right, and 0s below the 1s in the first and second columns. * 1 * * C* * * 3 * S * * * * * 1 * * C0 * * 3 * S * 0 * * Get 1 in the upper lefthand corner. Use the 1 in the first column to get 0s below it. * 1 * * C0 1 * 3 * S * 0 * * * 1 * * C0 1 * 3 * S * 0 0 * Get 1 in the second row, second column position. Use the 1 in the second column to get 0 below it. * 1 * * C0 1 * 3 * S * 0 0 1 Get 1 in the third row, third column position. 3. Write the system of linear equations corresponding to the matrix in step 2 and use back-substitution to find the system’s solution. P-BLTZMC08_805-872-hr 21-11-2008 13:26 Page 809 Section 8.1 Matrix Solutions to Linear Systems EXAMPLE 2 809 Gaussian Elimination with Back-Substitution Use matrices to solve the system: 3x + y + 2z = 31 c x + y + 2z = 19 x + 3y + 2z = 25. Solution Step 1 Write the augmented matrix for the system. Linear System Augmented Matrix 3x + y + 2z = 31 c x + y + 2z = 19 x + 3y + 2z = 25 3 C1 1 1 1 3 2 31 2 3 19 S 2 25 Step 2 Use matrix row operations to simplify the matrix to row-echelon form, with 1s down the main diagonal from upper left to lower right, and 0s below the 1s in the first and second columns. Our first step in achieving this goal is to get 1 in the top position of the first column. 3 C1 1 We want 1 in this position. 1 1 3 2 31 2 3 19 S 2 25 To get 1 in this position, we interchange row 1 and row 2: R1 4 R2 . (We could also interchange row 1 and row 3 to attain our goal.) 1 C3 1 1 1 3 2 19 2 3 31 S 2 25 This was row 2; now it’s row 1. This was row 1; now it’s row 2. Now we want to get 0s below the 1 in the first column. 1 C3 1 We want 0 in these positions. 1 1 3 2 19 2 3 31 S 2 25 To get a 0 where there is now a 3, multiply the top row of numbers by -3 and add these products to the second row of numbers: -3R1 + R2 . To get a 0 where there is now a 1, multiply the top row of numbers by - 1 and add these products to the third row of numbers: -1R1 + R3 . Although we are using row 1 to find the products, the numbers in row 1 do not change. Replace row 2 by −3R1 + R2. Replace row 3 by −1R1 + R3. 1 1 2 19 1 1 2 19 C –3(1)+3 –3(1)+1 –3(2)+2 3 –3(19)+31 S = C 0 –2 –4 3 –26 S 0 2 0 6 –1(1)+1 –1(1)+3 –1(2)+2 –1(19)+25 We want 1 in this position. We move on to the second column.To get 1 in the desired position, we multiply -2 by its reciprocal, - 12 . Therefore, we multiply all the numbers in the second row by - 12 : - 12 R2 . − 21 R2 1 1 2 C– 12 (0) – 12 (–2) – 12 (–4) 0 2 0 19 3 1 1 2 19 1 2 3 13 S . 0 2 0 6 – 12 (–26)S = C 0 6 We want 0 in this position. P-BLTZMC08_805-872-hr 21-11-2008 13:26 Page 810 810 Chapter 8 Matrices and Determinants So far, our matrix row operations have resulted in the matrix that we repeated in the margin. We are not yet done with the second column. The voice balloon shows that we want to get a 0 where there is now a 2. If we multiply the second row of numbers by -2 and add these products to the third row of numbers, we will get 0 in this position: -2R2 + R3 . Although we are using the numbers in row 2 to find the products, the numbers in row 2 do not change. 1 1 2 19 C 0 1 2 3 13 S 0 2 0 6 We want 0 in this position. The matrix from the bottom of the previous page (repeated) Replace row 3 by −2R2 + R3. 