Solution manual for linear algebra for engineers and scientists using matlab by hardy

Solution Manual for Linear Algebra for Engineers and Scientists Using Matlab by Hardy Full file at https://TestbankDirect.eu/ Chapter Linear Systems 1.1 Solving Linear Systems (x + 2)2 + (y − 1)2 = x2 + 4x + + y − 2y + = + x2 + y2 ⇒ 4x − 2y = −2 ⇒ 2x − y = −1 Linear (x − y)2 = x2 − 2xy + y = x2 + y2 ⇒ −2xy = ⇒ xy = Nonlinear (x − y)(x + y) = x2 + xy − xy − y = x2 + y2 ⇒ 2y = ⇒ y = Linear x + (x − 1)(y − 1) = x + xy − x − y + = xy + y − ⇒ −2y2 = −2 ⇒ y = Linear Pivot variables: x1 , x2 , x3 Free variable: x4 No variables uniquely determined Let x4 = t, where t is a real parameter Back-substituting, third equation: x3 = − 2t, second equation: x2 = −1+x3 −2x4 = −1+(3−2t)−2t = 2−4t, first equation: x1 = 2x2 +10x4 = 2(2−4t)+10t = 4+2t Solution: (x1 , x2 , x3 , x4 ) = (4 + 2t, − 4t, − 2t, t) Pivot variables: x1 , x2 , x4 Free variable: x3 Uniquely determined: x4 Let x3 = t, where t is a real parameter Back-substituting, third equation: x4 = 2.4, second equation: x2 = + x3 − 7x4 = 1+t−7(2.4) = −15.8+t, first equation: x1 = 2−2x2 −x3 +x4 = 2−2(−15.8+t)−t+2.4 = 36−3t Solution: (x1 , x2 , x3 , x4 ) = (36 − 3t, −15.8 + t, t, 2.4) Pivot variables: x1 , x3 Free variables: x2 , x4 No variables uniquely determined Let x2 = s and x4 = t, where s, t are real parameters Back-substituting, second equation: x3 = −1.5 + 0.5t, first equation: x1 = − 0.5s Solution: (x1 , x2 , x3 , x4 ) = (4 − 0.5s, s, −1.5 + 0.5t, t) Pivot variables: x1 , x3 , x4 Free variables: x2 , x5 Uniquely determined: x3 Let x2 = s and x5 = t, where s, t are real parameters Back-substituting, third equation: x4 = 2.0 − t, second Full file at https://TestbankDirect.eu/ Solution Manual for Linear Algebra for Engineers and Scientists Using Matlab by Hardy Full file at https://TestbankDirect.eu/ ISM Chapter Linear Systems equation: x3 = 2.1, first equation: x1 = 7.0+ x2 −2x3 −x4 = 7.0+s−2(2.1)− (2.0− t) = 0.8+s+t Solution: (x1 , x2 , x3 , x4 , x5 ) = (0.8 + s + t, s, 2.1, 2.0 − t, t) 3x1 = 6.0 E1 L1 x1 − x2 = 5.5 E2 L2 −2 Solve E1 : x1 = 2.0 Solve E2 : −x2 = 5.5 − x1 = 3.5 ⇒ x2 = −3.5 Solution: (x1 , x2 ) = (2, −3.5) −3.5 −4 Refer to Figure 1.1 [9] Equations E1 , E2 are represented, respectively, by lines L1 , L2 in R2 −6 −1 Figure 1.1 [9] 10 First equation: x1 = 32 , second equation: x2 = −x1 = − 23 , third equation: x3 = − 3x1 + 3x2 = − − = −3 Solution: (x1 , x2 , x3 ) = ( 32 , − 32 , −3) Geometrically, the first equation is represented by a plane P1 in R3 parallel to the x2 x3 -plane, the second equation is represented by a plane P2 passing through the x3 -axis, perpendicular to the x1 x2 -plane, the third equation is represented by a plane P3 that intercepts the x1 -, x2 -, x3 -axes at x1 = 13 , x2 = − 13 , x3 = 1, respectively Planes P1 , P2 intersect in a line L and L intersects P3 in the point ( 23 , − 23 , −3) 11  −1    Let (S) denote the linear system M = [A| b] =   In order to preserve 10 −4 integer coefficients, perform the following EROs on M: r2 − 2r1 → r2 , r3 − 3r1 → r3 , r1 ↔ r3 , r3 − 2r1 → r3 , r3 − r2 → r3 , to obtain an echelon form U for M and a linear system (S)′ equivalent to (S)    x + x2 − x3 = 1 1 −1     x2 + 3x3 = (S)′ 1 = U M s 0   0 −2 −2 − 2x3 = −2 Back-substituting, third equation: x3 = 1, second equation: x2 = − 3x3 = −2, first equation: x1 = − x2 + x3 = Solution: (x1 , x2 , x3 ) = (4, −2, 1) (S) is represented by three planes in R3 with a single point in common 12  3 −4.5    Let (S) denote the linear system M = [A| b] =  1 0.5  Perform the following −2 −2 5.0 EROs on M: r1 ↔ r2 , r2 − 3r1 → r2 , r3 + 2r1 → r3 , r3 + r2 → r3 , to obtain an echelon form U Full file at https://TestbankDirect.eu/ Solution Manual for Linear Algebra for Engineers and Scientists Using Matlab by Hardy Full file at https://TestbankDirect.eu/ ISM 1.1 for M and a linear system (S)′ equivalent to (S)   1 0.5   M s  0 −2 −6.0  = U 0 0 Solving Linear Systems  x + x2 + x3 = 0.5   (S)′ − 2x3 = −6.0 ,   0x1 + 0x2 + 0x3 = the zero equation occuring in (S)′ because some equation in (S) is a linear combination of other equations in (S) Pivot variables: x1 , x3 Free: x2 Uniquely determined: x3 Let x2 = t be a real parameter Back-substituting, second equation: x3 = 3.0, first equation: x1 = 0.5 − x2 − x3 = −2.5 − t Solution: (x1 , x2 , x3 ) = (−2.5 − t, t, 3.0) (S) is represented by three planes in R3 that intersect in a line 13   Let (S) denote the linear system M = [A| b] =    13  Perform the following −3 −2 EROs on M: r2 − 3r1 → r2 , r3 − 4r1 → r3 , r3 − r2 → r3 , to obtain an echelon form U for M and a linear system (S)′ equivalent to (S)    x + 2x2 − 3x3 = 1 −3     ′ (S) 10  = U 10x3 = 10 M s  0 10   0 −5 0x1 + 0x2 + 0x3 = − E3 Equation E3 in (S)′ indicates that (S) is inconsistent because (S)′ s (S) The solution set is empty (S) is represented by three planes in R3 that have no point in common 14 −3 −5 Perform the following 1 EROs on M: r1 ↔ r2 , to obtain an echelon form U for M and a linear system (S)′ equivalent to (S) x1 + 3x2 = 1 =U (S)′ Ms −5 −3 x2 − 5x3 = −3 Let (S) denote the linear system M = [A| b] = Pivot variables: x1 , x2 Free variables: x3 Let x3 = t be a real parameter Back-substituting, second equation: x2 = −3 + 5t, first equation: x1 = − 3x2 = − 3(−3 + 5t) = 10 − 15t Solution: (x1 , x2 , x3 ) = (10 − 15t, −3 + 5t, t) (S) is represented by two (nonparallel) planes in R3 that intersect in a line 15  3  Let (S) denote the linear system M = [A| b] =  1 −2  3   In order to preserve 2 11 integer coefficients, perform the following EROs on M: r1 ↔ r3 , r2 − 3r1 → r2 , r3 − 5r1 → r3 , r4 − 9r1 → r4 , r3 − 7r2 → r3 , r4 − 8r2 → r4 , r4 − r3 → r4 , to obtain an echelon form U for M Full file at https://TestbankDirect.eu/ Solution Manual for Linear Algebra for Engineers and Scientists Using Matlab by Hardy Full file at https://TestbankDirect.eu/ ISM Chapter Linear Systems and a linear system (S)′ equivalent to (S)  1  −1 −2  Ms 0 0 0  −3   =U 13  (S)′  x1 + x2 + 3x3 =      − x2 − 2x3 = −  0x1 + 0x2 + 0x3 =     0x1 + 0x2 + 0x3 = 13 E3 E4 Equation E4 in (S)′ shows that some equation in (S) is a linear combination of other equations in (S) and E3 indicates that (S) is inconsistent because (S)′ s (S) (S) is represented by four planes in R3 with no point in common 16  −1 1    −1  Perform the following Let (S) denote the linear system M = [A| b] =  2 −1 −2 EROs on M: r2 − 2r1 → r2 , r3 + r1 → r3 , r3 − 35 r2 → r3 , to obtain an echelon form U for M and a linear system (S)′ equivalent to (S)    x1 − x2 + x3 = 1 −1 1     ′ 5x2 − 3x3 = (S) M s 0 −3 2 = U   21 21 21 21 0 5 x3 = Pivot variables: x1 , x2 , x3 Free variables: none All variables uniquely determined Backsubstituting, third equation: x3 = 1, second equation: x2 = 51 (2 + 3x3 ) = 15 = 1, first equation: x1 = + x2 − x3 = + − = Solution: (x1 , x2 , x3 ) = (1, 1, 1) (S) is represented by three planes in R3 with a single point in common 17 Let (S) denote the linear system Then (S) is × and will be either inconsistent or consistent with at least one free variable, indicating infinitely many solutions Finding the reduced form M* of M = [A| b], we have     −4 1 0     M =  −1 −1  s  −1 1  = M* 1 −1 0 0 Pivot variables: x1 , x2 Free variables: x3 , x4 Uniquely determined: x1 Let x3 = s and x4 = t, where s, t are real parameters Then, reading the solution off from M*, we have x1 = 1, x2 = + s − t Solution: (x1 , x2 , x3 , x4 ) = (1, + s − t, s, t) 18 Let (S) denote the linear system Then (S) is × and any one of the three possibilities for solution may occur Finding the reduced form M* of M = [A| b], we have     −2 0  −5  6 0   0  M= s  = M* 1 2 4 0 2 Full file at https://TestbankDirect.eu/ 0 0 Solution Manual for Linear Algebra for Engineers and Scientists Using Matlab by Hardy Full file at https://TestbankDirect.eu/ ISM 1.1 Solving Linear