## anonymous one year ago Medal Solve by determinants. x + y + z = 21 x=y +z = 3 x +y - z = 1

1. jim_thompson5910

is there a typo in the second equation `x=y +z = 3` ??

2. jim_thompson5910

there are 2 equal signs there

3. anonymous

yes sorry its x - y + z = 3

4. jim_thompson5910

ok thanks

5. jim_thompson5910

were you able to set up the matrix?

6. anonymous

yes

7. jim_thompson5910

what matrix did you get

8. anonymous

1 1 1 1 -1 1 1 1 -1

9. anonymous

m i right?

10. jim_thompson5910

good

11. jim_thompson5910

12. jim_thompson5910

do you know how to take the determinant of that matrix?

13. anonymous

no

14. jim_thompson5910

|dw:1443401707454:dw|

15. jim_thompson5910

focus on the first two columns |dw:1443401787922:dw|

16. jim_thompson5910

copy those columns and write it off to the right of the vertical bar |dw:1443401811957:dw|

17. jim_thompson5910

with me so far?

18. anonymous

yes

19. jim_thompson5910

ok so we have this extended matrix |dw:1443401923170:dw|

20. jim_thompson5910

circle the elements that fall along the diagonals like this |dw:1443401944912:dw| we have 3 subgroups

21. anonymous

ohh now i see it

22. jim_thompson5910

multiply the numbers in the subgroups |dw:1443402013404:dw|

23. jim_thompson5910

then add up the products: 1+1+1 = 3

24. jim_thompson5910

next, circle the elements along the diagonals like this |dw:1443402083998:dw|

25. jim_thompson5910

multiply out the numbers in each subgroup |dw:1443402112490:dw|

26. jim_thompson5910

then add up the products -1+1+(-1) = -1

27. jim_thompson5910

The first sum was 3 the second sum was -1 subtract the sums 3 - (-1) = 3+1 = 4

28. jim_thompson5910

so in the end, the determinant of that 3x3 matrix you wrote above is 4

29. jim_thompson5910

there are a lot of steps here, so go back over them if you aren't sure what's going on. And you can ask about any step

30. anonymous

so for x we got 3

31. jim_thompson5910

we haven't found x or y yet

32. anonymous

oh ok

33. anonymous

whats the next step

34. jim_thompson5910

do you see how I got the determinant to be 4?

35. anonymous

36. jim_thompson5910

so do you see how to find the determinant of any 3x3 matrix?

37. anonymous

yes i see it

38. jim_thompson5910

ok we'll let D = 4 be the determinant of the coefficient matrix

39. jim_thompson5910

if we replace the first column with the numbers 21, 3, 1 (found on the right side of each equation ) we get |dw:1443402459893:dw| what is the determinant of this new matrix (shown above)?

40. anonymous

1 1 1 1 -1 1

41. anonymous

we need to add another column

42. jim_thompson5910

did you start off doing this? |dw:1443402694898:dw|

43. anonymous

i got confused with the ones

44. anonymous

so i got ( 21 ) ( -1) ( -1) = 21

45. anonymous

(1) (1) (1) = 1 (1) (3) (1) = 2

46. anonymous

sorry 3*

47. jim_thompson5910

good, add up those products to get ??

48. anonymous

then you add them 21 + 1 + 3

49. anonymous

25

50. jim_thompson5910

ok keep that sum in mind now move onto the next set of diagonals

51. anonymous

how do you make that

52. jim_thompson5910

circling them helps me keep all the numbers straight |dw:1443402998519:dw|

53. jim_thompson5910

1*(-1)*1 = ?? 1*1*21 = ?? -1*3*1 = ?? then add up the products

54. anonymous

-1

55. anonymous

21

56. anonymous

-3

57. anonymous

58. anonymous

m i right

59. jim_thompson5910

correct

60. jim_thompson5910

first sum = 25 second sum = 17 subtract the sums: first - second = 25 - 17 = 8

61. jim_thompson5910

So the determinant of this matrix (below) is 8 |dw:1443403306226:dw|

62. jim_thompson5910

since we replaced the first column, think of it as the x column, we call this matrix Dx

63. jim_thompson5910

to get the solution for x, we simply compute x = Dx/D

64. jim_thompson5910

Recall earlier we found D = 4 Dx = 8 so... x = Dx/D x = 8/4 x = 2

65. anonymous

and we do the same to find y and z

66. jim_thompson5910

yes you replace the second column with 23,3,1 and compute the determinant of that new matrix. That will give you Dy to find y, compute y = Dy/D

67. jim_thompson5910

same with z but you replace the third column with 23,3,1 z = Dz/D

68. anonymous

23 / 3 / 1

69. jim_thompson5910

compute the determinant of this |dw:1443403528965:dw| to get Dy

70. anonymous

1 * 3 *1

71. jim_thompson5910

|dw:1443403666260:dw|

72. jim_thompson5910

|dw:1443403686878:dw|

73. anonymous

-3 23 1

74. jim_thompson5910

75. anonymous

21

76. jim_thompson5910

good

77. jim_thompson5910

|dw:1443403784502:dw|

78. anonymous

3 1 -23

79. jim_thompson5910

80. anonymous

-19

81. jim_thompson5910

first sum = 21 second sum = -19 first - second = 21 - (-19) = 21+19 = 40

82. jim_thompson5910

hold on, I think I made a typo somewhere

83. jim_thompson5910

84. jim_thompson5910

I put 23 when I should have put 21

85. jim_thompson5910

86. anonymous

3 1 -21

87. anonymous

-17

88. jim_thompson5910

|dw:1443404234817:dw|

89. anonymous

-3 21 1

90. anonymous

19

91. jim_thompson5910

|dw:1443404265866:dw|

92. anonymous

3 1 -21= -17

93. jim_thompson5910

so first sum - second sum = 19 - (-17) = 19+17 = 36 making Dy = 36

94. jim_thompson5910

y = Dy/D y = 36/4 y = 9

95. anonymous

now we have to do the same for z

96. jim_thompson5910

yes, replace the third column with 21,3,1 and compute the determinant

97. anonymous

1*3*21 = 63

98. jim_thompson5910

|dw:1443404399145:dw|

99. jim_thompson5910

|dw:1443404416172:dw|

100. anonymous

-21 3 1

101. jim_thompson5910

|dw:1443404436727:dw|

102. anonymous

-1 3 21

103. anonymous

23

104. jim_thompson5910

|dw:1443404481727:dw|

105. anonymous

-21 3 1

106. anonymous

-17

107. jim_thompson5910

first sum = 23 second sum = -17 first - second = ???

108. anonymous

40

109. jim_thompson5910

good

110. jim_thompson5910

Dz = 40, z = Dz/D = ???

111. anonymous

40 / what was d

112. jim_thompson5910

D = 4 we got this when we computed the determinant of the first matrix we set up (the coefficient matrix)

113. anonymous

10

114. jim_thompson5910

z = 10, yes

115. jim_thompson5910

x = 2 y = 9 z = 10

116. jim_thompson5910

I recommend you check it with each original equation in the system

117. anonymous

yes i will i will look at the steps again thankyou soo much for your time

118. jim_thompson5910

btw all of this is called cramer's rule http://www.purplemath.com/modules/cramers.htm

119. anonymous

oh ok thank you again :)