1 1 2 19 1 C 3 S=C0 0 1 2 13 –2(0)+0 –2(1)+2 –2(2)+0 –2(13)+6 0 1 2 19 1 2 3 13 S 0 –4 –20 We want 1 in this position. We move on to the third column.To get 1 in the desired position, we multiply - 4 by its reciprocal, - 14 . Therefore, we multiply all the numbers in the third row by - 14 : - 14 R3 . − 41 R3 1 2 1 1 2 19 1 19 C 0 3 1 2 13 S = C 0 1 2 3 13 S 0 0 1 5 – 14 (0) – 14 (0) – 14 (–4) – 14 (–20) We now have the desired matrix in row-echelon form, with 1s down the main diagonal and 0s below the 1s in the first and second columns. Step 3 Write 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 1 C0 0 1 1 0 2 19 1x + 1y + 2z = 19 3 2 13 S : c 0x + 1y + 2z = 13 or 1 5 0x + 0y + 1z = 5 x + y + 2z = 19 c y + 2z = 13. z = 5 (1) (2) (3) We immediately see from equation (3) that the value for z is 5. To find y, we back-substitute 5 for z in the second equation. y + 2z y + 2152 y + 10 y = = = = 13 13 13 3 Equation (2) Substitute 5 for z. Multiply. Subtract 10 from both sides and solve for y. Finally, back-substitute 3 for y and 5 for z in the first equation. x + y + 2z x + 3 + 2152 x + 13 x Technology Most graphing utilities can convert an augmented matrix to row-echelon form, with 1s down the main diagonal and 0s below the 1s. However, row-echelon form is not unique. Your graphing utility might give a rowechelon form different from the one you obtained by hand. However, all row-echelon forms for a given system’s augmented matrix produce the same solution to the system. Enter the augmented matrix and name it A. Then use the 冷 REF 冷 (row-echelon form) command on matrix A. = = = = 19 19 19 6 Equation (1) Substitute 3 for y and 5 for z. Multiply and add. Subtract 13 from both sides and solve for x. With z = 5, y = 3, and x = 6, the solution set of the original system is 516, 3, 526. Check to see that the solution satisfies all three equations in the given system. Check Point 2 Use matrices to solve the system: 2x + y + 2z = 18 c x - y + 2z = 9 x + 2y - z = 6. Modern supercomputers are capable of solving systems with more than 600,000 variables. The augmented matrices for such systems are huge, but the solution using matrices is exactly like what we did in Example 2. Work with the augmented matrix, P-BLTZMC08_805-872-hr 21-11-2008 13:26 Page 811 Section 8.1 Matrix Solutions to Linear Systems 811 one column at a time. Get 1s down the main diagonal from upper left to lower right and 0s below the 1s. Let’s see how this works for a linear system involving four equations in four variables. Gaussian Elimination with Back-Substitution EXAMPLE 3 Use matrices to solve the system: 2w w d w -w + + x x x 2x + + - 3y 2y y 2y + - z 2z z z = 6 = -1 = -4 = - 7. Solution Step 1 Write the augmented matrix for the system. Linear System 2w w d w -w + + x x x 2x + + - Augmented Matrix 3y 2y y 2y + - z 2z z z = 6 = -1 = -4 = -7 2 1 D 1 -1 1 -1 -1 2 3 2 -1 -2 -1 6 -2 4 -1 T 1 -4 -1 -7 Step 2 Use matrix row operations to simplify the matrix to row-echelon form, with 1s down the main diagonal from upper left to lower right, and 0s below the 1s in the first, second, and third columns. Our first step in achieving this goal is to get 1 in the top position of the first column. To do this, we interchange row 1 and row 2: R1 4 R2 . 1 2 D 1 –1 We want 0s in these positions. –1 1 –1 2 2 3 –1 –2 –2 –1 –1 6 4 T 1 –4 –1 –7 This was row 2; now it’s row 1. This was row 1; now it’s row 2. Now we use the 1 at the top of the first column to get 0s below it. 1 0 D 0 0 Use the previous matrix and: Replace row 2 by −2R1 + R2. Replace row 3 by −1R1 + R3. Replace row 4 by 1R1 + R4. –1 3 0 1 2 –1 –3 0 –2 –1 3 8 4 T 3 –3 –3 –8 We want 1 in this position. We move on to the second column. We can obtain 1 in the desired position by multiplying the numbers in the second row by 13 , the reciprocal of 3. 