## shivam_bhalla 2 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. shivam_bhalla
2. UnkleRhaukus

can you factor all of the elements

3. shivam_bhalla

@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. shivam_bhalla

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. shivam_bhalla

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

9. shivam_bhalla

@apoorvk

10. inkyvoyd

can we just do it the foolish way?

11. apoorvk

Hmm, tried C1 - C2?

12. shivam_bhalla

@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. shivam_bhalla

@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. shivam_bhalla

@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. shivam_bhalla

LOL. @ "wikipediaing" . :)

25. apoorvk

wolfram!!!

26. shivam_bhalla

How to wolfram this ??

27. shivam_bhalla

28. experimentX

factor out the commons!!!

29. shivam_bhalla

@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. shivam_bhalla

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. shivam_bhalla

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. shivam_bhalla

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. shivam_bhalla

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. shivam_bhalla

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

46. shivam_bhalla

@experimentX

47. experimentX

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

48. shivam_bhalla

Sorry, didn't get you @experimentX

49. experimentX

|dw:1336588777520:dw|

50. experimentX

|dw:1336588827490:dw|

51. shivam_bhalla

|dw:1336588893052:dw|

52. eigenschmeigen

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. eigenschmeigen

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

55. shivam_bhalla

@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. eigenschmeigen

;D

57. eigenschmeigen

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

58. experimentX

Oo ... then i must have mistake!!

59. eigenschmeigen

no its correct

60. eigenschmeigen

if R1 = R2 then det(M) = 0

61. shivam_bhalla

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

62. experimentX

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

63. shivam_bhalla

@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. shivam_bhalla

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

66. shivam_bhalla

*exams

67. eigenschmeigen

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

68. shivam_bhalla

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. shivam_bhalla

|dw:1336590227531:dw|

75. experimentX

76. experimentX

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

77. shivam_bhalla

@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. eigenschmeigen

its B

80. eigenschmeigen

i'm just checking my working

81. eigenschmeigen

how do i write matrices on here?

82. shivam_bhalla

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

83. eigenschmeigen

ok gimme a sec

84. shivam_bhalla

*board.

85. eigenschmeigen

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. eigenschmeigen

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. eigenschmeigen

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

88. eigenschmeigen

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

89. eigenschmeigen

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

90. eigenschmeigen

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. eigenschmeigen

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

92. eigenschmeigen

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. shivam_bhalla

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

94. eigenschmeigen

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. eigenschmeigen

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!!