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Callisto

  • 2 years ago

Consider the system: x + 2z =1 x+2y+4z =1 2x-y+3z=2 For all solutions {x,y,z}, find the range of k such that \(k(x^2-3) > yz \)

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  1. abb0t
    • 2 years ago
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    I don't know!! :'(

  2. nubeer
    • 2 years ago
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    hmm i can find values of x , y and z but don;t know what to do after wards with them

  3. Callisto
    • 2 years ago
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    Same here...

  4. nubeer
    • 2 years ago
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    lol sorry guess not much help i am

  5. Callisto
    • 2 years ago
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    Wait, can you find the numerical values for x, y, z?

  6. nubeer
    • 2 years ago
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    sure , give me a moment.

  7. Jonask
    • 2 years ago
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    what about first finding the solution set of the system using algebra matrix

  8. abb0t
    • 2 years ago
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    Gaussian elimination method.

  9. Callisto
    • 2 years ago
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    |dw:1363361660822:dw|

  10. Callisto
    • 2 years ago
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    |dw:1363361796138:dw|

  11. Callisto
    • 2 years ago
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    That's what I can do and that's why I ask if you can find the numerical value for {x,y,z}

  12. nubeer
    • 2 years ago
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    k honestly i have done sole me this kind before where we get a who row zero but hard to explain the logic behind it..that whole zero row just made hard to find definite answer.. i normally would do this with the simultaneous method now. so we can get x, y , and z in terms of each other.

  13. nubeer
    • 2 years ago
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    if i solve simultaneously, i got z = -y and x = 1 -2z

  14. Callisto
    • 2 years ago
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    Yes, I got the same too, but how can we utilize it to find range of k?

  15. nubeer
    • 2 years ago
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    |dw:1363362152973:dw|

  16. nubeer
    • 2 years ago
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    so i think we can get value of x , y and z if we suppose any value of z .. according to it , we will get other values.

  17. Callisto
    • 2 years ago
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    \[k[(1-2z)^2-3]>-z^2\]\[k>\frac{-z^2}{(1-2z)^2-3}\]Hmm..

  18. nubeer
    • 2 years ago
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    hmm got same thing.. after this i did the long division.. but lol not possible way out

  19. Callisto
    • 2 years ago
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    I'm thinking of differetiating the fraction to find the min/max/range..

  20. nubeer
    • 2 years ago
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    can try tht

  21. Callisto
    • 2 years ago
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    Suppose wolf. is right, then we get \(\frac{z(z+1)}{(-2z^2+2z+1)^2}\) Put that derivative = 0, z=0 or z=-1

  22. Callisto
    • 2 years ago
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    Oh wait.. Why would I differentiate that!!!

  23. nubeer
    • 2 years ago
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    hmm thought u were trying to get max and min.

  24. Callisto
    • 2 years ago
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    But it won't work?

  25. nubeer
    • 2 years ago
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    yeah.. i still feel we using wrong approach.

  26. Azteck
    • 2 years ago
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    \[x + 2z =1[1]\] \[x+2y+4z =1[2]\] \[2x-y+3z=2[3]\] In Equation [1] \[x=1-2z\] In Equation [2] \[1-2z+2y+4z=1\] \[2y+2z=0[4]\] In Equation [3] \[2(1-2z)-y+3z=2\] \[2-4z-y+3z=2\] \[-y-z=0\] \[y+z=0[5]\] Therefore: \[x=1-2z\] or \[x=1+2y\] \[2z=1-x\] \[z=\frac{1-x}{2}\] \[2y=x-1\] \[y=\frac{x-1}{2}\] Therefore: \[yz=\frac{(1-x)(x-1)}{4}\] In: \[k(x^2−3)>yz\] \[k(x^2-3)>-\frac{(1-x)^2}{4}\]

  27. Azteck
    • 2 years ago
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    \[4k(x^2-3)>-(1-x)^2\] \[4kx^2-12k+(1-2x+x^2)>0\] \[4kx^2-12k+1-2x+x^2>0\] \[(4k+1)x^2-2x-(12k-1)>0\] Using the discriminant: \[b^2-4ac>0\] \[4+4(12k-1)(4k+1)>0\] Solve as quadratic and find the range of values for k.

