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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

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  1. shivam_bhalla Group Title
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    http://mypat.in/PaperImages/3/425/Q_4_004.gif

    • 2 years ago
  2. UnkleRhaukus Group Title
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    can you factor all of the elements

    • 2 years ago
  3. shivam_bhalla Group Title
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    @UnkleRhaukus ,how exactly??

    • 2 years ago
  4. UnkleRhaukus Group Title
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    \[\Delta=\left|\begin{array}\\1+a^2+a^4&1+ab+a^2b^2&1+ac+a^2c^2\\1+ab+a^2b^2&1+b^2+b^4&1+bc+b^2c^2\\1+ac+a^2c^2&1+bc+b^2c^2&1+c^2+c^4\end{array}\right|\]

    • 2 years ago
  5. UnkleRhaukus Group Title
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    yeah maybe not, i really dont want to do it the normal way

    • 2 years ago
  6. shivam_bhalla Group Title
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    Normal way-->never :D It would take ages :P

    • 2 years ago
  7. UnkleRhaukus Group Title
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    but it would work in theory, but i would make about ten mistakes

    • 2 years ago
  8. shivam_bhalla Group Title
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    I am sure you play around with determinant properties like Row1 = Row 1 + Row 2+Row3 . I mean some tweaking of that sort

    • 2 years ago
  9. shivam_bhalla Group Title
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    @apoorvk

    • 2 years ago
  10. inkyvoyd Group Title
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    can we just do it the foolish way?

    • 2 years ago
  11. apoorvk Group Title
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    Hmm, tried C1 - C2?

    • 2 years ago
  12. shivam_bhalla Group Title
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    @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

    • 2 years ago
  13. shivam_bhalla Group Title
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    @apoorvk , I did that. But nothing can be taken common out :(

    • 2 years ago
  14. inkyvoyd Group Title
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    I'm going to try this problem

    • 2 years ago
  15. inkyvoyd Group Title
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    Have no fear, for inky is here!

    • 2 years ago
  16. apoorvk Group Title
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    I got common stuff out through C1-C2 and C2-C3. but, i am still stuck with a trinomial in each ditch..

    • 2 years ago
  17. inkyvoyd Group Title
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    @shivam_bhalla I HAVE AN IDEA

    • 2 years ago
  18. inkyvoyd Group Title
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    aei+bfg+cdh-ceg-bdi-afh =a(ei-fh)+b(fg-dj)+c(dg-eg)

    • 2 years ago
  19. inkyvoyd Group Title
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    b=d f=h

    • 2 years ago
  20. inkyvoyd Group Title
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    c=g

    • 2 years ago
  21. inkyvoyd Group Title
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    aei+bfg+cdh-ceg-bdi-afh aei+bf^2+cbf-ec^2-ah^2

    • 2 years ago
  22. shivam_bhalla Group Title
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    @inkyvoyd , what is a,b,c,d,e,f,g?? Are you expanding the determinant??

    • 2 years ago
  23. inkyvoyd Group Title
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    I'm wikipediaing, and not making any progress.

    • 2 years ago
  24. shivam_bhalla Group Title
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    LOL. @ "wikipediaing" . :)

    • 2 years ago
  25. apoorvk Group Title
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    wolfram!!!

    • 2 years ago
  26. shivam_bhalla Group Title
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    How to wolfram this ??

    • 2 years ago
  27. shivam_bhalla Group Title
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    @radar sir, any clue ??

    • 2 years ago
  28. experimentX Group Title
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    factor out the commons!!!

    • 2 years ago
  29. shivam_bhalla Group Title
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    @experimentX , how exactly ?? I mean row wise or column wise. And moreover, what to factor out ??

    • 2 years ago
  30. experimentX Group Title
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    lol ... factoring out would lead to 3 more matrices ..thats ugly... must be some pattern!!

    • 2 years ago
  31. shivam_bhalla Group Title
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    Oh. Now get what you meant by factoring .:P I tried that too and observed what you said :)

    • 2 years ago
  32. experimentX Group Title
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    it seems transpose of itself!!

