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Math Help - Linear Algebra help Matrix

  1. #1
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    Linear Algebra help Matrix

    A dietitian is planning a meal containing 14 units of iron, 12 units of carbohy-
    drates and 50 units of protein. Five ingredients are available. One portion of
    each ingredient contains units of iron, carbohydrates and protein, as given in
    the following table

    I have attached, an image of the table

    Suppose xi portions of ingredient number i are used, for i = 1; 2; 3; 4; 5. Then
    three linear equations in   x_{1}; x_{2}; x_{3}; x_{4}; x_{5} must be satised. For example, the
    iron requirement gives  x_{1} + 3x_{2} + 6x_{3} + 5x_{4} + 4x_{5} = 14.


    (a) Write down the augmented matrix of this system of three equations and
    nd its reduced row-echelon form. Hence show that the solution can be
    expressed in terms of arbitrary parameters s and t as
    (x1; x2; x3; x4; x5) = (2; 4; 4; 0; 0) + s(1; 7; 7; 4; 0) + t(1; 19; 9; 0; 1).

    (b) The amount of any ingredient used cannot be less than 0. Use this fact
    to write down ve inequalities involving s and t. Show that t = 0 and
    deduce that there is only one possible value of s. How many portions of
    each ingredient should be used? (Fractions of a portion are allowed.)

    I have worked out the reduce row echelon form for the equations I got

    1 0 0 -1/4 1 |2
    0 1 0 -7/4 19|-4
    0 0 1 7/4 -9 |4

    so the equations now are

     x_{1}-\frac{1}{4}x_{4} + x_{5} = 2

     x_{2} - \frac{7}{4} + 19_{5} = -4

     x_{3} + \frac{7}{4}x_{5} = 4

    However I dont know how to get the solutions in the form of the parameter s and t? I am also stuck on part b,

    any help appreciated
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  2. #2
    Super Member ILikeSerena's Avatar
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    Re: Linear Algebra help Matrix

    Hi Tweety!

    Since you have 3 equations and 5 variables, you have a free choice for 2 of the variables.
    Let's pick x4 and x5 for that free choice, and let's also reparametrize them a bit.
    Suppose you substitute x_4 = 4s and x_5 = -t, what do you get?
    Thanks from Tweety
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    Re: Linear Algebra help Matrix

    Hello,

    Thank you for the reply,

    so using x4 =4s and x5=-t

     x_{1}-S-t=2

     x_{2} -7s-19t = -4

     x_{3} +7s +9t = 4

    But I dont see how this gives me (x1; x2; x3; x4; x5) = (2; 4; 4; 0; 0) + s(1; 7; 7; 4; 0) + t(1; 19; 9; 0; 1).
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  4. #4
    Super Member ILikeSerena's Avatar
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    Re: Linear Algebra help Matrix

    Quote Originally Posted by Tweety View Post
    Hello,

    Thank you for the reply,

    so using x4 =4s and x5=-t

     x_{1}-S-t=2

     x_{2} -7s-19t = -4

     x_{3} +7s +9t = 4

    But I dont see how this gives me (x1; x2; x3; x4; x5) = (2; 4; 4; 0; 0) + s(1; 7; 7; 4; 0) + t(1; 19; 9; 0; 1).
    It gives you:

     x_{1}=2 +s+t

     x_{2}  = -4 +7s+19t

     x_{3}  = 4 -7s -9t

     x_{4}  = 4s

     x_{5}  = -t

    or

     \begin{bmatrix}x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \end{bmatrix} = \begin{bmatrix}2 \\ -4 \\ 4 \\ 0 \\ 0 \end{bmatrix} + s \begin{bmatrix}1 \\ 7 \\ -7 \\ 4 \\ 0 \end{bmatrix}  + t \begin{bmatrix}1 \\ 19 \\ -9 \\ 0 \\ -1 \end{bmatrix}

    That is... almost what you're supposed to have...
    The difference would be caused either by a mistake in your calculations or by a mistake in the problem or supposed solution.
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  5. #5
    Super Member ILikeSerena's Avatar
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    Re: Linear Algebra help Matrix

    Here's what it should be based on your input data:

    x_1 = 2 + s - t

    x_2 = -4 + 7s - 19t

    x_3 = 4 - 7s + 9t

    x_4 = 4s

    x_5 = t

    See Wolfram|Alpha.

