The director of trust fund has $ 100000 to invest. The rules of the trust state that both a certificate of deposit (CD) and a long-term bond must be used. The director’s goalis to have he trust yield $7800 on its investments for the year. The CD chosen returns 5% per annum and the bond 9%. The director determines the amount x to invest in the CD and the amount y to invest in the bond as follows. Since the total investment is $100000, we must have x + y = 100,000. Since the desired return is $8700, we obtain the equation 0.05x + 0.09y = 7800. Thus, we have the linear system x + y = 100000 (3) 0.05x + 0.09y = 7800.To eliminate x, we add(-0.05) times the first equation to the second, obtaining x + y = 100000 0.04 y = 2800, where the second equation has no x term. We have eliminated the unknown x . Then solving for y in the second equation, we have y = 70000, and substituting y into the first equation of (3), we obtain x = 30000 .To check that x = 30000, y = 70000 is a solution to (3), we verify that these values of x and y satisfy each of the equations in the given linear system. Thus, the director of the trust should invest $ 30000 in the CD and $70000 in the long-term bond.Consider the linear system x – 3y = -7 (4) 2x – 6y = 7.Again, we decide to eliminate x. We add (-2) times the first equation to the second one, obtaining x – 3y = -7 0x – 0y = 21Whose second equation makes no sense. This means that the linear system (4) has no solution. We might have come to the same conclusion from observing that in (4) the left side of the second equation is twice the left side of the first equation, but the right side of the second equation is not twice the right side of the first equation. Consider the linear system x + 2y +3z = 6 2x - 3y +2z = 14 (5) 3x + y - z = -2To eliminate x, we add (-2) times the first equation to the second one and (-3) times the first equation to the third one, obtainingx + 2y +3z = 6 -7y - 4z = 2 (6) -5y -10z = -20We next eliminate y from the second equation in (6) as follows, Multiply the third equation of (6) by (-1/5), obtaining x + 2y +3z = 6 -7y - 4z = 2 y +2z = 4Next we interchange the second and third equations to give x + 2y +3z = 6 y + 2z = 4 (7) -7y -4z = 2We now add 7 times the second equation to the third one, to obtainx + 2y +3z = 6 y +2z = 4 10z = 30Multiplying the third equation by 1/10, we have x + 2y +3z = 6 y+ 2z = 4 (8) z = 3Substituting z = 3 into the second equation of (8), we find y = -2. Substituting these values of z and y into the first equation of (8), we have x =1. To check that x=1,y=-2,z=3 is a solution to (5), we verify that these values of x, solution to the linear system (5). The importance of the procedure lies in the fact that the linear systems (5) and (8) have exactly the same solutions. System (8) has the advantage that it can be solved quite sassily, given the foregoing values for x, y and z.Consider the linear system x+2y-3z=-4 2x+y -3z= 4 (9)Eliminating x, we add (-2) times the first equation to the second one, to obtain x+2y-3z=-4 -3y +3z= 12 (10)Solving the second equation in (10) for y, we obtain y=z-4,where z can be any real number. Then, from the first equation of (10) , x=-4-2y+3z=-4-2(z-4)+3z=z+4thus a solution to the linear system (9) is x=r+4, y= r-4, z=r,where r is any real number. This means that the linear system (9) has infinitely many solutions. Every time we assign a value to r, we obtain another solution to (9). Thus, if r=1,then x=5, y=-3 and z=1 is a solution, while if r=-2, then x=2, y=-6 and z=-2 is another solution.Consider the linear system x+2y=10 2x-2y=-4 (11) 3x+5y=20To eliminate x, we add(-2) times the first equation to the second one and (-3) times the first equation to the third one, to obtain x+2y=10 -6y=-24 -y=-10Multiplying the second equation by (-1/6) and the third one by (-1), we have the system x+2y=10 y=4 (12) y=10 ,which has no solution. Since (14) and (13) have the same solutions, we conclude that (13) has no solutions. These examples suggest that a linear system may have one solution (a unique solution), no solution, or infinitely many solutions.We have seen that the method of elimination consists of repeatedly perｆｏｒｍing the following operations:1. Interchange two equations.2. Multiply an equation by a nonzero constant.3. Add a multiple of one equation to another. It is not difficult to show(Exercises T.1 through T.3) that the method of elimination yields another linear system having exactly the same solutions as the given system. The new linear system can then be solved quite readily. As you have probably already observed, the method of elimination has been described, so far, in general terms. Thus we have not indicated any rules for selecting the unknowns to be eliminated. Before providing a systematic de脚本ion of the method of elimination, we introduce, in the next section, the notion of a matrix, which will greatly simplify our notation and will enable us to develop tools to solve many important problems.Consider a linear system of three equations in the unknowns x, y and z: (13)The graph of each of these equations is a plane, denoted by , and , respectively. As in the case of a linear system of two equations in two unknowns, the linear system in (16) can have a unique solution, no solution, or infinitely many solutions. These situations are illustrated in Figure 1.1. For a more concrete illustration of some of the possible cases, the walls(planes) of a room intersect in a unique point, a corner of the room, so the linear system has a unique solution. Next, think of the planes as pages of a book. Three pages of a book (when held open ) intersect in a straight line, the spine. Thus, the linear system has infinitely many solutions. On the other hand, when the book is closed, three pages of a book appear to be parallel and do not intersect, so the linear system has no solution.(Production Planning) A manufacturer makes three different 无效s of chemical products: A, B and C. Each product must go through two processing machines: X and Y. The products require the following times in machines X and Y:1. One ton of A requires 2 hours in machine X and 2 hours in machine Y.2. One ton of A requires 3 hours in machine X and 2 hours in machine Y.3. One ton of A requires 4 hours in machine X and 3 hours in machine Y.Machine X is available 80 hours per week and machine Y is available 60 hours per week. Since management does not want to keep the expensive machines X and Y idle, it would like to know how many tons of each product to make so that the machines are fully utilized. It is assumed that the manufacturer can sell as much of the products as is made. To solve this problem, we let , and denote the number of tons of products A, B and C, respectively, to be made. The number of hours that machine X will be used is 2 +3 +4 , which must equal 80. Thus we have 2 +3 +4 =80. Similarly, the number of hour that machine Y will be used is 60, so we have 2 +2 +3 =60.Mathematically, our problem is to find nonnegative values of , and so that 2 +3 +4 =80 2 +2 +3 =60. This linear system has infinitely many solutions. Following the method of Example 4, we see that all solutions are given by = =20- =any real number such that , since we must have , and . When =10, we have =5, =10, =-4 while =13/2, =13, and =7 when =7. The reader should observe that solution is just as good as the other. There is no best solution unless additional inｆｏｒｍation or restrictions are given.
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