    • 2 years ago
  33. apoorvk Group Title
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    |dw:1336584742255:dw| this is how far i got then..

    • 2 years ago
  34. experimentX Group Title
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    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!!!

    • 2 years ago
  35. shivam_bhalla Group Title
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    This problem really is trollling me very badly :( I will solve this problem before going to sleeep

    • 2 years ago
  36. experimentX Group Title
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    if you find let met know!!

    • 2 years ago
  37. experimentX Group Title
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    Oo ...we can drop all ones!!

    • 2 years ago
  38. experimentX Group Title
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    Oo ... i got zero!! LOL

    • 2 years ago
  39. experimentX Group Title
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    @shivam_bhalla you still there??

    • 2 years ago
  40. shivam_bhalla Group Title
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    Yes

    • 2 years ago
  41. experimentX Group Title
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    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}}

    • 2 years ago
  42. shivam_bhalla Group Title
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    LOL @experimentX , that is doing it the long way Atleast I know the answer know. :) Thanks for typing out the huge question on wolframalpha :)

    • 2 years ago
  43. experimentX Group Title
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    No .. i know the shortcut too!!

    • 2 years ago
  44. experimentX Group Title
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    actually i verified answer at wolframalpha!!

    • 2 years ago
  45. shivam_bhalla Group Title
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    Ok. Just tell me the method and not the steps because that would literally troll you

    • 2 years ago
  46. shivam_bhalla Group Title
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    @experimentX

    • 2 years ago
  47. experimentX Group Title
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    take out all ones!! sine they will have same values on row .. they will be zero ..

    • 2 years ago
  48. shivam_bhalla Group Title
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    Sorry, didn't get you @experimentX

    • 2 years ago
  49. experimentX Group Title
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    |dw:1336588777520:dw|

    • 2 years ago
  50. experimentX Group Title
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    |dw:1336588827490:dw|

    • 2 years ago
  51. shivam_bhalla Group Title
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    |dw:1336588893052:dw|

    • 2 years ago
  52. eigenschmeigen Group Title
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    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

    • 2 years ago
  53. experimentX Group Title
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    whenever matrix has same elements on row or column ... determinant is zero.

    • 2 years ago
  54. eigenschmeigen Group Title
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    and by symmetry (b-c) and (a-c) are

    • 2 years ago
  55. shivam_bhalla Group Title
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    @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 :)

    • 2 years ago
  56. eigenschmeigen Group Title
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    ;D

    • 2 years ago
  57. eigenschmeigen Group Title
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    proportional ->eg R1 = kR2 so k = 1 is a specific case

    • 2 years ago
  58. experimentX Group Title
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    Oo ... then i must have mistake!!

    • 2 years ago
  59. eigenschmeigen Group Title
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    no its correct

    • 2 years ago
  60. eigenschmeigen Group Title
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    if R1 = R2 then det(M) = 0

    • 2 years ago
  61. shivam_bhalla Group Title
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    @eigenschmeigen , there is no R1=R2 in @experimentX 's proof

    • 2 years ago
  62. experimentX Group Title
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    Oh my mistake ... i got the answer in fluke

    • 2 years ago
  63. shivam_bhalla Group Title
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    @experimentX , I wish these flukes come for me too in my exam :P

    • 2 years ago
  64. experimentX Group Title
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    LOL ... i haven't reviewed matrices in long time!! best of luck if you are having exams right now!!

    • 2 years ago
  65. shivam_bhalla Group Title
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    LOL. I have college entrance exam's right now :P

    • 2 years ago
  66. shivam_bhalla Group Title
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    *exams

    • 2 years ago
  67. eigenschmeigen Group Title
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    is this problem solved or still at large? i can have a proper go now i have a pen

    • 2 years ago
  68. shivam_bhalla Group Title
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    The problem as of now is unsolved @eigenschmeigen

    • 2 years ago
  69. experimentX Group Title
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    Okay ... i think i got this time!! you can use same technique to reduce matrix!!