Systems Pivot variables: x1 , x2 , x3 Free variables: none All variables uniquely determined Reading the solution off from M*, we have x1 = x2 = 0, x3 = Solution: (x1 , x2 , x3 ) = (0, 0, 2) 19 Let (S) denote the linear system Then (S) is × and will be either inconsistent or consistent with at least one free variable, indicating infinitely many solutions Finding the reduced form M* of M = [A| b], we have M= −1 −1 −4 1 −1 s − 21 1 − 32 11 −1 = M* Pivot variables: x1 , x2 Free variables: x3 , x4 Uniquely determined: none Let x3 = s and x4 = t, where s, t are real parameters Then, reading the solution off from M*, we have x1 = 5.5+0.5s+t, x2 = 2.5 + 1.5t Solution: (x1 , x2 , x3 ) = (5.5 + 0.5s + t, 2.5 + 1.5s, s, t) 20 Let (S) denote the linear system Then (S) is × and will be either inconsistent or consistent with at least one free variable, indicating infinitely many solutions Applying forward reduction on M = [A| b] using the sequence of EROs shown below: Clear below in column r2 − 2r1 → r2 , r4 − r1 → r4 Clear below in column r2 ↔ r3 , r3 − 2r2 → r3 , r4 + r2 → r4 Clear below in column − 14 r3 → r3 , r4 + 2r3 → r4 and we have  2  M= 0 2 2 −1 −1 −2 −1   10   14   s −1   0  −1   =U 1 −1 0 0 0 −6 and the last row in U indicates that (S) is inconsistent 21 Finding the reduced form M* of M = [A| b], we have    −2    M = 1 −2 4 s 0 −5 0 −0.6 −0.2 −1 2.8   −0.6  = M* 2.4 Pivot variables: x1 , x2 , x3 Free variable: x4 Let x4 = t, where t is a real parameter Then, reading the solution off from M*, we have x1 = 2.8 + 0.6t, x2 = −0.6 + 0.2t General solution: (x1 , x2 , x3 , x4 ) = (2.8 + 0.6t, −0.6 + 0.2t, 2.4 + t, t) We have x3 = 2.4 + t = ⇒ t = −2.4 and substituting gives the particular solution: (x1 , x2 , x3 ) = (1.36, −1.08, 0, −2.4) 22 Free variables: x2 , x3 Let x2 = s and x3 = t, where s, t are real parameters Then x1 = − 2s − 3t General solution: (x1 , x2 , x3 ) = (4 − 2s − 3t, s, t) Setting s = and t = −1 gives x1 = − + = 3, showing that (3, 2, −1) is a particular solution Setting x1 = p and Full file at https://TestbankDirect.eu/ Solution Manual for Linear Algebra for Engineers and Scientists Using Matlab by Hardy Full file at https://TestbankDirect.eu/ ISM Chapter Linear Systems x3 = q gives x2 = − p − 32 q, giving the general solution (x1 , x2 , x3 ) = (p, − 21 p − 32 q, q) For any particular solution corresponding to the ordered pair (s, t), for fixed s, t, the unique ordered pair (p, q) = (4 − 2s − 3t, t) corresponds to the same particular solution Conversely, for any particular solution corresponding to the ordered pair (p, q), for fixed p, q, the unique ordered pair (s, t) = (2 − 21 p − 23 q, q) corresponds to the same particular solution 23   M = [A| b] =  −1  28  −8 , where A is × 3, M is × and b is × Solving the 13 −2 1 16 system by Gauss–Jordan elimination corresponds to reducing M to its reduced form M*   12   4 M s 0 General solution: (x1 , x2 , x3 ) = (12 − 5t, − 3t, t), where t is a real parameter 0 0 Particular solutions in nonnegative integers must satisfy x1 = 12 − 5t ≥ 0, x2 = − 3t ≥ and t ≥ Hence 12 ≥ t ≥ and ≥ t ≥ so that ≥ t ≥ 0, where t is an integer Particular solutions: (x1 , x2 , x3 ) = (12, 4, 0), (7, 1, 1) 24 14 −2 −1 15 , where A is × 4, M is × and b is × Solving the system by Gauss–Jordan elimination corresponds to reducing M to its reduced form M* M = [A| b] = Ms 0 1 3 General solution: (x1 , x2 , x3 , x4 ) = (3 − 2s − t, s, − t, t), where s, t are real parameters Particular solutions in nonnegative integers must satisfy x1 = − 2s − t ≥ 0, x3 = − t ≥ 0, s ≥ 0, and t ≥ Hence ≥ t ≥ ⇒ t = 0, 1, 2, Particular solutions (x1 , x2 , x3 , x4 ) are: t = 0, t = 1, t = 2, t = 3, 25 x1 x1 x1 x1 = − 2s ≥ ⇒ 32 ≥ s ≥ = − 2s ≥ ⇒ ≥ s ≥ = − 2s ≥ ⇒ 12 ≥ s ≥ = − 2s ⇒ ⇒ ⇒ ⇒ s = 0, 1, s = 0, 1, s = 0, s=0 ⇒ ⇒ ⇒ (3, 0, 3, 0), (1, 1, 3, 0) (2, 0, 2, 1), (0, 1, 2, 1) (1, 0, 1, 2) (0, 0, 0, 3) Let (S) denote the system First, assume α = and apply forward reduction