1 1 3 (0) D 0 0 –1 1 3 (3) 0 1 –2 –1 1 3 (–1) 2 1 3 (3) 1 3 (8) –3 0 3 –3 4 1 0 T=D –3 0 –8 0 –1 2 1 – 13 0 –3 1 0 We want 0s in these positions. The top position already has a 0. –2 –1 8 1 3 4 T 3 –3 –3 –8 1 3 R2 P-BLTZMC08_805-872-hr 21-11-2008 13:26 Page 812 812 Chapter 8 Matrices and Determinants 1 0 D 0 0 –1 2 1 – 13 0 –3 1 0 So far, our matrix row operations have resulted in the matrix that we repeated in the margin. Now we use the 1 in the second row, second column position to get 0s below it. –2 –1 8 1 3 4 T 3 –3 –3 –8 Replace row 4 in the previous matrix by −1R2 + R4. We want 0s in these positions. The top position already has a 0. 1 0 D 0 0 –1 1 0 0 2 – 13 –3 1 3 –2 –1 8 1 4 3T 3 –3 –4 – 323 We want 1 in this position. We move on to the third column.We can obtain 1 in the desired position by multiplying the numbers in the third row by - 13 , the reciprocal of -3. The matrix from the bottom of the previous page (repeated) 1 –1 2 –2 –1 1 8 1 –3 0 1 1 0 3 D 1 4 T=D 0 – 3 (0) – 13 (0) – 13 (–3) – 13 (3) – 13 (–3) 32 1 0 0 –4 0 –3 3 –1 2 1 1 –3 0 1 1 0 3 –2 –1 8 1 3 4 T –1 1 –4 – 323 − 31 R3 We want 0 in this position. Now we use the 1 in the third column to get 0 below it. 1 0 D 0 0 Replace row 4 in the previous matrix by − 31 R3 + R4. –1 1 0 0 2 –2 –1 8 1 – 3 4 T 1 –1 1 0 – 113 –11 1 3 We want 1 in this position. We move on to the fourth column. Because we want 1s down the diagonal from upper left to lower right, we want 1 where there is now - 11 3 . We can obtain 1 in this 3 . position by multiplying the numbers in the fourth row by - 11 1 –1 2 –2 –1 8 0 1 1 – 13 3 D 4 T 0 0 1 –1 1 – 113 (0) – 113 (0) – 113 (0) – 113 (–113) – 113 (–11) 1 0 =D 0 0 –1 2 1 – 13 0 1 0 0 –2 –1 8 1 3 4 T –1 1 1 3 − 113 R4 We now have the desired matrix in row-echelon form, with 1s down the main diagonal and 0 s below the 1s. An equivalent row-echelon matrix can be obtained using a graphing utility and the 冷 REF 冷 command on the augmented matrix. Step 3 Write 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 1 0 D 0 0 -1 1 0 0 2 - 13 1 0 -2 -1 1w 0w + 1 4 83 T:d 1 0w + -1 3 0w + 1 1x 1x 0x 0x + + + 2y + 1y 0y + 1 3y 2z 1z 1z 1z = -1 = 83 or = 1 = 3 w - x + 2y - 2z = - 1 x - 13 y + z = 83 d y - z = 1 z = 3. We immediately see that the value for z is 3. We can now use back-substitution to find the values for y, x, and w. P-BLTZMC08_805-872-hr 21-11-2008 13:26 Page 813 Section 8.1 Matrix Solutions to Linear Systems These are the four equations from the last column. z=3 y-z=1 9 y=4 1 8 y+z= 3 3 w-x+2y-2z=–1 1 8 (4)+3= 3 3 w-1+2(4)-2(3)=–1 x- y-3=1 x- 813 9 x+ 8 9 5 = 3 3 w-1+8-6=–1 x=1 w+1=–1 w=–2 Let’s agree to write the solution for the system in the alphabetical order of the variables from left to right, namely 1w, x, y, z2. Thus, the solution set is 51 -2, 1, 4, 326. We can verify the solution by substituting the value for each variable into the original system of equations and obtaining four true statements. Check Point 3 Use matrices to solve the system: w 2w d 3w 5w � Use matrices and Gauss-Jordan elimination to solve systems. + 3x 7x 7x x + 2y y 3y 4y + + + - z 2z 3z 2z = -3 = 1 = -5 = 18. Solving Linear Systems Using Gauss-Jordan Elimination Using Gaussian elimination, we obtain a matrix in row-echelon form, with 1s down the main diagonal and 0s below the 1s. A second method, called Gauss-Jordan elimination, after Carl Friedrich Gauss and Wilhelm Jordan (1842–1899), continues the process until a matrix with 1s down the main diagonal and 0s in every position above and below each 1 is found. Such a matrix is said to be in reduced row-echelon form. For a system of three linear equations in three variables, x, y, and z, we must get the augmented matrix into the form 1 C0 0 0 1 0 0 a 0 3 b S. 1 c Based on this matrix, we conclude that x = a, y = b, and z = c. 