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Self-Assessment Exam C SYSTEMS, MATRICES AND DETERMINANTS 3x 1. Use the elimination method to solve the system x 5 x 3 3x 4y 2z 25 9x 12y 6z 4 x 3y 4z 11 4x 12y 16z 20x 12y 24z 4 5x 3y 6z 3 4 y 2 z 25 3 y 4z 11 3 y 6z 3 75 eliminate y 44 12 13x 10z 31 29x 30z 63 3 13x 10z 31 39x 30z 93 156 39 68x 156 x 68 17 29x 30z 63 29x 30z 63 2 39 13x 10z 31 13 10z 31 507 170z 527 170z 20 z 17 17 39 47 2 3y 4 11 3z 11 47 51z 187 17 17 17 234 78 51y 234 y 51 17 39 78 2 x , y , z 17 17 17 x 3y 4z 11 3z 68 5 x 2. Use the elimination method to solve the system 2 y 5 z 71 7 x 3 y 25 3z 68 3z 68 3z 68 5x 5x 5x 2y 5z 71 3 2y 5z 71 6y 15z 213 7 x 3y 14x 6y 25 2 7 x 3y 25 50 5x 3z 68 5 5x 3z 68 25x 15z 340 263 14x 15z 263 14x 15z 14x 15z 263 11x 77 x 7 5x 3z 68 35 3z 68 3z 33 z 11 6y 15z 213 6y 165 213 6y 48 y 8 x 7, y 8, z 11 -1 3. The sum of three numbers is 86. If the third number is subtracted from twice the sum of the first two numbers, the result is 148. The first number is 14 less than three times the second number. Find the numbers. Show the work. Solution: let the numbers be: first = x, second = y and third = z x + y + z = 86 2(x + y) –z = 148 x = 3y – 14 x y z 86 x y z 86 3x 3y 234 2x 2y z 148 2x 2y z 148 x 3y 14 x 3y x 3y 14 14 4x 220 x 55 x 3y 14 55 3y 14 3y 69 y 23 x y z 86 55 23 z 86 78 z 86 z 8 The numbers are 55, 23 and 8 4. Write the augmented matrix for the system: 3 y z 4w 10 x 3x 2 y 3z 10w 10 5y 7z 8w 26 3x y 5 z 4 answer: 3 1 1 4 10 3 2 3 10 10 0 5 7 8 26 5 0 4 3 1 5. The row-reduced echelon form for the matrix above is 1 0 0 0 2 0 1 0 0 3 0 0 1 0 1 , give the solution of the system 1 0 0 0 1 2 x 2 ___ y 3 ___ z 1 ___ 1 w ____ 2 6. Find the missing numbers for the given product of matrices. 4 1 2 3 5 3 1 31 12 4 4 3 4 3 4 3 2 27 1 2 1 7 5 3 5 7. IF A= and B= , show that A and B are inverse matrices. 4 3 4 7 7 5 3 5 1 0 4 3 4 7 = 0 1 They are inverses because their product is the identity matrix. -2 8. Write the augmented matrix for the following system of equation: x y 2 z w 11 3x y z 2w 16 2 y 3z w 9 2 x y 3w 14 1 1 1 2 3 1 1 2 Answer: 0 2 3 1 0 3 2 1 11 16 9 14 9. Use the elementary row transformations to convert the matrix in exercise 8 to row reduced echelon form. Give the solution of the system. 1 11 1 11 1 1 2 1 1 2 3 1 1 2 16 first row 3 and add to second row 0 4 7 1 17 0 2 3 1 9 first row 2 and add to fourth row 0 2 3 1 9 0 3 14 2 1 0 3 4 1 8 1 11 1 1 2 0 1 3 2 9 fourth row by 1 and add to second row 0 2 3 1 9 0 3 4 1 8 1 0 1 1 2 add second row to first row 0 1 3 2 9 second row by 2 and add to third row 0 0 3 3 9 second row by 3 and add to fourth row 7 19 0 0 5 1 0 1 1 2 0 1 3 2 9 divide third row by 3 0 0 1 1 3 7 19 0 0 5 1 0 0 0 5 1 0 0 Add third row to first row 0 1 0 1 