Problem Set 2

See the Guidelines. I will post on ASULearn answers to select questions I receive via messaging there or in office hours. I am always happy to help!

Mathematics, you see, is not a spectator sport. [George Polya, How to Solve it]


Problem 1:  1.3 #30
Note that this asks about general vectors, so do not define the vectors to be ones with specific numbers - leave them general. Do use the general definition of center of gravity as written in #29 and rewrite it as a linear combination. The definition already gives you a version very close to that, but you'll need to do a tiny bit of algebra of vectors and solve for the c_i's.

Problem 2: 
Look at Vector([1,-2,3]) and let L=Span of (1,-2,3) = {t (1,-2,3) where t is real}. Notice L is a line through the origin in R3 (1 free variable) and we can graph the vector in Maple (without the arrowhead) as:
with(LinearAlgebra): with(plots):
a:=spacecurve([t,-2*t,3*t],t=0..1,color=red,linestyle=solid):
display(a);
Part A: Find a vector w1 so that {Vector([1,-2,3]), w1} spans a plane P1 (and not just the [t,-2t,3t] line) and describe how you chose it.

Part B: Set up the augmented matrix resulting from the spanning equation (with the vector (b1, b2, b3) as the equals column) and use Gaussian in Maple to find an equation involving b1, b2 and b3 that represents the plane that Vector([1,-2,3]) and w1 lie in (set any inconsistencies equal to 0).
AugmentedPr2:=Matrix([Vector([1,-2,3]),Vector([your vector w1]),Vector([b1,b2,b3])]);
GaussianElimination(AugmentedPr2);

Part C: Find a vector w2 not on the same line through the origin as w1 so that {Vector([1,-2,3]), w2} also spans the same plane P1 (linear combinations are in the same plane) and describe how you chose w2 (Hint: the diagonal of the parallelogram would be a natural choice).

Part D: In Maple, use a spacecurve command for each vector, like
a:=spacecurve([t,-2,3*t],t=0..1,color=red, linestyle=solid): to show that all three vectors lie in the same plane but no 2 are on the same line. (Hint: you can use different colors like black, blue, green..., and one display command like display(a,b,c); and turn the plane "head on" to show this-just be sure you have executed with(plots): Another option for plotting is: spacecurve({[4*t,7*t,3*t,t=0..1],[-1*t,2*t,6*t,t=0..1]}, color=red, linestyle=solid); )

Part E: Describe all the vectors w for which {Vector([1,-2,3]), w} spans the same plane P1 as a linear combination
a Vector([1,-2,3]) + b w1 where you specify a non-zero condition on a or b or both (Hint: which is it? - linear combinations are in the same geometric space so think about what a's and b's in the linear combinations you can use that will still span the plane with Vector([1,-2,3]) - and not just the the [t,-2*t,3*t] line - something can't be 0 in the linear combination - specify and annotate).

Part F: Find a vector u so that {Vector([1,-2,3]), u} spans a different plane P2 through the origin. Describe how you chose u.

Part G: Create a spacecurve graph for u in Maple, then display all 4 vectors on one graph, like display(a,b,c,d); and turn the other vectors "head on" to show u lies outside the plane P1.

Problem 3: 
Concrete mix, which is used in jobs as varied as making sidewalks and building bridges, is composed of five main materials: cement, water, sand, gravel, and fly ash. By varying the percentages of these materials, mixes of concrete can be produced with differing characteristics. For example, the water-to-cement ratio affects the strength of the final mix, the sand-to-gravel ratio affects the "workability" of the mix, and the fly-ash-to-cement ratio affects the durability. Since different jobs require concrete with different characteristics, it is important to be able to produce custom mixes.

Assume you are the manager of a building supply company and plan to keep on hand three basic mixes of concrete from which you will formulate custom mixes for your customers. The basic mixes have the following characteristics:
  Super-Strong Type S   All-Purpose Type A   Long-Life Type L  
Cement 20 1812
Water 10 1010
Sand 20 2515
Gravel 10 515
Fly ash 0 28
Each measuring scoop of any mix weighs 60g, and the numbers in the table above give the breakdown by grams of the components of the mix. Custom mixes are made by combining the three basic mixes. For example, a custom mix might have 10 scoops of Type S, 14 of Type A, and 7.5 of Type L. We can represent any mixture by a vector [c,w,s,g,f] in R5 representing the amounts of cement, water, sand, gravel, and fly ash in the final mix.

