## anonymous 4 years ago Ok. A determinants question(Link below). What I have done? Well, I tried Row1 = Row 1+Row 2+row 3 but nothing great. Can you guys guide me to get the answer??

1. anonymous
2. UnkleRhaukus

can you factor all of the elements

3. anonymous

@UnkleRhaukus ,how exactly??

4. UnkleRhaukus

$\Delta=\left|\begin{array}\\1+a^2+a^4&1+ab+a^2b^2&1+ac+a^2c^2\\1+ab+a^2b^2&1+b^2+b^4&1+bc+b^2c^2\\1+ac+a^2c^2&1+bc+b^2c^2&1+c^2+c^4\end{array}\right|$

5. UnkleRhaukus

yeah maybe not, i really dont want to do it the normal way

6. anonymous

Normal way-->never :D It would take ages :P

7. UnkleRhaukus

but it would work in theory, but i would make about ten mistakes

8. anonymous

I am sure you play around with determinant properties like Row1 = Row 1 + Row 2+Row3 . I mean some tweaking of that sort

9. anonymous

@apoorvk

10. inkyvoyd

can we just do it the foolish way?

11. apoorvk

Hmm, tried C1 - C2?

12. anonymous

@inkyvoyd Sorry, please, In my exam I have only 3 min(max) to solve this question @apoorvk ,let me give it a try and get back to you

13. anonymous

@apoorvk , I did that. But nothing can be taken common out :(

14. inkyvoyd

I'm going to try this problem

15. inkyvoyd

Have no fear, for inky is here!

16. apoorvk

I got common stuff out through C1-C2 and C2-C3. but, i am still stuck with a trinomial in each ditch..

17. inkyvoyd

@shivam_bhalla I HAVE AN IDEA

18. inkyvoyd

aei+bfg+cdh-ceg-bdi-afh =a(ei-fh)+b(fg-dj)+c(dg-eg)

19. inkyvoyd

b=d f=h

20. inkyvoyd

c=g

21. inkyvoyd

aei+bfg+cdh-ceg-bdi-afh aei+bf^2+cbf-ec^2-ah^2

22. anonymous

@inkyvoyd , what is a,b,c,d,e,f,g?? Are you expanding the determinant??

23. inkyvoyd

I'm wikipediaing, and not making any progress.

24. anonymous

LOL. @ "wikipediaing" . :)

25. apoorvk

wolfram!!!

26. anonymous

How to wolfram this ??

27. anonymous

28. experimentX

factor out the commons!!!

29. anonymous

@experimentX , how exactly ?? I mean row wise or column wise. And moreover, what to factor out ??

30. experimentX

lol ... factoring out would lead to 3 more matrices ..thats ugly... must be some pattern!!

31. anonymous

Oh. Now get what you meant by factoring .:P I tried that too and observed what you said :)

32. experimentX

it seems transpose of itself!!

33. apoorvk

|dw:1336584742255:dw| this is how far i got then..

34. experimentX

try making these types to a^4 + a^2 + 1 => a^6 - 1 (ab)^2 + (ab) + 1 => (ab)^3 - 1 ... not sure if it can help ... too tired!!!

35. anonymous

This problem really is trollling me very badly :( I will solve this problem before going to sleeep

36. experimentX

if you find let met know!!

37. experimentX

Oo ...we can drop all ones!!

38. experimentX

Oo ... i got zero!! LOL

39. experimentX

@shivam_bhalla you still there??

40. anonymous

Yes

41. experimentX

looks like it got it correct http://www.wolframalpha.com/input/?i=det+ {{1%2Ba^2%2Ba^4%2C+1+%2B+%28ab%29+%2B+%28ab%29^2%2C+1+%2B+%28ac%29+%2B+%28ac%29^2}%2C++%09{1+%2B+%28ab%29+%2B+%28ab%29^2%2C+1%2Bb^2%2Bb^4%2C+1+%2B+bc+%2B+%28bc%29^2}%2C++%09{1+%2B+%28ac%29+%2B+%28ac%29^2%2C+1+%2B+bc+%2B+%28bc%29^2%2C+1+%2B+c^2+%2B+c^4}}

42. anonymous

LOL @experimentX , that is doing it the long way Atleast I know the answer know. :) Thanks for typing out the huge question on wolframalpha :)

43. experimentX

No .. i know the shortcut too!!

44. experimentX

actually i verified answer at wolframalpha!!

45. anonymous

Ok. Just tell me the method and not the steps because that would literally troll you

46. anonymous

@experimentX

47. experimentX

take out all ones!! sine they will have same values on row .. they will be zero ..

48. anonymous

Sorry, didn't get you @experimentX

49. experimentX

|dw:1336588777520:dw|

50. experimentX

|dw:1336588827490:dw|

51. anonymous

|dw:1336588893052:dw|

52. anonymous

i'm not sure how far everyone has got so far but first thoughts: setting a = b we can see that the determinant is zero as R1 = R2 hence (a-b) must be a factor

53. experimentX

whenever matrix has same elements on row or column ... determinant is zero.