0 0 1 0 divide fourth third row by 3 and add to second row 0 0 1 1 3 0 0 1 row by 2 third row by 5 and add to fourth row 0 0 0 2 4 0 0 0 1 fourth row by 1 and add to second row 0 fourth row by 1 and add to third row 0 0 -3 5 x5 y 2 1 0 0 2 0 1 0 1 z 1 w 2 0 0 1 2 0 0 0 0 5 1 0 1 3 1 2 10. Use a calculator to find the product of the following matrices: 1 3 2 3 4 20 14 7 2 1 2 4 0 24 3 5 2 8 11. Use a calculator to find the inverse of the matrix 1 6 Without a calculator: 2 8 Step 1. Augment the by the 2 2 identity matrix 1 6 2 8 1 0 1 6 0 1 Step 2. Convert the left 2 2 matrix to the identity matrix by using the elementary row transformations. 2 8 1 0 Interchange the two rows of the augmented matrix to obtain 1 6 0 1 1 6 0 1 1 6 0 1 first row by 2 and add to second row 2 8 1 0 0 4 1 2 1 6 0 1 1 1 Divide the second row by 4 , 0 1 4 2 3 1 0 2 the second row by 6 and add to first row 1 0 1 4 3 The inverse matrix is 2 1 4 3 2 8 2 2 Check : 1 6 1 1 4 2 2 1 2 1 0 0 1 -4 2 1 2 12. Use a calculator to find the value of the determinant of the following matrix: 5 1 2 1 3 1 0 2 3 5 7 2 4 3 1 1 6 3 2 4 2 1 5 0 2 answer: ___2162____________ 1 2 1 2 13. Find the inverse of the matrix 3 1 2 1 0 1 2 1 3 1 2 2 1 0 1 1 2 15 15 4 2 15 15 1 1 3 3 1 3 1 3 1 3 14. Find the value of the determinant 3 5 3 7 3 5 3 7 = (-3)(7) – (3)(-5) = -21 + 15 = -6 . Do not use a calculator. Show the work. Answer: -6 15. Use Cramer’s rule to find the value of z in the system: 8 2 x 3 y z x 4 y 2 z 3 2 x 4z 1 Use a calculator to find the values of the determinants involved. 2 3 1 z 4 8 3 2 0 1 Dz 2 3 1 D 1 4 2 2 0 35 48 4 -5 16. Find the product of the following matrices. You can use a calculator. 8 21 2 3 1 0 1 3 11 12 1 4 3 4 2 5 11 16 7 17 17. Find the inverse of the matrix A. Give answer in fractional form. Use a calculator. 2 1 0 A= 1 3 2 1 0 1 2 1 0 1 3 2 1 0 1 1 1 1 3 9 1 2 9 3 1 1 3 9 2 9 4 9 7 9 18. Use elementary row transformations to change the given matrix to an equivalent matrix with a11 1 and zeros in every entry below a11 1 1 3 6 12 3 4 2 1 0 Add the third row to first row 4 2 2 1 2 4 1 first row by 2 and add to third row 4 0 7 10 2 1 15 18 1 7 first row by 4 and add to second row 0 26 0 15 10 41 18 7 10 2 1 1 2 0 18 7 28 18 7 8 2 x 3 y z 19. Write the augmented matrix for the system: x 4 y 2 z 3 2 x 4z 1 8 2 3 1 ANSWER: 1 4 2 3 2 0 4 1 -6 7 0 4 20. Use a calculator to find the solution of the system below by using the row reduced echelon form of the matrix. x 2 y z 2w 3 2x y 3z 4w 9 2 3 x 2 y z 2 x 3 y 2 z 4w 10 a) Augmented matrix: 2 1 2 3 1 2 1 3 4 9 R 3 2 1 0 2 3 2 4 10 2 Row reduced echelon form of the matrix in part a) x 3 1 0 0 0 3 0 1 0 0 4 4 y 0 0 1 0 1 z 1 1 1 w 0 0 0 1 2 2 20. Use determinants to find the solution of the following system: 5 3x 2 y 