Joseph HeavnerHonors Linear Algebra (MAT-208H-01)

Honors Assignment (Chapter 1.10 Problems)September 30, 2015

1.10.4:

[M] The Cambridge Diet supplies .8 of calcium per day, in addition to the nutrients listed in the Table1 for Example 1. The amounts of calcium per unit (100 g) supplied by the three ingredients in theCambridge Diet are as follows: 1.26 g from nonfat milk, .19 g from soy flour, and .8 g from whey.Another ingredient in the diet mixture is isolated soy protein, which provides the following nutrientsin each unit: .08 g of protein, 0 g of carbohydrate, 3.4 g of fat, and .18 g of calcium.

a.) Set up a matrix equation whose solution determines the amounts of nonfat milk, soy flour, whey,and isolated soy protein necessary to supply the precise amounts of protein, carbohydrate, fat, andcalcium in the Cambridge Diet. State what the variables in the equation represent.

b.) Solve the equation in (a) and discuss the answer.

a.) Let x1, x2, x3, x4 be the units of nonfat milk, soy flour, whey, and isolated soy protein. Then, we havethat:

36 51 13 8052 34 74 00 7 1.1 3.4

1.26 .19 .8 .18

x1x2x3x4

=

33453.8

b.) Utilizing a TI-83 calculator, or similar calculators such as Wolfram Alpha or a programming lan-guage such as MATLAB, to row-reduce yields:

36 51 13 80 3352 34 74 0 450 7 1.1 3.4 3

1.26 .19 .8 .18 .8

∼ · · · ∼

1 0 0 0 .640 1 0 0 .540 0 1 0 −.090 0 0 1 −.21

Therefore, this suggests the solution is x1 = .64, x2 = .54, x3 = −.09, and x4 = −.21, but this is not

a realistic solution, because one cannot have negative quantities of whey or isolated soy protein. Thus,the best one can do is mix .64 g of nonfat milk and .54 g of soy flour to get 50.6 g of protein, 51.6 g ofcarbohydrate, 3.8 g of fat, and 0.9 g of calcium. (Note that some of these values are very far from theirdesired values.)

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1.10.5:

[M] Write a matrix equation that determines the loop currents. If MATLAB or another matrix programis available, solve the system for the loop currents.

The resistance vector in the first, second, third, and fourth loops are, respectively:

R̃1 =

11−500

, R̃2 =

−510−10

, R̃3 =

0−19−2

, R̃4 =

00−210

Bringing that all together yields the matrix:

R =

11 −5 0 0−5 10 −1 00 −1 9 −20 0 −2 10

To determine the currents we need an equation of the form R~I = ~V where R is the resistance matrixabove, ~I is the desired current vector, and ~V is the voltage vector. The voltage vector can be found tobe, by inspection:

~V =

50−4030−30

Thus, we have that:

R~I = ~V ⇔

11 −5 0 0−5 10 −1 00 −1 9 −20 0 −2 10

I1I2I3I4

=

50−4030−30

We can solve this using computer assistance and the result is:

~I =

I1I2I3I4

=

3.68−1.902.57−2.49

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1.10.14:

[M] Study how changes in boundary temperatures on a steel plate affect the temperatures at the inte-rior of the plate.

a.) Begin by estimating the temperatures T1, T2, T3, T4 at each of the sets of four points on the steel plateshown in the figure. In each case, the value of Tk is approximated by the average of the temperaturesat the four closest points. See Exercises 33 and 34 in Section 1.1, where the values (in degrees) turn outto be (20, 27.5, 30, 22.5). How is this list of values related to your results for the points in set (a) and set(b)?

b.) Without making any computations, guess the interior temperatures in (a) when the boundarytemperatures are all multiplied by 3. Check your guess.

c.) Finally, make a general conjecture about the correspondence from the list of eight boundary tem-peratures to the list of four interior temperatures.

Due to the very lengthy nature of this problem in particular, I have taken the liberty of truncating thework and whatnot by a non-trivial amount to save time in writing this document. If clarification isneeded, feel free to ask.

a.) The following image is referenced:

The equations for (a) are:

4T1 = 0 + T2 + 20 + T4

4T2 = T1 + 20 + T3 + 04T3 = 0 + T2 + 20 + T4

4T4 = T1 + 0 + T3 + 20

Putting that in a matrix and row-reducing using software yields:1 0 0 0 100 1 0 0 100 0 1 0 100 0 0 1 10

Similarly, for (b) we have:

4T1 = 10 + T2 + 0 + T4

4T2 = T1 + 0 + T3 + 404T3 = 40 + T2 + 10 + T4

4T4 = T1 + 10 + T3 + 10

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Putting that in a matrix and row-reducing using software yields:1 0 0 0 100 1 0 0 17.50 0 1 0 200 0 0 1 12.5

So, we have solutions to all three of the plates. Simply aligning these solutions one will see that thesolution from Exercises 33 and 34 is the sum of those from (a) and (b). A related observation is that theboundary temperatures on the plates are related via a linear combination. In particular, the plate fromSection 1.1’s boundary temperatures are obtained by summing the boundary temperatures of (a) and(b).

b.) When the boundary temperatures are multiplied by a particular constant, say 3, then so are theinterior temperatures. This is hopefully intuitive enough.

c.) The correspondence is a linear transformation. To obtain the interior temperatures one solves anequation of the form Ax = b where A is given by the interior points’ arrangement and b ∈ R4 is givenby the boundary temperatures. We also have a superposition principle, as was hopefully revealedin part (a) of this question. A more thorough demonstration of this fact is possible but somewhatsuperfluous.

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