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 a row-equivalent matrix in reduced row-echelon form, with 1s down the main diagonal from upper left to lower right, and 0s above and below the 1s. a. Get 1 in the upper left-hand corner. b. Use the 1 in the first column to get 0s below it. c. Get 1 in the second row, second column. d. Use the 1 in the second column to make the remaining entries in the second column 0. e. Get 1 in the third row, third column. f. Use the 1 in the third column to make the remaining entries in the third column 0. g. Continue this procedure as far as possible. 3. Use the reduced row-echelon form of the matrix in step 2 to write the system’s solution set. (Back-substitution is not necessary.) P-BLTZMC08_805-872-hr 21-11-2008 13:26 Page 814 814 Chapter 8 Matrices and Determinants Technology EXAMPLE 4 Most graphing utilities can convert a matrix to reduced row-echelon form. Enter the system’s augmented matrix and name it A. Then use the 冷 RREF 冷 (reduced row-echelon form) command on matrix A. Using Gauss-Jordan Elimination Use Gauss-Jordan elimination to solve the system: 3x + y + 2z = 31 c x + y + 2z = 19 x + 3y + 2z = 25. Solution In Example 2, we used Gaussian elimination to obtain the following matrix: We want 0s in these positions. 1 C0 0 1 1 0 2 19 2 3 13 S . 1 5 To use Gauss-Jordan elimination, we need 0s both above and below the 1s in the main diagonal. We use the 1 in the second row, second column to get a 0 above it. This is the augmented matrix for the system in Example 4. Replace row 1 in the previous matrix by −1R2 + R1. 1 C0 0 0 1 0 0 6 2 3 13 S 1 5 We want 0s in these positions. We use the 1 in the third column to get 0s above it. 1 C0 0 This is the matrix in reduced rowechelon form we obtained in Example 4. 0 1 0 0 6 0 3 3S 1 5 Replace row 2 in the previous matrix by −2R3 + R2. This last matrix corresponds to x = 6, y = 3, z = 5. As we found in Example 2, the solution set is 516, 3, 526. Check Point 4 Solve the system in Check Point 2 using Gauss-Jordan elimination. Begin by working with the matrix that you obtained in Check Point 2. Exercise Set 8.1 Practice Exercises In Exercises 1–8, write the augmented matrix for each system of linear equations. 2x + y + 2z = 2 1. c 3x - 5y - z = 4 x - 2y - 3z = - 6 2. c 3x - 2y + 5z = 31 x + 3y - 3z = - 12 - 2x - 5y + 3z = 11 3. c x - y + z = 8 y - 12z = - 15 z = 1 4. c x - 2y + 3z = 9 y + 3z = 5 z = 2 5. c 5x - 2y - 3z = 0 x + y = 5 2x - 3z = 4 6. c x - 2y + z = 10 3x + y = 5 7x + 2z = 2 2w + 5x - 3y 3x 7. d w - x 5w - 5x + + + - z y 5y 2y = = = = 2 4 9 1 4w + 7x - 8y + z 5x + y 8. d w - x - y 2w - 2x + 11y In Exercises 9–12, write the system of linear equations represented by the augmented matrix. Use x, y, and z, or, if necessary, w, x, y, and z, for the variables. 