(a) If the distance of the flight is d, p the plane's speed (assumed constant), and w the wind speed (assumed constant) then, for the outward journey the net speed of the plane is v+w and for the return journey v-w. 36 mins=36/60=0.6hr.
d=(p+w)t1 and d=(p-w)t2 where t1=3hr, t2=3.6hr. Since the distance flown is the same 3(p+w)=3.6(p-w).
To solve algebraically: 3p+3w=3.6p-3.6w, (3.6-3)p=(3+3.6)w, 0.6p=6.6w, p=(6.6/0.6)w=11w.
Part (a) doesn't contain sufficient information to find the speeds of the plane and the wind; the best we could do was to relate the two speeds p=11w; nor does it help to determine the distance flown d. (A windspeed of 50mph implies a plane speed of 550mph and a distance of 1800 miles.)
More generally, d=(t2-t1)p=(t2+t1)w from which any one of the variables (including the distance) can be found, knowing the values of the others. This is a clue in producing matrix A.
(b) There are two destinations, Europe and Africa. If Shirley sends a total of 35 letters abroad, then:
E+A=35 where E is the number of European letters and A that for Africa.
The cost for sending the letters is 0.45E+0.65A=$18.55.
To solve algebraically: E=35-A, so 0.45(35-A)+0.65A=15.75-0.45A+0.65A=18.55, (0.65-0.45)A=18.55-15.75=2.8, 0.20A=2.8, A=2.8/0.2=14, so E=35-14=21.
To solve using matrices first write the equations E+A=35 and 0.45E+65A as one 2×2 matrix, call it V=
⎛ 1 1 ⎞
⎝ 0.45 0.65 ⎠
In this matrix the first column is the Europe postage information and the second column that for Africa.
The second matrix (call it Z) is a 2×1 matrix containing the variables, Z=
⎛ E ⎞
⎝ A ⎠
Call the product of these W=
⎛ 35 ⎞
⎝ 18.55 ⎠
So VZ=W, Z (which is what we want)=V-1W, for which we need the inverse of V.
Assume we don't have matrix calculator. Since V-1V=I, the identity matrix, we can write V-1Y=I as:
⎛ v11 v12 ⎞⎛ 1 1 ⎞= ⎛ 1 0 ⎞
⎝ v21 v22 ⎠⎝ 0.45 0.65 ⎠ ⎝ 0 1 ⎠
v11+0.45v12=1, v11+0.65v12=0⇒-0.65v12+0.45v12=1, -0.2v12=1, v12=-5⇒v11=3.25;
v21+0.45v22=0, v21+0.65v22=1⇒-0.45v22+0.65v22=1, 0.2v22=1, v22=5⇒v21=-2.25.
V-1W=
⎛ 3.25 -5 ⎞⎛ 35 ⎞
⎝ -2.25 5 ⎠⎝ 18.55 ⎠=
⎛ 21 ⎞
⎝ 14 ⎠
Therefore E=21 and A=14 (same as algebraic result).
21 letters were sent to Europe and 14 to Africa.
More generally, if pA=cost of postage to Africa and pE=cost of postage to Europe, and L=the total number of half-ounce letters: pE(L-A)+ApA=P (total postage), pEL-pEA+ApA=P, A(pA-pE)=P-pEL, A=(P-pEL)/(pA-pE); E=L-A=L-(P-pEL)/(pA-pE)=(pAL-P)/(pA-pE)=(P-pAL)/(pE-pA).
To express this situation using a matrix, it will be necessary to examine the expected final result for:
(AB)ij=k=1∑k=2aikbkj=[ai1b1j+ai2b2j]ij.
where i and j need to be determined for each matrix A and B.
So we need to relate the matrix (AB)ij to A (for part (a)) and B (for part (b)) and work out what this matrix is telling us in relation to the algebraic solutions of each part of the question. The clue is in the composition of each matrix element ai1b1j+ai2b2j. It's the sum of two products.
MATRIX A
Consider the matrix Y=
⎛ 1 1 ⎞
⎝ 1 -1 ⎠
and the 1×2 matrix X=[ p w ].
XY=[ p+w p-w ].
A=XY is a possible matrix for (a).
The matrix product A is a 1×2 (one row by two columns) matrix, the first element (a11) being the net speed on the outward flight (tail wind) and the second element (a12) the net speed on the return flight (head wind). So the product could be written [ nT nH ] for the net speeds in the outward (T=Tailwind) and return (H=Headwind) flights . Therefore a11=nT=p+w, a12=nH=p-w. In this case i=1 and j=1 or 2.
This is just an example of a couple of matrices. Other matrices will deliver different results. The elements of matrix A could, for instance, be the time of the outward flight and time of the return flight (which would bring in the distance d and utilise the net speeds, tT=d/(v+w) and tH=d/(v-w), that is, tT=d/nT, tH=d/nH). The most suitable matrix A would be the one that produces the most relevant meaning to the matrix product AB.
MATRIX B
Unlike part (a), we have determined the individual number of letters Shirley sends to Europe (21) and Africa (14).
Matrix B could simply be [ 21 4 ], the number of letters sent to each continent. Matrix B could also be what was called V in the above solution for (b).
MATRIX PRODUCT AB
There must be as many rows in B as there are columns in A, otherwise multiplication is not possible. The matrix product must have as many rows as A and as many columns as B. Since k has only two values, 1 and 2, then B must have 2 rows and A must have 2 columns. If A has i rows then the product also has i rows, and B and the product both have j columns. If B is identified as V (extracted from (b)) then j=2. If A is identified as Y (extracted from (a)---see under MATRIX A), then the product is:
⎛ 1 1 ⎞⎛ 1 1 ⎞
⎝ 1 -1 ⎠⎝ 0.45 0.65 ⎠=
⎛ 1.45 1.65 ⎞
⎝ 0.55 0.35 ⎠
This product doesn't seem to have any meaning in itself, but, as stated before, it's not clear which matrices are to be extracted from (a) and (b).