Part A: Compute 3S+5A+2L.

Part B: Compare the strength of 3S+5A+2L to that of Types S, A and L - rank them in order of strength (low water to cement ratio). (Note that this is one example of a practical interpretation of vector coordinates).

Part C: What does Span{S,A,L} = {a S + b A + c L where a, b, and c are real numbers} represent in this context - relate your answer to the word mixes?

Part D: A customer requests a custom mix with the following proportions of cement, water, sand, gravel, and fly ash: 16:10:21:9:4 . Is it possible to make using only S, A and L? If so, find the weights (in scoops/parts of scoops) of the basic mixes (S, A, and L) needed to create this mix. If you choose to do this by-hand, don't forget to show the by-hand work, like row reduction. If you choose to use Maple, you can use commands like:
S := Vector([20,10,20,10,0]):
A := Vector([18,10,25,5,2]):
L := Vector([12,10,15,15,8]):
V := Vector([16,10,21,9,4]);
N := Matrix([S,A,L,V]); ReducedRowEchelonForm(N);

Part E: If there is a solution in Part D, is the solution unique? Explain.

Part F: Let V=Vector([16,10,21,9,4]). Explain why any linear combination of S, A, L and V can also be achieved by a combination of just S,A, and L, ie that V is redundant (think about substitution and using your answer from part D).

Part G: What is the definition of linear independence using vectors? Using a matrix with the vectors as its columns?

Part H: Is S, A, L, V linearly independent? Why or why not?

Part I: Let U=Vector([12,12,12,12,12]). Show that {S,A,L,U} is a linearly independent set of vectors by using the definition of l.i. If you choose to do this by hand, then show your reduction work and steps. If you choose to use Maple, and have already defined the vectors S, A and L, as in the sample Maple commands in part D, then you can use commands like:
U := Vector([12,12,12,12,12]): zero:= Vector([0,0,0,0,0]):
SALUzero:=Matrix([S,A,L,U,zero]); ReducedRowEchelonForm(SALUzero);

Part J: What practical advantage does {S,A,L,U} being linearly independent have?

Part K: Compute the mix S+A+L+U.

Part L: Next modify the first entry of the vector S+A+L+U to define a fifth basic mix W to add to {S,A,L,U} such that any custom mixture can be expressed as a linear combination of the set of mixes {S,A,L,U,W}. Set up the augmented system {S,A,L,U,W, generic vector} and use GaussianElimination in Maple to show that your vectors span R5
Once you define W in Maple using a Vector command, you can execute:
SALUWspan:=Matrix([S,A,L,U,W,Vector([b1,b2,b3,b4,b5])]); GaussianElimination(SALUWspan);

Part M: In real-life, there will still be mixes that cannot be physically produced from this set of five basic mixes. Give an example of a custom mix where at least one weight (of S,A,L,U,W) is negative and in your annotations address why we can't use negative weights in mixing cement in real-life.

A Review of Various Maple Commands:
> with(LinearAlgebra): with(plots):
> A:=Matrix([[-1,2,1,-1],[2,4,-7,-8],[4,7,-3,3]]);
> ReducedRowEchelonForm(A);
> GaussianElimination(A); (only for augmented matrices with unknown variables like k or a, b, c in the augmented matrix)
> Vector([1,2,3]);
> A.B;
> A+B;
> B-A;
> 3*A;
> A^3;
> evalf(M) (decimal approximation of M)
> spacecurve({[4*t,7*t,3*t,t=0..1],[-1*t,2*t,6*t,t=0..1]},color=red, linestyle=solid); plot vectors as line segments in R3 (ie the columns of matrices) to show whether the the columns are in the same plane, etc.
> implicitplot({2*x+4*y-2,5*x-3*y-1}, x=-1..1, y=-1..1);
> implicitplot3d({x+2*y+3*z-3,2*x-y-4*z-1,x+y+z-2},x=-4..4,y=-4..4,z=-4..4); plot equations of planes in R^3 (rows of augmented matrices) to look at the geometry of the intersection of the rows (ie 3 planes intersect in a point, a line, a plane, or no common points)
> display(a,b,c); if plots are set to variables (careful to end those with a colon instead of semi-colon) like a:=implicitplot(...): then this will display them on the same plot