Linear Algebra help Matrix

**Tweety** · February 19th 2013, 11:53 AM

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

**ILikeSerena** · February 19th 2013, 12:18 PM

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?

**Tweety** · February 19th 2013, 12:30 PM

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).

**ILikeSerena** · February 19th 2013, 01:18 PM

Originally Posted by Tweety

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.

**ILikeSerena** · February 19th 2013, 01:30 PM

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.

**Tweety** · February 19th 2013, 01:48 PM

Do you mean I cant do part 'b', cause I did something wrong in part a?

**ILikeSerena** · February 19th 2013, 01:54 PM

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.

You should go on from the equations you already had.

**Tweety** · February 20th 2013, 05:13 AM

(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,

**ILikeSerena** · February 20th 2013, 05:21 AM

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?

**Tweety** · February 20th 2013, 05:37 AM

Originally Posted by ILikeSerena

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?

**ILikeSerena** · February 20th 2013, 07:08 AM

Originally Posted by Tweety

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).

**Tweety** · February 20th 2013, 11:38 AM

is it 0<t<2?

**ILikeSerena** · February 20th 2013, 11:40 AM

Originally Posted by Tweety

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?

**Tweety** · February 20th 2013, 11:56 AM

Originally Posted by ILikeSerena

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,

**ILikeSerena** · February 20th 2013, 11:59 AM

This is what you already had:

Originally Posted by Tweety

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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