on M = [A| b]  α  1 α 1 1 α   1 s r1 ↔ r3 r2 − r1 → r2 r3 − αr1 → r3 s r3 + r2 → r3 Full file at https://TestbankDirect.eu/   0  α−1 1−α 1  0 α 1−α 1−α 1−α α α−1 1−α (1 − α)(2 + α)  (1.1)      1−α (1.2) Solution Manual for Linear Algebra for Engineers and Scientists Using Matlab by Hardy Full file at https://TestbankDirect.eu/ ISM 1.1 Solving Linear Systems If α = 1, then the right side of (1.1) shows that (S) has infinitely many solutions Now assume α = and reduce further to obtain (1.2), noting that − α2 + − α = − α − α2 = (1 − α)(2 + α) If α = −2, then (S) has a unique solution because A s I3 and if α = −2, the (S) is inconsistent If α = 0, then the reduced form of M is   0 0.5   M* =  0.5  0 0.5 Summary (a) α = −2, no solutions (b) α = and α = −2, unique solution (c) α = 1, infinitely many solutions 26 Let (S) denote the system Apply forward reduction   2 s   1  α r3 + r1 → r3 −1 α α s r3 − 3r2 → r3 on M = [A| b]   1 2   α  0 α+2 α+1   1 2   α  0 0 2(1 − α) α−2 (a) α = 1, no solution, (b) α = 1, unique solution, (c) No value α gives infinitely many solutions 27 Apply forward and backward reduction on the augmented matrix M = [A| b] of (S)     1 −1 −1 1 0     M = 1 −1 0 s 0  = M* 2 −2 0 1 Pivot variables: x1 , x3 , x4 Free variable: x2 Let x2 = t, where t is a real parameter Using M*, the general solution is: (x1 , x2 , x3 , x4 ) = (1 − t, t, 3, 1) Any particular solution in positive integers must satisfy x1 = − t > 0, where t is an integer But then > t and so there are none 28 Substituting x1 = −1, x2 = 1, x3 = into E1 gives −3(−1) + 5(1) − 7(0) = = −2 The particular solution fails to satisfy E1 and so is not a particular solution to (S) Apply forward and backward reduction on the augmented matrix M = [A| b] of (S) M= −3 −7 −6 −2 s 0 −1 −2 −1 −1 = M* Pivot variables: x1 , x2 Free variable: x3 Let x3 = t, where t is a real parameter Using M*, the general solution is: (x1 , x2 , x3 ) = (−1 + t, −1 + 2t, t) For a particular solution in positive integers we must have: x1 = −1 + t > ⇒ t > and x2 = −1 + 2t > ⇒ t > 0.5 Hence t = 2, 3, determines all particular solutions in positive integers The condition x3 = t < ⇒ t = giving the single solution (1, 3, 2) Full file at https://TestbankDirect.eu/ Solution Manual for Linear Algebra for Engineers and Scientists Using Matlab by Hardy Full file at https://TestbankDirect.eu/ 29 ISM Chapter Linear Systems M is × 4, A is × (S)  x1 + 2x2 + 4x3 =      2x1 + 4x2 + 8x3 =  −x1 + 6x2 + 8x3 = −15     4x1 + 2x2 + 7x3 = 21 Perform forward reduction on M, maintaining integer coefficients    M=  −1  6   −15  4 8 21 s r2 − 2r1 → r2 r3 + r1 → r3 r4 − 4r1 → r4  s r2 ↔ r4 − 31 r3 → r3 r3 − 4r2 → r3 0   0 0   0  −6 12 −9 0  0   −12   −3   =U 0 0 Pivot variables: x1 , x2 Free variables: x3 Let x3 = t, where t is a real parameter Reading off solutions from U, we have 2x2 = −3 − 3t ⇒ x2 = −1.5 − 1.5t and x1 = − 2x2 − 4t = − 2(−1.5 − 1.5t) − 4t = − t General solution: (x1 , x2 , x3 ) = (6 − t, −1.5 − 1.5t, t) 30 M is × and A is × (S)      x1 − x2 + x4 + x5 = −x1 + 2x2 − x3 + x4 + x5 = 5x1 − 8x2 + 3x3 − x4 − x5 = Perform forward reduction on M: Eliminate below eliminate below in column 2: r3 + 3r2 → r3   −1 1   −1 1 2 s M =  −1 −8 −1 −1 s Row of U show that (S) is inconsistent 31 in column 1: r2 + r1 → r2 , r3 − 5r1 → r3 ,   0   0 −1 (S) Perform forward reduction on M Full file at https://TestbankDirect.eu/      2x2 + 2x3 1 −3 −1 −6 −6 −1 1 −1 2 0  4 = U = 3x3 + 3x4 = x1 −5x1 − 5x2  M is × 5, A is × We have       − x4 = = 15   4 −10 Solution Manual for Linear Algebra for Engineers and Scientists Using Matlab by Hardy Full file at https://TestbankDirect.eu/ ISM Solving Linear Systems 1.1  2   M=  0 −1 −5 −5 0  3   2 15  s r1 ↔ r3 r4 + 5r1 → r4 r2 ↔ r3 0   0 0  0   0 s r4 + 52 r2 → r4 r4 − 53 r3 → r4 −5 0 2 0  −1 −5 25  0   3 3 0   3 −1 −10 20 Back-substitution: −10x4 = 20 ⇒ x4 = −2, 3x3 = − 3x4 = − 3(−2) = ⇒ x3 = 3, 2x2 = −x3 = −3, x1 = + x4 = General solution: (x1 , x2 , x3 , x4 ) = (0, −3, 3, −2) 32 x1 − x2 M is × 5, A is × (S) + x4 + x5 = −x1 + 2x2 − x3 + x4 + x5 = Performing forward reduction using r2 + r1 → r2 , we have M= −1 −1 −1 1 2 s −1 1 −1 2 Pivot variables: x1 , x2 Free variables: x3 , x4 , x5 Let x3 = r, x4 = s, x5 = t, where r, s, t are real parameters Back-substituting, we have x2 = + x3 − 2x4 − 2x5 = + r − 2s − 2t, x1 = + x2 − x4 − x5 = + (4 + r − 2s − 2t) − s − t = + r − 3s − 3t General solution: (x1 , x2 , x3 , x4 , x5 ) = (6 + r − 3s − 3t, + r − 2s − 2t, r, s, t) Note that (S) is a subsystem of the linear system in Exercise 30 This problem illustrates the fact that deleting a single equation from an inconsistent linear system may (or may not) result in a consistent subsystem See Exercise 60 33 Let (S) denote the linear system Performing forward and backward elimination on (S) using augmented matrices, we have   M =  −3 −9 16   6 16 s r2 + 3r1 → r2 r3 − 2r1 → r3 s r3 + r2 → r3 r2 → r2 r1 − 5r2 → r1      0 6  12  12  −8  2 Unique solution: (x, y) = (−8, 2) Refer to Figure 1.1 [33] Equations E1 , E2 , E3 in (S) are represented, respectively, by lines L1 , L2 , L3 in R2 that intersect in a single point Full file at https://TestbankDirect.eu/  L3 L2 L1 −1 −10 −8 −6 −4 −2 Figure 1.1 [33] Solution Manual for Linear Algebra for Engineers and Scientists Using Matlab by Hardy Full file at https://TestbankDirect.eu/ 10 Chapter 34 Rearranging the system into standard form, we have 2x + 4y = − E1 Perform forward and back(S) 7x + 5y = E2 ward elimination using augmented matrices The EROs are chosen to preserve integer coefficients M= ISM Linear Systems −6 L2 −1 s r2 − 3r1 → r2 r1 ↔ r2 s r 18 → r2 r1 + 7r2 → r1 −7 18 24 −54 −2 L1 0 −3 −3 −4 Unique solution: (x, y) = (3, −3) Refer to Figure 1.1 [34] Equations E1 , E2 in (S) are represented, respectively, by lines L1 , L2 in R2 that intersect in a single point 35 2y = The system is (S) 2 −4 s r1 ↔ r2 r1 + 2r2 → r1 s r1 → r1 r2 → r2 −6 The system is (S) 6x − 9y = 0 L1 L2 1 1.5 E1 Perform for−x + 1.5y = 2.5 E2 ward and backward elimination using augmented matrices   −1 −4 s 2.5 −1 1.5   2 M = 1 r1 ↔ r2 0 15 c r2 + 6r1 → r2 Row in the last matrix shows that (S) is inconsistent Refer to Figure 1.1 [36] Equations E1 , E2 in (S) are represented, respectively, by parallel lines L1 , L2 in R2 with no point in common Full file at https://TestbankDirect.eu/ 3 0 Unique solution: (x, y) = (1.5, 2) Refer to Figure 1.1 [35] Equations E1 , E2 in (S) are represented, respectively, by lines L1 , L2 in R2 that intersect in a single point 36 E1 Perform for2x − 4y = − E2 ward and backward elimination using augmented matrices M= Figure 1.1 [34] 1.5 Figure 1.1 [35] L2 L1 0 Figure 1.1 [36] Solution Manual for Linear Algebra for Engineers and Scientists Using Matlab by Hardy Full file at https://TestbankDirect.eu/ ISM 1.3 Applications 39 N! Let xi , ≤ i ≤ 6, denote the number of trucks leaving each depot for the other two depots as designated in the digraph (directed graph, see Section 2.4.1) opposite Each xi is constrained to be a nonnegative integer A N# N N N" B N$ C Translating the given information, the problem is modeled by a sparse1 × linear system (S), namely  x1 + x2 = (Total leaving A)       x3 + x4 = 10 (Total leaving B)   (S) x5 + x6 = (Total leaving C)     x4 + x5 = (Total arriving at A)     x1 + x6 = 11 (Total arriving at B) Solving by Gauss-Jordan elimination  1 0  0 1  0 0   0 0 1 0 0 (check using M A     10      7  s 0   8 0 11 T LA B), we have 0 0 0 0 0 Let x6 = t, where t is a real parameter Then x1 = 11 − t, x2 = −6 + t, x3 = − t,  11  −1 −6   9   −1 1 1 x4 = + t, x5 = − t We note that x1 ≥ and x5 ≥ implies ≥ t ≥ and x2 ≥ implies t ≥ The only possible integer solutions correspond to t = 6, The value t = gives (x1 , x2 , x3 , x4 , x5 , x6 ) = (5, 0, 3, 7, 1, 6), which is impossible because we are given that x2 > x5 The value t = gives the unique (integer) solution (x1 , x2 , x3 , x4 , x5 , x6 ) = (4, 1, 2, 8, 0, 7) Let xi , ≤ i ≤ 4, denote the number of units of coal burned by A, B, C, D, respectively We will interpret a unit as an indivisible quantity so that each xi is a nonnegative integer Translating the information given, the problem is modeled by the × linear system (S) shown Solving by Gauss-Jordan elimination, we have      x + x3 + 7x4 = 250 0 100 250 1       550  s  (S) 2x1 + x2 + 2x3 + 9x4 = 550 ⇒  2 −5 50    2 400 0 150 2x2 + 2x3 = 150 Let x4 = t, where t is a real parameter Then x1 = 100 − 2t, x2 = 50 + 5t, x3 = 150 − 5t The constraint x1 = 100−2t ≥ implies 50 ≥ t and the constraint x4 = t ≥ implies x2 = 50+5t ≥ 50 The constraint x3 = 150−5t ≥ implies 30 ≥ t The fact that x3 ≤ 45 (given) implies 150−5t ≤ 45 1A lin ear sy stem is called spar se if it tain s a lar ge p er cen tage of coeffi cien ts eq u al to zer o S ee, for ex am p le, th e lin ear sy stem s in S ection 2.4.3 Full file at https://TestbankDirect.eu/ Solution Manual for Linear Algebra for Engineers and Scientists Using Matlab by Hardy Full file at https://TestbankDirect.eu/ 40 Chapter ISM Linear Systems implies 21 ≤ t and so 21 ≤ t ≤ 30 Conclusions: B broke the law, D did not break the law and A may or may not have broken the law because 21 ≤ t ≤ 30 implies 40 ≤ x1 ≤ 58 and so x1 is not uniquely determined Let xi , ≤ i ≤ 3, denote the dollar cost per pound of, respectively, Spanish Nuts (SN), Cocktail Nuts (CN), Cashews (C) Then each xi is constrained to be a nonnegative real number Equating cost per pound of each type of mixed nut to the cost of its ingredients gives the × linear system (S) shown Multiplying (S) throughout by 10 and solving by Gauss-Jordan elimination, we have      39 0.4x1 + 0.6x2 = 1.80 (MP) 0 18  10      , 40  s  (S) 0.4x1 + 0.4x2 + 0.2x3 = 4.00 (CP) ⇒  4    57 44 0 0.2x1 + 0.5x2 + 0.3x3 = 4.40 (AS) giving the unique solution (x1 , x2 , x3 ) = (3.90, 0.40, 11.4) dollars Let xi , ≤ i ≤ 3, denote the number of bundles formed of type A, B, C, respectively We assume that a bundle is indivisible and so each xi is a nonnegative integer Equating the total number of lengths required of each gauge of steel to the numbers found in each bundle, we obtain a × linear system (S), which is solved by Gauss-Jordan elimination      2x1 + 3x2 + 5x3 = 37 (G1 )  1 37     x1 + x2 + 2x3 = 14 (G2 ) 1  14  1 9     (S) ⇒  s       24 0 0 3x + x + 4x = 24 (G )  3    33 0 0 3x1 + 2x2 + 5x3 = 33 (G4 ) Let x3 = t, where T is a real parameter Then x1 = − t, x2 = − t We have x1 ≥ and x2 ≥ implies ≥ t ≥ because x3 = t ≥ Hence there are six solutions corresponding to the values t = 0, 1, , Computing the cost of each solution, we have t=0 t=1 t=2 t=3 t=4 t=5 (x1 , x2 , x3 ) (5, 9, 0) (4, 8, 1) (3, 7, 2) (2, 6, 3) (1, 5, 4) (0, 4, 5) Cost 5(18) + 9(16) + 0(21) = 234 4(18) + 8(16) + 1(21) = 221 3(18) + 7(16) + 2(21) = 208 2(18) + 6(16) + 3(21) = 195 1(18) + 5(16) + 4(21) = 182 0(18) + 4(16) + 5(21) = 169 Cheapest There are two ways of filling the order with exactly two bundle types and the solution with t = is cheapest at $169.00 10 Let xi , ≤ i ≤ 4, denote the number of bags of type Bi required We will assume that a unit is an indivisible quantity and so each xi is a nonnegative integer Equating total number of units of Po, N, Pho required to the number of units in each bag, we obtain a × linear system (S), which is Full file at https://TestbankDirect.eu/ Solution Manual for Linear Algebra for Engineers and Scientists Using Matlab by Hardy Full file at https://TestbankDirect.eu/ ISM 1.3 Applications 41 solved by Gauss-Jordan elimination  2x + x2 + x4 = 12 (Po)   (S) x1 + x2 + 2x3 = (N)   x2 + x3 + 2x4 = (Pho)   1 1 ⇒  2 0  s 0  12  8 −1 −1   4 Let x4 = t, where t is a real parameter Then x3 = t, x2 = − 3t, x1 = + t The constraint x2 = − 3t ≥ implies 43 ≥ t implies ≥ t ≥ There are two solutions in integers corresponding to t = 0, 1, namely (x1 , x2 , x3 , x4 ) = (4, 4, 0, 0), (5, 1, 1, 1) 11 Let xi , ≤ i ≤ 4, denote, respectively, the number of batches of rolls, cookies, buns, bread to be produced We will assume that a batch is an indivisible quantity and so each xi is a nonnegative integer Equating the total number of ounces of yogurt, wheat germ and butter to the total amounts in all batches, we have  10x1 + 20x4 = 6400 (yogurt)   (S) 10x1 + 15x2 = 5400 (wheat germ)   10x1 + 15x2 + 10x3 + 15x4 = 9000 (butter) Solving (S) by Gauss-Jordan elimination (check by    6400 10 0 20    5400  s   10 15 10 15 10 15 9000 M A T LA B), we have 0 0 − 43 640   − 200  360 Let x4 = t, where t is a real parameter Then x1 = 640 − 2t, x2 = − 200 + t, x3 = 360 − t To obtain integer values for the variable x2 and x3 , we require t to be divisible by (t even) and −200 + 4t to be divisible by However, x2 ≥ implies −200 + 4t ≥ implies t ≥ 50 The minimum value t = 50 gives x1 = 540 as the maximum batches of rolls Note that x3 = in this case, which is divisible by 1.3.2 General Networks 12 Referring Example and Figure 1.3, the new constraint x1 + x4 = x2 + x3 adds a fifth equation x1 − x2 − x3 + x4 = to (S) Solving using augmented matrices, we have     0 150 0 450     0 250   0 400   −1       −1 350  50   s ,     1 350    0 300  −1 −1 0 0 0 resulting in a unique (integer) solution (x1 , x2 , x3 , x4 ) = (150, 400, 50, 300) Full file at https://TestbankDirect.eu/ Solution Manual for Linear Algebra for Engineers and Scientists Using Matlab by Hardy Full file at https://TestbankDirect.eu/ 42 Chapter ISM 13 (a) Equating input to output at each node, we have Linear Systems A B C 100 x1 x2 + x3 = x1 + x2 = 20 + x3 = 80 ⇒ Solving by Gauss-Jordan elimination, we have    100 1    20  s   −1 1 80  x + x2 = 100   (S) x1 − x3 = 20   x2 + x3 = 80 −1 20 0  80   Let x3 = t, where t is a real parameter Then x1 = 20 + t, x2 = 80 − t and x2 = 80 − t ≥ implies 80 ≥ t ≥ because the flow in BC is nonnegative (b) The minimum flow in AB is x1 = 20 when t = (c) Given x1 = kx2 , we have 20 + t = k(80 − t) and solving for t, we have t = Substituting for t in x1 and x2 gives (x1 , x2 , x3 ) = 14 (100k, 100, 20(4k − 1)) k+1 80k − 20 k+1 (a) Equating input to output at each node, we have A 800 = x1 + x2 B x2 + x5 = 500 + x3 C x3 + x4 = 600 D 300 + x1 = x4 + x5 ⇒ (S) Solving (S) by Gauss-Jordan elimination, we have      −1 1 0 0 0 1 1   800   500   s 600   300       x1 + x2 = 800 x2 − x3 + x5 = 500 x3 + x4      −x1 0 0 = 600 + x4 + x5 = 300 −1 −1 1 0 −300  1100    600  Let x4 = s and x5 = t, where s and t are real parameters Then x1 = −300+s+t, x2 = 1100−s−t, x3 = 600 − s (b) Given x3 = 0, we have s = 600 and then x2 = 500 − t is nonnegative implies 500 ≥ t giving x5 = t = 500 units as the maximum flow in BD (c) Given t = 2s, we have x1 = −300 + 3s ≥ and x2 = 1100 − 3s ≥ so that 1100 ≥ s ≥ 100 The minimum and maximum flows in DC are, respectively, 100 and 366 units Full file at https://TestbankDirect.eu/ Solution Manual for Linear Algebra for Engineers and Scientists Using Matlab by Hardy Full file at https://TestbankDirect.eu/ ISM 15 1.3 Applications 43 (a) Equating input to output at each node, we have A x1 + x2 = 700 B 200 + x6 = x2 + x4 C 600 + x3 = 200 + x6 D 250 + x7 = 450 + x3 E x2 + x5 = 250 + x7 F 550 = x1 + x5 ⇒ (S)                      The sparse linear system (S) can be solved easily by Jordan elimination, we have    700 0 0    1 −1 200          0 0 −1 −400 s    0 0  −200 −1      0 0 −1 250    0 0 550 x1 + x4 x2 + x4 x3 = 700 − x6 = 200 − x6 = − 400 x3 − x7 = − 200 x2 x1 + x5 − x7 = 250 + x5 = 550 hand (or using M A T LA B) Applying Gauss0 0 1 0 1 0 −1 −1 0 0 0 0 0 550  250   −200    150    200  0 −1 0 −1  Let x5 = s and x7 = t, where s and t are real parameters Then x1 = 550 − s, x2 = 250 − s + t, x3 = −200 + t, x4 = 150 + s, x6 = 200 + t units (b) The only constraint on x7 is x2 = −200+t ≥ 0, which implies that t ≤ 200 giving x7 = t = 200 units as the