  28. Callisto
    • 2 years ago
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    Almost there, but still not correct.

  29. mathslover
    • 2 years ago
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    Man at work.

  30. AravindG
    • 2 years ago
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    am I late? :)

  31. mathslover
    • 2 years ago
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    I dont' think so.

  32. linshan789
    • 2 years ago
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    Can you just solve for x,y,and z (using a system of equations) then plug the variables back in?

  33. Callisto
    • 2 years ago
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    I've tried solving the system above...

  34. linshan789
    • 2 years ago
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    I ended up with x=1, y=0, z=0. So, \[k(1 ^{2}-3)>(0)(0)\] =>\[k(-2)>0\] =>k<0

  35. Callisto
    • 2 years ago
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    It shouldn't have a unique solution.

  36. linshan789
    • 2 years ago
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    Oh. What level of math is this?

  37. Callisto
    • 2 years ago
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    High school

  38. mathslover
    • 2 years ago
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    k belongs to ( - infinity , -1/4 ] union (-1/6 , 0 ) ?

  39. Callisto
    • 2 years ago
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    No, almost there. Can you tell us how you get there?

  40. linshan789
    • 2 years ago
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    wait, how is k<0 unique? (-inf, 0)

  41. mathslover
    • 2 years ago
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    k belongs to ( - infinity , -1/4 ] union (-1/6 , 0 ] ?

  42. Callisto
    • 2 years ago
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    No....

  43. mathslover
    • 2 years ago
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    can I post here after 30 minutes please? I have to go for dinner ?

  44. Callisto
    • 2 years ago
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    Sure, take your time :)

  45. mathslover
    • 2 years ago
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    I am back to work.

  46. Callisto
    • 2 years ago
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    Any guideline?

  47. mathslover
    • 2 years ago
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    Well first of all I solved for x and then put it into the eqn and then took diff. conditions .

  48. mathslover
    • 2 years ago
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    k belongs to ( - infinity , -1/4 ] union [-1/6 , 0 ] ?

  49. Callisto
    • 2 years ago
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    No.. diff = different/differentiate?

  50. mathslover
    • 2 years ago
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    different.

  51. aajugdar
    • 2 years ago
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    looks easy wait

  52. mathslover
    • 2 years ago
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    I am getting diff. answers. Wait!

  53. Callisto
    • 2 years ago
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    I have some patience, so I keep waiting, and trying :)

  54. mathslover
    • 2 years ago
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    That is good.

  55. shubhamsrg
    • 2 years ago
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    is it [0, -1/6] ?

  56. Callisto
    • 2 years ago
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    Very very close!!! How do you get it??

  57. shubhamsrg
    • 2 years ago
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    -(x-1)^2 /(4x^2 -12) tried to work with this :|

  58. mathslover
    • 2 years ago
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    k belongs to [-1/6,0] union (-infinity ,1/4] .. i am still getting it

  59. mathslover
    • 2 years ago
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    is it wrong^ ?

  60. mathslover
    • 2 years ago
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    @Callisto ?

  61. Callisto
    • 2 years ago
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    It is wrong.

  62. shubhamsrg
    • 2 years ago
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    Did you get what I did ?

  63. shubhamsrg
    • 2 years ago
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    Coz I am not very confident myself! :P

  64. Callisto
    • 2 years ago
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    I didn't get what you did :(

  65. Callisto
    • 2 years ago
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    Did you start from this:\[k(x^2-3)>-\frac{(1-x)^2}{4}\]?

  66. shubhamsrg
    • 2 years ago
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    I am calling that f(x) k> f(x) whatever be the range of f(x), k's range will be opposite of that, i.e. maxima of f(x) is min of k and vice versa. so, is the ans (0,-1/6) ?

  67. Callisto
    • 2 years ago
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    The answer is correct, but I don't understand your explanation.

  68. mathslover
    • 2 years ago
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    But aren't 0 and -1/6 satisfying the equation ?

  69. Callisto
    • 2 years ago
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    The inequality itself is > , not >=

  70. mathslover
    • 2 years ago
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    k(x^2 - 3)>yz Put, x = 1-2z and y = -z Then k{1+4z^2 -4z - 3} > -z^2 => k(4z^2 - 4z - 2) > -z^2 => (4k+1)z^2 - 4kz - 2k>0 Now put, k = 0 you get z^2 > 0 Which is true ...... So k = 0 satisfies your inequality ....

