Concrete applicaiton (part 2). This problem follows up on Problem 6. Below is the table that describes the composition of the three basic mixtures of concrete, Type S, Type A, and Type L. The corresponding vectors for these mixtures and for two custom mixtures, Type U and Type V have also been entered below.

	Super-Strong Type S	All-Purpose Type A	Long-Life Type L	Type U	Type V
Cement	20	18	12	12	15
Water	10	10	10	12	10
Sand	20	25	15	12	20
Gravel	10	5	15	12	10
Fly ash	0	2	8	12	5

Let
S=[20,10,20,10,0].
A=[18,10,25,5,2]
L=[12,10,15,15,8]
U=[12,12,12,12,12]
V=[15,10,20,10,5]

Part A: Show that we can make the custom mix V but not the custom mix U from the three basic mixes S,A,L. Explain why just this calculation shows that {S, A, L, V} is not linearly independent, but does not (yet) say anything about the linearly independence of {S, A, L, U}.

Part B: Explain why, in general, any combination of S,A,L and V can also be achieved by a combination of just S,A, and L. In addition, show how to make the specific custom mix 3S +4A +2L +3V using only S,A, and L.

Part C: Show that {S,A,L,U} is a linearly independent set of vectors. What practical advantage does this have?

Part D: 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} (and show this). What does this say about the span of {S,A,L,U,W}?

Part E: Why will there still be real-life mixes that cannot ever be physically produced from this set of five basic mixes? Give an example where this happens - where the real-life mix has non-negative entries, but is impossible to make from {S,A,L,U,W}.