## shivam_bhalla Group Title 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?? 2 years ago 2 years ago

1. shivam_bhalla Group Title
2. UnkleRhaukus Group Title

can you factor all of the elements

3. shivam_bhalla Group Title

@UnkleRhaukus ,how exactly??

4. UnkleRhaukus Group Title

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

5. UnkleRhaukus Group Title

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

6. shivam_bhalla Group Title

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

7. UnkleRhaukus Group Title

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

8. shivam_bhalla Group Title

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 Group Title

@apoorvk

10. inkyvoyd Group Title

can we just do it the foolish way?

11. apoorvk Group Title

Hmm, tried C1 - C2?

12. shivam_bhalla Group Title

@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 Group Title

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

14. inkyvoyd Group Title

I'm going to try this problem

15. inkyvoyd Group Title

Have no fear, for inky is here!

16. apoorvk Group Title

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

17. inkyvoyd Group Title

@shivam_bhalla I HAVE AN IDEA

18. inkyvoyd Group Title

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

19. inkyvoyd Group Title

b=d f=h

20. inkyvoyd Group Title

c=g

21. inkyvoyd Group Title

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

22. shivam_bhalla Group Title

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

23. inkyvoyd Group Title

I'm wikipediaing, and not making any progress.

24. shivam_bhalla Group Title

LOL. @ "wikipediaing" . :)

25. apoorvk Group Title

wolfram!!!

26. shivam_bhalla Group Title

How to wolfram this ??

27. shivam_bhalla Group Title

28. experimentX Group Title

factor out the commons!!!

29. shivam_bhalla Group Title

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

30. experimentX Group Title

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

31. shivam_bhalla Group Title

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

32. experimentX Group Title

it seems transpose of itself!!

33. apoorvk Group Title

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

34. experimentX Group Title

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 Group Title

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

36. experimentX Group Title

if you find let met know!!

37. experimentX Group Title

Oo ...we can drop all ones!!

38. experimentX Group Title

Oo ... i got zero!! LOL

39. experimentX Group Title

@shivam_bhalla you still there??

40. shivam_bhalla Group Title

Yes

41. experimentX Group Title

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 Group Title

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 Group Title

No .. i know the shortcut too!!

44. experimentX Group Title

actually i verified answer at wolframalpha!!

45. shivam_bhalla Group Title

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

46. shivam_bhalla Group Title

@experimentX

47. experimentX Group Title

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

48. shivam_bhalla Group Title

Sorry, didn't get you @experimentX

49. experimentX Group Title

|dw:1336588777520:dw|

50. experimentX Group Title

|dw:1336588827490:dw|

51. shivam_bhalla Group Title

|dw:1336588893052:dw|

52. eigenschmeigen Group Title

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 Group Title

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

54. eigenschmeigen Group Title

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

55. shivam_bhalla Group Title

@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 Group Title

;D

57. eigenschmeigen Group Title

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

58. experimentX Group Title

Oo ... then i must have mistake!!

59. eigenschmeigen Group Title

no its correct

60. eigenschmeigen Group Title

if R1 = R2 then det(M) = 0

61. shivam_bhalla Group Title

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

62. experimentX Group Title

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

63. shivam_bhalla Group Title

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

64. experimentX Group Title

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

65. shivam_bhalla Group Title

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

66. shivam_bhalla Group Title

*exams

67. eigenschmeigen Group Title

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

68. shivam_bhalla Group Title

The problem as of now is unsolved @eigenschmeigen

69. experimentX Group Title

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

70. experimentX Group Title

|dw:1336589808195:dw|

71. experimentX Group Title

and then it's simplification!!

72. experimentX Group Title

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

73. experimentX Group Title

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

74. shivam_bhalla Group Title

|dw:1336590227531:dw|

75. experimentX Group Title

76. experimentX Group Title

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

77. shivam_bhalla Group Title

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

78. experimentX Group Title

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 Group Title

its B

80. eigenschmeigen Group Title

i'm just checking my working

81. eigenschmeigen Group Title

how do i write matrices on here?

82. shivam_bhalla Group Title

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

83. eigenschmeigen Group Title

ok gimme a sec

84. shivam_bhalla Group Title

*board.

85. eigenschmeigen Group Title

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 Group Title

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 Group Title

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

88. eigenschmeigen Group Title

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

89. eigenschmeigen Group Title

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

90. eigenschmeigen Group Title

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 Group Title

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

92. eigenschmeigen Group Title

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 Group Title

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

94. eigenschmeigen Group Title

no problem, love matrices :)

95. experimentX Group Title

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 Group Title

what else does linear algebra cover?

97. experimentX Group Title

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!!