  71. mathslover
    • 2 years ago
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    @Callisto

  72. aajugdar
    • 2 years ago
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    lol x=1 btw I was having dinner z=0 and y =-1/2

  73. aajugdar
    • 2 years ago
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    do this, eliminate 1 from lowest row in matrix you get two equations of x and z,solve them then u wil get answers of x,y,z

  74. mathslover
    • 2 years ago
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    And MY z = -1/6 have a look: (4k+1)z^2 - 4kz - 2k>0 Put k =-1/6 We get z^2 + 2z + 1 >0 OOPS that is true toooooo

  75. mathslover
    • 2 years ago
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    So k =-1/6 also satisifies your inequality

  76. Callisto
    • 2 years ago
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    What if z=0?

  77. mathslover
    • 2 years ago
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    Well if z = 0 then go for it: k < 1/2

  78. aajugdar
    • 2 years ago
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    yes z=0

  79. Callisto
    • 2 years ago
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    @aajugdar I don't mind if you check my matrix work above. I can't find a unique solution to the system.

  80. mathslover
    • 2 years ago
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    Well the solutions given by you and shubhamsrg have many cases left out .....

  81. Callisto
    • 2 years ago
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    But when z=0, the inequality fails.

  82. aajugdar
    • 2 years ago
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    put them in equation dude,the values are correct ok hold on,i wil tell ya

  83. aajugdar
    • 2 years ago
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    perform R3- R2/2 on R3 and then again arrange them in equation form,u will get x+2z =1 and then 3/2x+z =3/2

  84. mathslover
    • 2 years ago
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    Well ok in that case we get (by putting in my quadratic inequality in terms of z and k), k < 1/2 What about that ?

  85. aajugdar
    • 2 years ago
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    solve them,u will get values of x and z,from that values,find value of y by putting it in equation 2

  86. Callisto
    • 2 years ago
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    @mathslover That is not even the answer @aajugdar Check this |dw:1363447547599:dw|

  87. aajugdar
    • 2 years ago
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    what operation did you perform on 3rd row????

  88. Callisto
    • 2 years ago
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    -2 x R1 + R3 -> R3

  89. Callisto
    • 2 years ago
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    @mathlover When I learnt inequality, my teacher told me that if (1) ax^2+bx+c>0, where a>0, then the solution would be like (alpha)<x or x>(beta) (2) ax^2+bx+c<0, where a>0, then the solution would be like (alpha)<x<(beta)

  90. aajugdar
    • 2 years ago
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    nope man you didn't solve it as it should be solved,you see by makng complete 1 row 0 you wil never get solution you shouldn't keep a row NULL its a rule

  91. mathslover
    • 2 years ago
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    @Callisto very correct

  92. aajugdar
    • 2 years ago
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    remember 3 2 unknowns,2 equations,3 unknowns 3 eqautions.

  93. mathslover
    • 2 years ago
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    (4k+1)z^2 - 4kz - 2k>0 I used that (please have a look at my solution)

  94. Callisto
    • 2 years ago
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    I am not a man This system has no unique solution, that's what the determinant of coefficient matrix tell us.

  95. aajugdar
    • 2 years ago
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    2***

  96. aajugdar
    • 2 years ago
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    well,solve it correctly,it has a solution I found it

  97. Callisto
    • 2 years ago
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    @aajugdar You can try using WolframAlpha, you can't get a unique solution. @mathslover Can you guarantee that (4k+1)>0?