5 9. C 0 7 1 -1 11. D 2 0 0 1 2 3 - 11 - 4 3 12 S 0 3 1 1 0 0 4 -1 0 12 1 3 0 4 7 T 5 11 4 5 7 10. C 0 2 0 1 7 4 1 12. D 3 0 1 -1 0 0 4 - 13 - 5 3 11 S 0 6 5 0 0 11 1 -1 7 5 6 4 8T 4 3 In Exercises 13–18, perform each matrix row operation and write the new matrix. = = = = 3 2 5 17 13. C 1 3 4 -6 5 0 4 10 -5 3 0 S 4 7 1 2 R1 P-BLTZMC08_805-872-hr 21-11-2008 13:26 Page 815 Section 8.1 Matrix Solutions to Linear Systems -12 -4 0 1 15. C 3 2 -3 1 -2 2 0 -1 3 7 S 1 3 1 16. C 3 1 -1 3 3 5 -6 - 1 3 10 S 2 5 1 0 17. D 2 5 -1 1 0 1 1 -2 3 2 1 0 18. D 3 -4 -5 1 0 1 6 4 7 1 3 R1 3 14. C 1 2 9 3 0S 4 2 -3 2 4 - 3R1 + R2 - 3R1 + R2 1 3 -1 4 0 T 4 11 4 6 - 2R1 + R3 - 5R1 + R4 4 -2 -1 4 0 T -1 6 2 -3 - 3R1 + R3 4R1 + R4 In Exercises 19–20, a few steps in the process of simplifying the given matrix to row-echelon form, with 1s down the diagonal from upper left to lower right, and 0s below the 1s, are shown. Fill in the missing numbers in the steps that are shown. 1 19. C 2 3 1 20. C 2 -3 -1 3 -2 -2 1 4 1 8 1 -1 3 -2 S : C 0 -9 9 0 -1 5 1 1 8 n 3 nS n n 1 : C0 0 -1 1 1 1 8 n 3 nS n n 815 x + 2y = z - 1 29. c x = 4 + y - z x + y - 3z = - 2 2x + y = z + 1 30. c 2x = 1 + 3y - z x + y + z = 4 3a - b - 4c = 3 31. c 2a - b + 2c = - 8 a + 2b - 3c = 9 3a + b - c = 0 32. c 2a + 3b - 5c = 1 a - 2b + 3c = - 4 2x + 2y + 7z = - 1 33. c 2x + y + 2z = 2 4x + 6y + z = 15 3x + 2y + 3z = 3 34. c 4x - 5y + 7z = 1 2x + 3y - 2z = 6 w 2w 35. d w 3w + + + x x 2x 2x + + y 2y y y + + z z 2z 3z = 4 = 0 = -2 = 4 w w 36. d w 2w + + - x 2x 3x x + + y y 3y 2y + - z 2z z z = 5 = -1 = -1 = -2 3w w 37. d 2w -w + + + 4x x x 2x + + + y y 4y y + - z z 2z 3z = = = = 2w w 38. d 3w w + y - 3z = 8 - x + 4z = - 10 + 5x - y - z = 20 + x - y - z = 6 9 0 3 3 Practice Plus 39. Find the quadratic function f1x2 = ax2 + bx + c for which f1-22 = - 4, f112 = 2, and f122 = 0. 3 4 1 -4 3 3 S : C 0 -1 - 2 0 -2 5 -2 3 4 n 3 nS n n 40. Find the quadratic function f1x2 = ax2 + bx + c for which f1- 12 = 5, f112 = 3, and f122 = 5. 1 : C0 0 -2 1 -2 3 4 n 3 nS n n 42. Find the cubic function f1x2 = ax3 + bx2 + cx + d for which f1- 12 = 3, f112 = 1, f122 = 6, and f132 = 7. 41. Find the cubic function f1x2 = ax3 + bx2 + cx + d for which f1- 12 = 0, f112 = 2, f122 = 3, and f132 = 12. 43. Solve the system: In Exercises 21–38, solve each system of equations using matrices. Use Gaussian elimination with back-substitution or Gauss-Jordan elimination. x + y - z = -2 21. c 2x - y + z = 5 - x + 2y + 2z = 1 x - 2y - z = 2 22. c 2x - y + z = 4 - x + y - 2z = - 4 x + 3y = 0 23. c x + y + z = 1 3x - y - z = 11 24. c 2x - y - z = 4 25. c x + y - 5z = - 4 x - 2y = 4 x - 3z = - 2 26. c 2x + 2y + z = 4 3x + y - 2z = 5 x + y + z = 4 27. c x - y - z = 0 x - y + z = 2 3x + y - z = 0 28. c x + y + 2z = 6 2x + 2y + 3z = 10 3y - z = - 1 x + 5y - z = - 4 -3x + 6y + 2z = 11 2 ln w 4 ln w d ln w ln w + ln x + 3 ln x + ln x + ln x + 3 ln y - 2 ln z = - 6 + ln y - ln z = - 2 + ln y + ln z = - 5 - ln y - ln z = 5. (Hint: Let A = ln w, B = ln x, C = ln y, and D = ln z. Solve the system for A, B, C, and D. Then use the logarithmic equations to find w, x, y, and z.) 44. Solve the system: ln w -ln w d ln w -ln w + ln x + 4 ln x - 2 ln x - 2 ln x + + + + ln y ln y ln y ln y + ln z - ln z - 2 ln z + 2 ln z = -1 = 0 = 11 = - 3. (Hint: Let A = ln w, B = ln x, C = ln y, and D = ln z. Solve the system for A, B, C, and D. Then use the logarithmic equations to find w, x, y, and z.) P-BLTZMC08_805-872-hr 21-11-2008 13:26 Page 816 816 Chapter 8 Matrices and Determinants Application Exercises 45. A ball is thrown straight upward. A position function s1t2 = 12 at2 + v0t + s0 can be used to describe the ball’s height, s1t2, in feet, after t seconds. Height above Ground (feet) s(t) 60 50 If a meal consisting of the three foods allows exactly 660 calories, 25 grams of