54. anonymous

and by symmetry (b-c) and (a-c) are

55. anonymous

@experimentX , I never studied that. I only studied that if two rows or 2 colums are proportional to each other, then the determinant of the matrix is zero @eigenschmeigen , Very logical approach which eliminates option (a) for us :)

56. anonymous

;D

57. anonymous

proportional ->eg R1 = kR2 so k = 1 is a specific case

58. experimentX

Oo ... then i must have mistake!!

59. anonymous

no its correct

60. anonymous

if R1 = R2 then det(M) = 0

61. anonymous

@eigenschmeigen , there is no R1=R2 in @experimentX 's proof

62. experimentX

Oh my mistake ... i got the answer in fluke

63. anonymous

@experimentX , I wish these flukes come for me too in my exam :P

64. experimentX

LOL ... i haven't reviewed matrices in long time!! best of luck if you are having exams right now!!

65. anonymous

LOL. I have college entrance exam's right now :P

66. anonymous

*exams

67. anonymous

is this problem solved or still at large? i can have a proper go now i have a pen

68. anonymous

The problem as of now is unsolved @eigenschmeigen

69. experimentX

Okay ... i think i got this time!! you can use same technique to reduce matrix!!

70. experimentX

|dw:1336589808195:dw|

71. experimentX

and then it's simplification!!

72. experimentX

apply operation as above!!! |dw:1336589933207:dw|

73. experimentX

same as you had in your second place!! -- minus --- one

74. anonymous

|dw:1336590227531:dw|

75. experimentX

76. experimentX

|dw:1336590458004:dw| It can be proved!!

77. anonymous

@experimentX , I will try and let you know in 5-10 minutes @experimentX

78. experimentX

okay ... it's just simplification after that. factoring out common factor ... then again you will have 1's and then use same technique ... in the end you will get two linearly dependent columns!! then it will be zero!!

79. anonymous

its B

80. anonymous

i'm just checking my working

81. anonymous

how do i write matrices on here?

82. anonymous

@eigenschmeigen , your answer is right. You can use drawing doard :)

83. anonymous

ok gimme a sec

84. anonymous

*board.

85. anonymous

i like the equation writer, i'm quicker that way: $\left[\begin{matrix}1+a^2 + a^4 & 1+ ab+a^2b^2 & 1 +ac +a^2c^2 \\ 1 + ab +a^2b^2 & 1+b^2 + b^4 & 1 +ac +a^2c^2 \\ 1 +ac +a^2c^2 & 1 +bc+b^2c^2 & 1 + c^2 + c^4\end{matrix}\right]$

86. anonymous

letting a = b we get R1 = R2 letting b = c we get R2 = R3 so the operations to do are R1- R2 = R1' and R2 - R3 = R2'

87. anonymous

$\left[\begin{matrix}a^4 + a^2 - ab - a^2b^2 & a^2b^2 + ab -b^2 -b^4 & a^2c^2 + ac - bc - b^2c^2 \\ a^2b^2 + ab -ac -a^2c^2 & b^4 + b^2 - bc - b^2c^2 & b^2c^2 + bc -c^2 - c^4 \\ 1 +ac +a^2c^2 & 1 +bc+b^2c^2 & 1 + c^2 + c^4\end{matrix}\right]$

88. anonymous

factor out (a-b)(b-c)

89. anonymous

$(a-b)(b-c)\left[\begin{matrix}a^3 + a +ba^2 & b^3 + b + ab^2 & c(ac+bc+1) \\ a(ac+ab+1) & b^3 + b +cb^2 & c^3 + c + bc^2 \\ 1 +ac +a^2c^2 & 1 +bc+b^2c^2 & 1 + c^2 + c^4\end{matrix}\right]$

90. anonymous

letting a = b we now have C1 = C2 so we can see that (a-b) is again a factor. this narrows it down to B as we will have (a-b)^2 in our answer

91. anonymous

you can continue to factorise if you want, but this is sufficient to get the right option.

92. anonymous

its not said in many textbooks but i find using factor theorem is really useful for factorising determinants, as it tells us an operation that will result in something we can factor

93. anonymous

Thank you @eigenschmeigen and @experimentX for both of your solutions :D Thanks for taking your time :)

94. anonymous

no problem, love matrices :)

95. experimentX

looks like i didn't get that right ... i'm embarrassed. looks like 'linear algebra' is the first thing i should do after i'm done with calc

96. anonymous

what else does linear algebra cover?

97. experimentX

well ... all forgot!! let's see ... i have a book here. Linear algebra - Serge Lang Vectors, Vector spaces, Matirces, Linear mappings, Linear mappings and matrices, .... Groups, Rings, Convex Sets ... Appendix I thought $$\Delta$$ on wolfram was the answer!!