  98. shubhamsrg
    • 2 years ago
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    k > -(x-1)^2 /(4x^2 -12) the derivatives are 0 at x=3 and 1 respective values of y are -1/6 (local maxima) and 0(local minima) so k > -1/6 and k<0

  99. aajugdar
    • 2 years ago
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    -facewall-

  100. shubhamsrg
    • 2 years ago
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    only in that range x>f(x)

  101. shubhamsrg
    • 2 years ago
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    I mean k>f(x)

  102. mathslover
    • 2 years ago
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    Case - 1 : 4k + 1 > 0 => k>-1/4 Now, that means the inequality has a graph upward. Now for the inequality >0. b^2 - 4ac < 0 => k(6k+1) < 0 => If k< 0 then 6k+1>0 i.e. k>-1/6 And if k>0 then 6k+1 < 0 i.e. k<-1/6 (This option is not possible) So, k<0 ; k>-1/6 ; and k>-1/4 Take intersection: k belongs to (-1/6,0) @Callisto I am thus taking cases

  103. mathslover
    • 2 years ago
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    Case -2 : (4k+1)<0 => k <-1/4 i .e. the graph is bent downwards. Now for inequality> 0. b^2 - 4ac > 0 => k(6k+1) >0 => k> 0 and k>-1/6 or k<0 and k<-1/6 => k>0 or k<-1/6 But we took k<-1/4 So, k>0 case is eliminated. So, k belongs to (-infinity, -1/4)

  104. mathslover
    • 2 years ago
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    Case - 3: 4k+1 = 0 => k = -1/4 So, inequality reduces to z>-1/2

  105. mathslover
    • 2 years ago
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    Which obviously consitutes part of real values of z.

  106. mathslover
    • 2 years ago
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    So, I said answer should be: k belongs to (-infinity, -1/4] union (-1/6,0)........ Kindly please tell if I am correct or If I went wrong somewere

  107. mathslover
    • 2 years ago
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    Well I agree with you that k = -1/4 and -1/6 do not constitute the solution so modify it k belongs to (-infinity, -1/4) union (-1/6,0) But I still don't know that how will I then deal with 4k+1 = 0 case

  108. Callisto
    • 2 years ago
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    I'm thinking.. for the inequality, (4k+1)z^2 - 4kz - 2k>0, it doesn't imply that (4k+1) >0 or (4k+1)<0. It can be 4k+1 =0 Oh, you've pointed the problem out.

  109. mathslover
    • 2 years ago
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    Think about that

  110. mathslover
    • 2 years ago
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    I hope I didn't annoy you?

  111. Callisto
    • 2 years ago
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    Nope, it just the exchange of ideas.

  112. Callisto
    • 2 years ago
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    *it's

  113. stamp
    • 2 years ago
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    it seems the solutions are not numbers, but rather variables (see attachment) also, http://tutorial.math.lamar.edu/Classes/Alg/AugmentedMatrixII.aspx

    1 Attachment
  114. Callisto
    • 2 years ago
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    The solutions to the system are not numbers, that is it has no unique solution. The link doesn't wok.

  115. Callisto
    • 2 years ago
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    *work

  116. mathslover
    • 2 years ago
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    @stamp read the question again ;)

  117. genius12
    • 2 years ago
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    The system is easy to solve for x, y, z. It's just a little tedious. I am in a rush as it's nearly 4 am so I'll give you a brief solution to this problem. First rearrange x + 2z = 1 to z = (1 - x) / 2 Now you have z in terms of x. Find the value of y in terms of x from the second equation and plug in (1 - x)/2 for z in it. Then plug these y and z values in terms of x in the third equation and solve for x. Then solve for z and then y. You can of course use Gaussian-Elimination Method but that can be annoying at times. Anyway, after you got the solution for x,y,z, just plug the values in to the inequality and rearrange it to k > yz/(x^2 - 3). And that will give you the range for k. G2g now. @Callisto

  118. Callisto
    • 2 years ago
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    1. There is not unique solution to the system. I have solved it using Gaussian elimination and have already put what I have got above. 2. I have put x, y, z into the inequality (in terms of z), but it just leaves me a range with unknown z. 3. I have tried differentiate z to get min/max, but the values I got (z=0/z=-1) does not match with the answer.

  119. genius12
    • 2 years ago
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    So does the system have infinite solutions or a finite number of solutions?

  120. Callisto
    • 2 years ago
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    Infinitely many solution.

  121. genius12
    • 2 years ago
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    If it has infinitely many solutions then which solution gives the smallest value for yz / x^2-3?

  122. some_someone
    • 2 years ago
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    I would say range is all real numbers?

  123. Callisto
    • 2 years ago
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    No.

  124. some_someone
    • 2 years ago
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    no? do you know the answer then?