5 z 4y 1 x 2 x 6 y 3z 12 2 5 3 D 1 0 112 , D x 1 3 12 4 2 6 3 5 5 Dy 1 1 0 56, 2 12 3 x 2 5 5 4 0 336 6 3 3 Dz 1 2 2 5 4 1 6 12 112 D y 56 D x 336 D 1 112 3, y , z x 1 D 112 D 112 2 D 112 21. Add the following 2 by 4 matrices 6 12 3 5 3 4 3 5 3 8 6 10 6 2 5 0 6 2 5 0 12 4 10 0 -7 For the following two word problems, write the system of equations, do not solve the system. Leave answer in standard form 22. Pablo has 25 coins: nickels, dimes and quarters. The value of all the coins is $2.80 and the number of nickels is twice the number of quarters. Find how many nickels (x), dimes (y) and quarters (z) he has. Coins: n: nickels, d: dimes, q: quarters Equation 1: 25 coins Equation 2: value = $2.80 Equation 3: The number of Each nickel = 5 cents nickels is twice the number of Each dime = 10 cents quarters. Each quarter = 25cents Each dollar = 100 cents n + d + q =25 5n + 10d +25q = 280 n = 2q d q 25 n System : 5n 10d 25q 280 n 2q 0 10 n d q 25 10n 10d 10q 250 5n 15q 30 10d 25q 280 5n 10d 25q 280 5n 2q 0 n n n 2q 0 2q 0 n 3q 6 Divide the first equation by 5 q6 n 2q 0 n 2q n 12 and n d q 25 12 d 6 25 d 7 Answer: Pablo has 12 nickels, 7 dimes and 6 quarter. 23. A certain job can be done in 5 days by three men, John, David and Carlos, if they work together. John and Carlos can do the job by themselves in 8 days. If Carlos and David work by themselves on the job for 6 days, then John can finish the job in 2 more days working alone. Find how long it will take John working alone (x), how long it will take David working along (y) and how long it will take Carlos working alone (z)? John can do the job in x hours David can do the job in y hours Carlos can do the job in z hours 1 of the job per hour x 1 David is able to do of the job per hour y 1 Carlos is able to do of the job per hour z John is able to do -8 1)All three together do the job in 5 days Together they do Therefore : first equation is 1 of job per hour 5 1 1 1 1 5 5 5 or 1 where 1 represents the whole job x y z 5 x y z 2)John and Carlos can do the job by themselves in 8 days John and Carlos do 1 of 8 the job per hour 1 1 1 8 8 or 1 x z 8 x z 3) Carlos and David work alone for 6 days, then John finishes in 2 days by himself Therefore : second equation is Therefore : third equation is 6 6 2 1 y z x 5 5 5 1 x y z 1 x a 5a 5b 5c 1 1 8 8 System : 1 8z 1 Let b 8a x z y 2a 6b 6c 1 6 6 2 1 1 c y z x z 6 5a 5b 5c 1 30a 30b 30c 6 1 20a 8z 1 8a 8z 1 8a 8a 8c 1 10a 30b 30c 5 5 2a 6b 6c 1 1 8 2 3 3 20a 1 a , 8a 8c 1 8c 1 8c 1 8c c 20 20 5 5 40 5 15 25 15 3 5a 5b 5c 1 5b 1 5b 1 5b b 20 40 40 8 40 1 a 20 x 20 days 3 40 1 y 13 days b 40 3 3 3 40 c z days 40 3 John can do the job by himself in 20 days . 1 David can do the job by himself in 13 days . 3 1 Carlos can do the job by himself in 13 days . 3 -9