Linear Algebra help Matrix

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• Feb 19th 2013, 11:53 AM
Tweety
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 $\displaystyle x_{1}; x_{2}; x_{3}; x_{4}; x_{5}$ must be satised. For example, the
iron requirement gives $\displaystyle x_{1} + 3x_{2} + 6x_{3} + 5x_{4} + 4x_{5} = 14.$

(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

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

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

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

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
• Feb 19th 2013, 12:18 PM
ILikeSerena
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 $\displaystyle x_4 = 4s$ and $\displaystyle x_5 = -t$, what do you get?
• Feb 19th 2013, 12:30 PM
Tweety
Re: Linear Algebra help Matrix
Hello,

so using x4 =4s and x5=-t

$\displaystyle x_{1}-S-t=2$

$\displaystyle x_{2} -7s-19t = -4$

$\displaystyle x_{3} +7s +9t = 4$

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).
• Feb 19th 2013, 01:18 PM
ILikeSerena
Re: Linear Algebra help Matrix
Quote:

Originally Posted by Tweety
Hello,

so using x4 =4s and x5=-t

$\displaystyle x_{1}-S-t=2$

$\displaystyle x_{2} -7s-19t = -4$

$\displaystyle x_{3} +7s +9t = 4$

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:

$\displaystyle x_{1}=2 +s+t$

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

$\displaystyle x_{3} = 4 -7s -9t$

$\displaystyle x_{4} = 4s$

$\displaystyle x_{5} = -t$

or

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

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.
• Feb 19th 2013, 01:30 PM
ILikeSerena
Re: Linear Algebra help Matrix
Here's what it should be based on your input data:

$\displaystyle x_1 = 2 + s - t$

$\displaystyle x_2 = -4 + 7s - 19t$

$\displaystyle x_3 = 4 - 7s + 9t$

$\displaystyle x_4 = 4s$

$\displaystyle x_5 = t$

See Wolfram|Alpha.

Hence you did not make any mistakes, but your supposed solution does not match your problem.
• Feb 19th 2013, 01:48 PM
Tweety
Re: Linear Algebra help Matrix
Do you mean I cant do part 'b', cause I did something wrong in part a?
• Feb 19th 2013, 01:54 PM
ILikeSerena
Re: Linear Algebra help Matrix
Quote:

Originally Posted by Tweety
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.

• Feb 20th 2013, 05:13 AM
Tweety
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,
• Feb 20th 2013, 05:21 AM
ILikeSerena
Re: Linear Algebra help Matrix
Since each ingredient cannot be less than zero, you get a system of 5 inequalities:

$\displaystyle x_1 \ge 0$

$\displaystyle x_2 \ge 0$

$\displaystyle x_3 \ge 0$

$\displaystyle x_4 \ge 0$

$\displaystyle x_5 \ge 0$

Substitute what you have?
• Feb 20th 2013, 05:37 AM
Tweety
Re: Linear Algebra help Matrix
Quote:

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

$\displaystyle x_1 \ge 0$

$\displaystyle x_2 \ge 0$

$\displaystyle x_3 \ge 0$

$\displaystyle x_4 \ge 0$

$\displaystyle x_5 \ge 0$

Substitute what you have?

oh, I though it was a lot more complicated than that!

$\displaystyle 2+s-t \ge 0$

$\displaystyle -4 + 7s-19t \ge 0$

$\displaystyle 4-7s+9t \ge 0$

$\displaystyle 4s \ge 0$

$\displaystyle t \ge 0$

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?
• Feb 20th 2013, 07:08 AM
ILikeSerena
Re: Linear Algebra help Matrix
Quote:

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

$\displaystyle 2+s-t \ge 0$

$\displaystyle -4 + 7s-19t \ge 0$

$\displaystyle 4-7s+9t \ge 0$

$\displaystyle 4s \ge 0$

$\displaystyle t \ge 0$

Good! ;)

Quote:

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?

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

Do you see why t has to be zero?

Afterward, do the same thing for $\displaystyle s$ and substitute t=0 (which you should have just found).
• Feb 20th 2013, 11:38 AM
Tweety
Re: Linear Algebra help Matrix
is it 0<t<2?
• Feb 20th 2013, 11:40 AM
ILikeSerena
Re: Linear Algebra help Matrix
Quote:

Originally Posted by Tweety
is it 0<t<2?

No... from the 1st and 5th inequality you can only say that $\displaystyle 0 \le t \le 2 + s$.
Which are the 4 inequalities for t that you can get?
• Feb 20th 2013, 11:56 AM
Tweety
Re: Linear Algebra help Matrix
Quote:

Originally Posted by ILikeSerena
No... from the 1st and 5th inequality you can only say that $\displaystyle 0 \le t \le 2 + s$.
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,
• Feb 20th 2013, 11:59 AM
ILikeSerena
Re: Linear Algebra help Matrix

Quote:

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

$\displaystyle 2+s-t \ge 0$

$\displaystyle -4 + 7s-19t \ge 0$

$\displaystyle 4-7s+9t \ge 0$

$\displaystyle 4s \ge 0$

$\displaystyle t \ge 0$

You can rewrite it into inequalities for t as follows.

$\displaystyle t \le 2+s$

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

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

$\displaystyle t \ge 0$

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