protein, and 425 milligrams of vitamin C, how many ounces of each kind of food should be used? 48. A furniture company produces three types of desks: a children’s model, an office model, and a deluxe model. Each desk is manufactured in three stages: cutting, construction, and finishing. The time requirements for each model and manufacturing stage are given in the following table. (2, 48) (1, 40) Children’s model Office model Deluxe model Cutting 2 hr 3 hr 2 hr Construction 2 hr 1 hr 3 hr Finishing 1 hr 1 hr 2 hr 40 30 (3, 24) 20 10 1 2 3 4 Time (seconds) 5 t a. Use the points labeled in the graph to find the values of a, v0 , and s0 . Solve the system of linear equations involving a, v0 , and s0 using matrices. b. Find and interpret s13.52. Identify your solution as a point on the graph shown. Each week the company has available a maximum of 100 hours for cutting, 100 hours for construction, and 65 hours for finishing. If all available time must be used, how many of each type of desk should be produced each week? 49. Imagine the entire global population as a village of precisely 200 people. The bar graph shows some numeric observations based on this scenario. c. After how many seconds does the ball reach its maximum height? What is its maximum height? can be used to describe the ball’s height, s1t2, in feet, after t seconds. Height above Ground (feet) s(t) 300 (5, 246) 250 200 (2, 198) 100 (8, 6) 0 5 Time (seconds) 125 100 75 60 50 Asian w Under age 15 African Over age 65 x 25 14 Unable to read or write European American (U.S.) y 32 z Eat at McDonald’s each day 1 Source: Gary Rimmer, Number Freaking, The Disinformation Company Ltd., 2006 150 50 150 Number of People 46. A football is kicked straight upward. A position function s1t2 = 12 at2 + v0t + s0 Earth’s Population as a Village of 200 People 10 t a. Use the points labeled in the graph to find the values of a, v0 , and s0 . Solve the system of linear equations involving a, v0 , and s0 using matrices. b. Find and interpret s172. Identify your solution as a point on the graph shown. c. After how many seconds does the ball reach its maximum height? What is its maximum height? Combined, there are 183 Asians, Africans, Europeans, and Americans in the village. The number of Asians exceeds the number of Africans and Europeans by 70. The difference between the number of Europeans and Americans is 15. If the number of Africans is doubled, their population exceeds the number of Europeans and Americans by 23. Determine the number of Asians, Africans, Europeans, and Americans in the global village. 50. The bar graph shows the number of rooms, bathrooms, fireplaces, and elevators in the U.S. White House. The U.S. White House by the Numbers 150 w Write a system of linear equations in three or four variables to solve Exercises 47–50. Then use matrices to solve the system. 125 47. Three foods have the following nutritional content per ounce. 100 Calories Protein (in grams) Vitamin C (in milligrams) Food A 40 5 Food B 200 2 10 Food C 400 4 300 30 Rooms 75 Bathrooms 50 x 25 Fireplaces y Elevators z Source: The White House P-BLTZMC08_805-872-hr 21-11-2008 13:26 Page 817 Section 8.1 Matrix Solutions to Linear Systems Combined, there are 198 rooms, bathrooms, fireplaces, and elevators. The number of rooms exceeds the number of bathrooms and fireplaces by 69. The difference between the number of fireplaces and elevators is 25. If the number of bathrooms is doubled, it exceeds the number of fireplaces and