National Universit y of Science National Sciences s & Technology School of Ele Electri ctri cal Engineering Engineering and Computer Science Department De partment of Computin g MATH-12 MATH -121: 1: Linea Lin earr Alg ebra &Ordinary Dif ferenti al Equation s (3+0): (3+0): BEE-8 BEE-8ABC ABC Fall Fall 2016 2016 Assi As si gnmen gn mentt 1 CLO: NA Maximu m Ma Maximu Marks: rks: 80 Anno An noun un cem cement ent Date: 14 th Octo October ber 201 2016 6
Instructo r: Dr. Na Naila ila Amir st Due Date: 21 October 2016
Instructions: 1. Understanding Understanding t he questio questio n is part of t he assign assign ment and copying is not allowed. 2. It It is a group assig nment. Tasks: Tasks: Attempt all questions . Q - 1: [10 marks ] A nut n ut di st ri buto bu torr w ants ant s to t o kno k now w t he nut n utri ri tion ti onal al con c onten tentt of o f vari v ariou ous s m ix tures tu res of alm onds on ds , cashews, and pecans. Her Her supplier has provid ed the following nutri tion info rmation:
Her Her fir st mi xture, a protein bl end, consists of 6 cups of almonds, 3 cups of cashews, and and 1 cup of pecans. Her Her second mixtur e, a low fat mix, consist s of 3 cups of almon ds, 6 cups of cashews, and 1 cup of pecans. Her Her third mi xture, a low carb mix consi sts of 3 cups of almonds, 1 cup o f cashews, and 6 cups of pecans. Dete Determine rmine the amount of pr otein, carbs, and fats in a 1 cup serving of each each of the mixtur es. Solution: It makes makes sense to put t he first gro up of data into a matrix wit h Almond s, Cashews, Cashews, and Peca Pecans ns as columns , and and then put th e second second gr oup of d ata into a matrix wi th info rmation about Almonds , Cashe Cashews, ws, and and Pecans Pecans as as rows. This way the column s of the first matri x lines up with the rows of the second second matrix, and we can can perform matrix mult iplic ation. This way way we get rid of the number of c ups o f Almon ds, Cashews, Cashews, and Peca Pecans ns , which we don’t need. So here is the information we have in table/matrix table/matrix f orm:
Then we can multi ply the matric es since we want to end up with the amount of Protein, Carbs, and Fat in each of the mixtures. The produc t of the matrices consist s of rows of Protein, Carbs, and Fat, and column s of the Protein, Low Fat, and Low Carb mixtu res:
But we have to be careful, since these amounts are for 10 cups. Also, notice how t he cups unit “canceled out” when we did the matrix multiplication (grams/cup time cups = grams). Therefore, to get the answers, we have to di vide each answer by 10 to get grams p er cup. So the numbers in bo ld are our answers:
Q - 2: [10 marks ] An ou tbreak of Chicken Pox h it the local publ ic school s. Appr ox im ately 15% of th e male and female juniors and 25% of the male and female seniors are currently healthy, 35% of the male and female juniors and 30% of th e male and f emale seniors are currently s ick, and 50% of the male and female juniors and 45% of th e male and f emale seniors are carriers of Chicken Pox. There are 100 male juni ors , 80 male senior s, 120 female jun ior s, and 100 female senio rs. Using two m atrices and one matrix equation, find out how m any males and how many females (don’t need to divi de by cl ass) are healthy, si ck, and carriers. Solution: The best way to approach these types of probl ems is to s et up a few manual calculations and see what we’re doing. For example, to find o ut how many healthy m ales we would have, we’d set up the following equation and do the calcu latio n: .15(100) + .25(80) = 35. Lik ewise, to find out how many females are carr iers , we can calc ulate: .50(120) + .45(100) = 105. We can tell that this looks l ike matrix multi plic ation. And since we want to end up with a matrix t hat has males and females by healthy, sick and carriers, we know it will be either a 2 x 3 or a 3 x 2. But si nce we know th at we have both j unior s and seniors wi th males and females, the first matri x will pro bably be a 2 x 2. That means, in order to do matrix mu ltip lication, the second matrix that holds the %’s of students will have to be a 2 x 3, since there are 3 types of students, healthy (H), sick (S), and carriers (C). Notice how th e percentages in the rows in the second matrix add up to 100%. Also noti ce that if we add up the number of st udents in the firs t matrix and t he last matri x, we come up with 400.
For the present case we have follow ing:
So there wi ll be 35 healthy m ales, 59 sic k males, and 86 carri er males, 43 healthy f emales, 72 sick females, and 95 carrier females. Q - 3: [20 marks ] Solve the follow ing lin ear system using Gauss-Jordan elimination: 2x1 + 5x2 - 8x3 + 2x4 + 2x5 = 0 6x1 + 2x2 -10x3 + 6x4 + 8x5 = 6 3x1 + 6x2 + 2x3 + 3x4 + 5x 5 = 6 3x1 + 1x2 - 5x3 + 3x4 + 4x5 = 3 6x1 + 7x2 - 3x3 + 6x4 + 9x5 = 9 What wil l be rank for th e above matrix? Solution:
Q - 4: [10 marks ] For which values of will the following system have no solutions? Exactly one solution? Infinitely many solu tions? +−= −+= + + ( − ) = +
Solution:
Q - 5: [20 marks ] Determine whether the foll owing are vector spaces or not? J ustif y your answers. () = {(, ) ∈ ℝ : = + } () = {(, ) ∈ ℝ : ≥ , ≥ } () = {(,,) ∈ ℝ : = = −}
Solution:
Q - 6: [10 marks ] Consider the vector s pace ℙ -- the space of polyno mials with d egree less than or equal to 4 . Let ’s consider the following four vectors in this vector space: = − , = − + − ,
= − + , = − + − + .
Check whether the above set of vector s for m basis fo r ℙ or not? Solution: The set of vectors fr om a vector s pace V is called basis if it is: Linearly independent and It is a spanning s et for V. . Spanning set:
To check that these vectors fo rm a spanning set for ℙ , we should t ake an arbitrary vector from ℙ and try to express it as a linear comb ination of t he vectors from t he basis. Let's take an arbitrary pol ynomial = + + + + ∈ ℙ , such t hat w c1u1
ax
4
c2u 2
bx
3
c3u 3 c 4u 4
cx
2
dx
2
c1 ( x 2) c2 x
c4 x
4
(1)
e
4 x 4 c3 x 3 6 x 2
12 x 8 c4 x 4
8 x 3 24 x 2
32 x 16
8c4 c3 x3 24c4 6c3 c2 x 2 32c4 12c3 4c2 c1 x 16c4 8c3 4c2 2c1
Comparing the coefficients of the powers of x on both si des we get the followi ng system of equations:
a
b
c
d e
c4
8c4
c3
24c4
6c3
32c4
112a
8a
c3
c2
12c3
8b
4c
4c2
2d
c1
By solvi ng the above system, we get the follow ing sol ution: c4
a
c3
b
8a
c2
c
24a
6b
d
32a
12b 4c
c1
Note that the values of the unknow n quantities c1 , c2 , c3 and c4 do not satisfy equation (1) i.e., the vecto r cannot be expressed as a linear combination of t he given vectors. Thus, we deduce, that this s ystem of vectors is not a spanning set. Since one of the condit ions f or basis is no t satisfi ed, therefore, we concl ude that the given set of vectors do not f orm basis for ℙ .