  125. Callisto
    • 2 years ago
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    I know the answer.

  126. Callisto
    • 2 years ago
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    I don't know how to get the answer though.

  127. some_someone
    • 2 years ago
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    oh ok got you

  128. genius12
    • 2 years ago
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    Do you know what the smallest solution is for x,y,z such that yz / x^2 - 3 is the smallest possible value? Try to find such a solution.

  129. Callisto
    • 2 years ago
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    @genius12 Would you mind checking me steps of solving the system?|dw:1363507811134:dw|

  130. Callisto
    • 2 years ago
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    *my

  131. genius12
    • 2 years ago
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    Did you do this through gaussian elimination?

  132. Callisto
    • 2 years ago
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    Yes.

  133. genius12
    • 2 years ago
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    Have you tried solving this algebraically? It will be tedious, but could be promising. It's easy to mess up on matrices, but not as much when you solve algebraically.

  134. genius12
    • 2 years ago
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    I know for sure that you will be able to solve this algebraically, and I am pretty confident that you won't get "infinitely many solutions", hopefully. 4:16 AM. Dark room, people sleeping in here, I think I'll head to sleep as well. Good luck Callisto.

  135. Callisto
    • 2 years ago
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    x + 2z =1 ---(1) x+2y+4z =1 --- (2) 2x-y+3z=2 ---(3) -(1)+(2) 2y + 2z = 0 => y = -z --- (4) Sub. (4) into (3) 2x - (-z) + 3z =2 x+2z = 1 --- (5) , which is also (1) Pretty obvious, huh?

  136. Callisto
    • 2 years ago
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    If it is not infinitely many solutions, then it will be no solution.

  137. Callisto
    • 2 years ago
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    You'd better sleep now. Good night.

  138. genius12
    • 2 years ago
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    Then there is infinitely many solutions.

  139. Callisto
    • 2 years ago
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    What's next?

  140. some_someone
    • 2 years ago
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    Wouldn't the system of equations be inconsistent?

  141. Callisto
    • 2 years ago
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    The system should be consistent.

  142. genius12
    • 2 years ago
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    Yes, for you to solve the inequality and get the range for k, it should be consistent which it isn't.

  143. some_someone
    • 2 years ago
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    @genius12 yes i agree it is inconsistent.

  144. Callisto
    • 2 years ago
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    Please point out my mistakes then.

  145. Callisto
    • 2 years ago
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    FYI: http://www.wolframalpha.com/input/?i=x+%2B+2z+%3D1+and+x%2B2y%2B4z+%3D1+and+2x-y%2B3z%3D2

  146. genius12
    • 2 years ago
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    I could pick any value for z and get the same three equations. This gives us no way to come up with a range for k. =.=

  147. Callisto
    • 2 years ago
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    That's why we need to think how to get a way to solve it.

  148. genius12
    • 2 years ago
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    Something wrong with the question? lol

  149. Callisto
    • 2 years ago
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    No way! It is a past exam question.

  150. genius12
    • 2 years ago
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    What unit does this question cover?

  151. Callisto
    • 2 years ago
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    Matrix and determinant.

  152. genius12
    • 2 years ago
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    If it's a past exam, isn't there a given solution?

  153. Callisto
    • 2 years ago
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    The final answer - Yes.

  154. genius12
    • 2 years ago
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    Check it? lol

  155. Callisto
    • 2 years ago
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    I know what the final answer is. I just don't know how to get the final answer.

  156. Callisto
    • 2 years ago
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    When z=-1, (max) => k=min (x, y, z) = (3, 1, -1) \[k>\frac{yz}{x^2-3}\]\[k>\frac{(1)(-1)}{3^2-3}\]\[k>-\frac{1}{6}\] When z = 0, (min) => k = max (x, y, z) = (1, 0, 0) \[k<\frac{yz}{x^2-3}\]\[k<\frac{(0)(0)}{1^2-3}\]\[k<0\] @shubhamsrg Your method works!! Too intuitive for me though :(

  157. shubhamsrg
    • 2 years ago
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    I am not too sure how to put my words more properly. :| Hey, why don't you try to understand this with the graph, you may use wolfram for the graphing.

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