Solutions

Solutions to Try Its

1. AB=[14−1−3][−3−411]=[1(−3)+4(1)1(−4)+4(1)−1(−3)+−3(1)−1(−4)+−3(1)]=[1001]BA=[−3−411][14−1−3]=[−3(1)+−4(−1)−3(4)+−4(−3)1(1)+1(−1)1(4)+1(−3)]=[1001]

2. A−1=[3515−2515]

3. A−1=[11224−336−5]

4. X=[43858]

Solutions to Odd-Numbered Exercises

1. If A−1 is the inverse of A, then AA−1=I, the identity matrix. Since A is also the inverse of A−1,A−1A=I. You can also check by proving this for a 2×2 matrix.

3. No, because ad and bc are both 0, so ad−bc=0, which requires us to divide by 0 in the formula.

5. Yes. Consider the matrix [0110]. The inverse is found with the following calculation: A−1=10(0)−1(1)[0−1−10]=[0110].

7. AB=BA=[1001]=I

9. AB=BA=[1001]=I

11. AB=BA=[100010001]=I

13. 129[92−13]

15. 169[−2793]

17. There is no inverse

19. 47[0.51.51−0.5]

21. 117[−55−320−3121−14]

23. 1209[47−57691019−12−2438−13]

25. [1860−168−56−1404484080−280]

27. (−5,6)

29. (2,0)

31. (13,−52)

33. (−23,−116)

35. (7,12,15)

37. (5,0,−1)

39. 134(−35,−97,−154)

41. 1690(65,−1136,−229)

43. (−3730,815)

45. (10123,−1,25)

47. 12[21−1−1011−10−11101−11]

49. 139[321−718−53321024−36219−946−16−5]

51. [100000010000001000000100000010−1−1−1−1−11]

53. Infinite solutions.

55. 50% oranges, 25% bananas, 20% apples

57. 10 straw hats, 50 beanies, 40 cowboy hats

59. Tom ate 6, Joe ate 3, and Albert ate 3.

61. 124 oranges, 10 lemons, 8 pomegranates