elevators by 39. Determine the number of rooms, bathrooms, fireplaces, and elevators in the U.S. White House. Writing in Mathematics 817 61. When I use matrices to solve linear systems, I spend most of my time using row operations to express the system’s augmented matrix in row-echelon form. 62. Using row operations on an augmented matrix, I obtain a row in which 0s appear to the left of the vertical bar, but 6 appears on the right, so the system I’m working with has no solution. In Exercises 63–66, determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement. 51. What is a matrix? 63. A matrix row operation such as - 45 R1 + R2 is not permitted because of the negative fraction. 52. Describe what is meant by the augmented matrix of a system of linear equations. 64. The augmented matrix for the system x - 3y = 5 y - 2z = 7 is 2x + z = 4 53. In your own words, describe each of the three matrix row operations. Give an example with each of the operations. 54. Describe how to use row operations and matrices to solve a system of linear equations. 55. What is the difference between Gaussian elimination and Gauss-Jordan elimination? Technology Exercises 56. Most graphing utilities can perform row operations on matrices. Consult the owner’s manual for your graphing utility to learn proper keystrokes for performing these operations. Then duplicate the row operations of any three exercises that you solved from Exercises 13–18. 57. If your graphing utility has a 冷 REF 冷 (row-echelon form) command or a 冷 RREF 冷 (reduced row-echelon form) command, use this feature to verify your work with any five systems that you solved from Exercises 21–38. 58. Solve using a graphing utility’s 冷 REF 冷 or 冷 RREF 冷 command: 2x1 x1 e x1 - x1 x1 + + + - 2x2 + 3x3 2x2 - x3 x3 x2 - x3 x2 - + + - x4 2x4 x4 2x4 x4 + x5 5x5 3x5 x5 = 12 = -7 = 1 = 0 = 4. Critical Thinking Exercises Make Sense? In Exercises 59–62, determine whether each statement makes sense or does not make sense, and explain your reasoning. 59. Matrix row operations remind me of what I did when solving a linear system by the addition method, although I no longer write the variables. 60. When I use matrices to solve linear systems, the only arithmetic involves multiplication or a combination of multiplication and addition. 1 C1 2 -3 -2 1 5 3 7 S. 4 65. In solving a linear system of three equations in three variables, we begin with the augmented matrix and use row operations to obtain a row-equivalent matrix with 0s down the diagonal from left to right and 1s below each 0. 66. The row operation kRi + Rj indicates that it is the elements in row i that change. 67. The table shows the daily production level and profit for a business. x (Number of Units Produced Daily) y (Daily Profit) 30 50 100 $5900 $7500 $4500 Use the quadratic function y = ax2 + bx + c to determine the number of units that should be produced each day for maximum profit. What is the maximum daily profit? Preview Exercises Exercises 68–70 will help you prepare for the material covered in the next section. In each exercise, refer to the following system: 3x - 4y + 4z = 7 c x - y - 2z = 2 2x - 3y + 6z = 5. 68. Show that 112z + 1, 10z - 1, z2 satisfies the system for z = 0. 69. Show that 112z + 1, 10z - 1, z2 satisfies the system for z = 1. 70. a. Select a value for z other than 0 or 1 and show that 112z + 1, 10z - 1, z2 satisfies the system. b. Based on your work in Exercises 68–70(a), how does this system differ from those in Exercises 21–34?