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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??

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  1. shivam_bhalla
    • 2 years ago
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    http://mypat.in/PaperImages/3/425/Q_4_004.gif

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

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

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

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

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

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

  8. shivam_bhalla
    • 2 years ago
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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

  9. shivam_bhalla
    • 2 years ago
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    @apoorvk

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

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

  12. shivam_bhalla
    • 2 years ago
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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

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

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

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

  16. apoorvk
    • 2 years ago
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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..

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

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

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

  20. inkyvoyd
    • 2 years ago
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    c=g

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

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

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

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

  25. apoorvk
    • 2 years ago
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    wolfram!!!

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

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

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

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

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

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

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

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

  34. experimentX
    • 2 years ago
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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!!!

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

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

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

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

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

  40. shivam_bhalla
    • 2 years ago
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    Yes

  41. experimentX
    • 2 years ago
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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}}

  42. shivam_bhalla
    • 2 years ago
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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 :)

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

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

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

  46. shivam_bhalla
    • 2 years ago
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    @experimentX

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

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

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

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

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

  52. eigenschmeigen
    • 2 years ago
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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

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

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

  55. shivam_bhalla
    • 2 years ago
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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 :)

  56. eigenschmeigen
    • 2 years ago
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    ;D

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

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

  59. eigenschmeigen
    • 2 years ago
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    no its correct

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

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

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

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

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

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

  66. shivam_bhalla
    • 2 years ago
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    *exams

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

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

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

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

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

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

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

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

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

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

  78. experimentX
    • 2 years ago
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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!!

  79. eigenschmeigen
    • 2 years ago
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    its B

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

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

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

  83. eigenschmeigen
    • 2 years ago
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    ok gimme a sec

  84. shivam_bhalla
    • 2 years ago
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    *board.

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

  86. eigenschmeigen
    • 2 years ago
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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'

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

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

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

  90. eigenschmeigen
    • 2 years ago
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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

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

  92. eigenschmeigen
    • 2 years ago
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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

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

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

  95. experimentX
    • 2 years ago
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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

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

  97. experimentX
    • 2 years ago
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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!!

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