    Hence you did not make any mistakes, but your supposed solution does not match your problem.
    Thanks from Tweety
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  6. #6
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    Re: Linear Algebra help Matrix

    Do you mean I cant do part 'b', cause I did something wrong in part a?
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  7. #7
    Super Member ILikeSerena's Avatar
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    Re: Linear Algebra help Matrix

    Quote Originally Posted by Tweety View Post
    Do you mean I cant do part 'b', cause I did something wrong in part a?
    You can do part (b).
    Actually, the fact that all coefficients in the solution of (a) are positive, confirms that it is wrong.
    It would make no sense to do (b) with the given solution of (a).
    Apparently a couple of minus signs were dropped.

    You should go on from the equations you already had.
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  8. #8
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    Re: Linear Algebra help Matrix

    (b) The amount of any ingredient used cannot be less than 0. Use this fact
    to write down five inequalities involving s and t. Show that t = 0 and
    deduce that there is only one possible value of s. How many portions of
    each ingredient should be used? (Fractions of a portion are allowed.)

    Thank you,

    Still not sure how to go about part 'b' from my three equations, how do i form inequalities in s and t?

    each ingredient cannot be less than zero, but not sure how to use this fact to form 5 inequalities in s and t?

    Any help appreciated,
    thank you,
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  9. #9
    Super Member ILikeSerena's Avatar
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    Re: Linear Algebra help Matrix

    Since each ingredient cannot be less than zero, you get a system of 5 inequalities:

    x_1 \ge 0

    x_2 \ge 0

    x_3 \ge 0

    x_4 \ge 0

    x_5 \ge 0

    Substitute what you have?
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  10. #10
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    Re: Linear Algebra help Matrix

    Quote Originally Posted by ILikeSerena View Post
    Since each ingredient cannot be less than zero, you get a system of 5 inequalities:

    x_1 \ge 0

    x_2 \ge 0

    x_3 \ge 0

    x_4 \ge 0

    x_5 \ge 0

    Substitute what you have?
    oh, I though it was a lot more complicated than that!


      2+s-t \ge 0

     -4 + 7s-19t \ge 0

     4-7s+9t \ge 0

     4s \ge 0

     t \ge 0

    to show that t = 0, do i just choose values for s and t?

    example let t = 1, s = 2,

    than the third inequality does not work casue it gives -29, which less than 0,

    how do i show s can only have one value?
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  11. #11
    Super Member ILikeSerena's Avatar
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    Re: Linear Algebra help Matrix

    Quote Originally Posted by Tweety View Post
    oh, I though it was a lot more complicated than that!


      2+s-t \ge 0

     -4 + 7s-19t \ge 0

     4-7s+9t \ge 0

     4s \ge 0

     t \ge 0
    Good!


    to show that t = 0, do i just choose values for s and t?

    example let t = 1, s = 2,

    than the third inequality does not work casue it gives -29, which less than 0,

    how do i show s can only have one value?
    No, you can't just choose values.

    Can you rewrite the inequalities to the following form?

    t \ge ... \qquad \text{ and/or }\qquad t \le ...

    Does anything catch your attention?
    Do you see why t has to be zero?

    Afterward, do the same thing for s and substitute t=0 (which you should have just found).
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  12. #12
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    Re: Linear Algebra help Matrix

    is it 0<t<2?
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  13. #13
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    Re: Linear Algebra help Matrix

    Quote Originally Posted by Tweety View Post
    is it 0<t<2?
    No... from the 1st and 5th inequality you can only say that 0 \le t \le 2 + s.
    Which are the 4 inequalities for t that you can get?
    Last edited by ILikeSerena; February 20th 2013 at 12:43 PM.
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  14. #14
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    Re: Linear Algebra help Matrix

    Quote Originally Posted by ILikeSerena View Post
    No... from the 1st and 5th inequality you can only say that 0 \le t \le 2 + s.
    Which are the 4 inequalities for t that you can get?
    I am acutally very confused, but do I just put t = 1, t = 2+s, and t = 0, in each inequality, ?

    so for 4-7s-19t>0

    i get -34-26t,
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  15. #15
    Super Member ILikeSerena's Avatar
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    Re: Linear Algebra help Matrix

    This is what you already had:

    Quote Originally Posted by Tweety View Post
    oh, I though it was a lot more complicated than that!

      2+s-t \ge 0

     -4 + 7s-19t \ge 0

     4-7s+9t \ge 0

     4s \ge 0

     t \ge 0

    You can rewrite it into inequalities for t as follows.

      t \le 2+s

     t \le (-4 + 7s) / 19

     t \ge (-4+7s) / 9

     t \ge 0

    Do you notice anything strange about the 2nd and 3rd inequality?
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