    • 2 years ago
  70. experimentX Group Title
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    |dw:1336589808195:dw|

    • 2 years ago
  71. experimentX Group Title
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    and then it's simplification!!

    • 2 years ago
  72. experimentX Group Title
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    apply operation as above!!! |dw:1336589933207:dw|

    • 2 years ago
  73. experimentX Group Title
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    same as you had in your second place!! -- minus --- one

    • 2 years ago
  74. shivam_bhalla Group Title
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    |dw:1336590227531:dw|

    • 2 years ago
  75. experimentX Group Title
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    • 2 years ago
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  76. experimentX Group Title
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    |dw:1336590458004:dw| It can be proved!!

    • 2 years ago
  77. shivam_bhalla Group Title
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    @experimentX , I will try and let you know in 5-10 minutes @experimentX

    • 2 years ago
  78. experimentX Group Title
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    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!!

    • 2 years ago
  79. eigenschmeigen Group Title
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    its B

    • 2 years ago
  80. eigenschmeigen Group Title
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    i'm just checking my working

    • 2 years ago
  81. eigenschmeigen Group Title
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    how do i write matrices on here?

    • 2 years ago
  82. shivam_bhalla Group Title
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    @eigenschmeigen , your answer is right. You can use drawing doard :)

    • 2 years ago
  83. eigenschmeigen Group Title
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    ok gimme a sec

    • 2 years ago
  84. shivam_bhalla Group Title
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    *board.

    • 2 years ago
  85. eigenschmeigen Group Title
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    i like the equation writer, i'm quicker that way: \[\left[\begin{matrix}1+a^2 + a^4 & 1+ ab+a^2b^2 & 1 +ac +a^2c^2 \\ 1 + ab +a^2b^2 & 1+b^2 + b^4 & 1 +ac +a^2c^2 \\ 1 +ac +a^2c^2 & 1 +bc+b^2c^2 & 1 + c^2 + c^4\end{matrix}\right]\]

    • 2 years ago
  86. eigenschmeigen Group Title
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    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'

    • 2 years ago
  87. eigenschmeigen Group Title
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    \[\left[\begin{matrix}a^4 + a^2 - ab - a^2b^2 & a^2b^2 + ab -b^2 -b^4 & a^2c^2 + ac - bc - b^2c^2 \\ a^2b^2 + ab -ac -a^2c^2 & b^4 + b^2 - bc - b^2c^2 & b^2c^2 + bc -c^2 - c^4 \\ 1 +ac +a^2c^2 & 1 +bc+b^2c^2 & 1 + c^2 + c^4\end{matrix}\right]\]

    • 2 years ago
  88. eigenschmeigen Group Title
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    factor out (a-b)(b-c)

    • 2 years ago
  89. eigenschmeigen Group Title
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    \[(a-b)(b-c)\left[\begin{matrix}a^3 + a +ba^2 & b^3 + b + ab^2 & c(ac+bc+1) \\ a(ac+ab+1) & b^3 + b +cb^2 & c^3 + c + bc^2 \\ 1 +ac +a^2c^2 & 1 +bc+b^2c^2 & 1 + c^2 + c^4\end{matrix}\right]\]

    • 2 years ago
  90. eigenschmeigen Group Title
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    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

    • 2 years ago
  91. eigenschmeigen Group Title
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    you can continue to factorise if you want, but this is sufficient to get the right option.

    • 2 years ago
  92. eigenschmeigen Group Title
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    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

    • 2 years ago
  93. shivam_bhalla Group Title
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    Thank you @eigenschmeigen and @experimentX for both of your solutions :D Thanks for taking your time :)

    • 2 years ago
  94. eigenschmeigen Group Title
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    no problem, love matrices :)

    • 2 years ago
  95. experimentX Group Title
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    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

    • 2 years ago
  96. eigenschmeigen Group Title
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    what else does linear algebra cover?

    • 2 years ago
  97. experimentX Group Title
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    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!!

    • 2 years ago
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