required minimum (c) x7 = 200 implies x3 = (d) Given x5 = s = 0, we have x4 = 150 units 16 Equating input to output at each node, we have A 200 = x1 + x5 B x1 + x6 = 100 + x2 C x2 + x4 = 100 D 100 + x3 = x4 + x6 E 200 + x5 = 300 + x3 ⇒  x1       x − x2   x2 (S)         + x5 + x4 x3 − x4 − x3 = 200 + x6 = 100 = 100 − x6 = − 100 + x5 = 100 The sparse linear system (S) can be solved easily by hand (or using M A T LA B) Applying GaussJordan elimination, we have     0 200 0 200     0 100   0 −1 100   −1      0 1 0 100  −1 −100   s      −1 −1 −100   0 0 −1 0 0 −1 100 0 0 0 Full file at https://TestbankDirect.eu/ Solution Manual for Linear Algebra for Engineers and Scientists Using Matlab by Hardy Full file at https://TestbankDirect.eu/ 44 Chapter ISM Linear Systems Let x5 = s and x6 = t, where s and t are real parameters Then x1 = 200 − s, x2 = 100 − s + t, x3 = −100 + s, x4 = s − t units If x1 = 0, then s = 200 and so x2 = −100 + t ≥ implies t ≥ 100 and x4 = 200 − t ≥ implies 200 ≥ t showing that x6 = 200 units is the maximum flow in BD If x1 = x2 = 0, then x5 = s = 200, x6 = t = 100, x3 = x4 = 100 units 1.3.3 Electrical Networks 17 Applying Kirchhoff’s laws, we obtain the × linear system (S), which is solved by elimination     5I + 5I3 = 10 (Loop ABDA) 10 5      (S) 5I2 − 5I3 = (Loop BCDB) ⇒  −5 6 s    −1 −1 0 I1 − I2 − I3 = (Nodes B, D) Gauss-Jordan Applying Kirchhoff’s laws, we obtain the × linear system (S), which is solved by elimination     12I1 + 3I3 = (Loop ACBA) 12      (S) 6I2 − 3I3 = (Loop BCDB) ⇒  −3 0 s    −1 −1 0 I1 − I2 − I3 = (Nodes B, C) Gauss-Jordan The unique solution is (I1 , I2 , I3 ) = 18 15 (26, 22, 4) 0 0 amps 0 0 5 , 42 , 21 ) amps The unique solution is (I1 , I2 , I3 ) = ( 14 19 Applying Kirchhoff’s laws, we obtain the × linear system (S)  10I1 + 10I2 = 20 (Loop ABGA)       − 10I2 + 10I3 + 10I4 = (Loop BCF GB)       − 10I4 + 20I5 = (Loop BCF GB)   (S) I1 − I2 − I3 = (Node B)     I3 − I4 − I5 = (Node C)       I4 + I5 − I6 = (Node F )     I1 − I2 − I6 = (Node G) Solving (S) by Gauss-Jordan elimination, we have    10 10 0 0 20    10 0    −10 10      0 −10 20 0        −1 −1 0 0 s       −1 −1 0        0 1 −1    −1 0 −1 0 Full file at https://TestbankDirect.eu/ 26 15 22 15 15 0 0 0 0 0 0 0 0 0 0 0 0 16 13 10 13 13 13 13 13              14 42 21       Solution Manual for Linear Algebra for Engineers and Scientists Using Matlab by Hardy Full file at https://TestbankDirect.eu/ ISM 1.3 Applications 45 The unique solution is (I1 , I2 , I3 , I4 , I5 , I6 ) = 20 13 (16, 10, 6, 4, 2, 6) amps Applying Kirchhoff’s laws, we obtain the × linear system (S)  2I4 + 2I6 = 10 (Loop ABF DA)       4I2 − 2I4 − 3I5 = (Loop BCF B)       6I3 + 3I5 − 2I6 = (Loop CDF C)   (S) I1 − I2 − I4 = (Node B)    − I + I  − I5 = (Node C)       I1 − I3 − I6 = (Node D)     I4 − I5 − I6 = (Node F ) Solving (S) by Gauss-Jordan elimination, we have    10 0 2    −2 −3 0  0    0  −2 0       −1 0 0 s   −1     −1  −1 0       −1 0 −1 0  1 0 −1 −1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 64 (225, 70, 60, 155, −10, 165) The unique solution is (I1 , I2 , I3 , I4 , I5 , I6 ) = current I5 is negative, indicating that the direction of flow is C → F 225 64 35 32 15 16 155 64 − 32 165 64              amps Note that the 1.3.4 Balancing Chemical Equations 21 Equating numbers of atoms before and after reaction gives a × homogeneous which we solve by reducing the coefficient matrix to its reduced echelon form     2x + x2 − 2x3 − 6x4 = (O) −2 −6      (S) x1 − x4 = (C) ⇒  0 −6  s    −12 2x2 − 12x4 = (H) linear system (S) 0 −6   −6  1 −6 Let x4 = t, where t is a real parameter Then x1 = x2 = x3 = 6t Choosing t = gives the smallest solution (6, 6, 6, 1) and the reaction is LIGH T 6CO2 + 6H2 O 22 =⇒ 6O2 + C6 H12 O6 Equating numbers of atoms before and after combustion gives a × homogeneous (S) which we solve by reducing the coefficient matrix to its reduced echelon form     8x1 − x3 = (C) −1      (S) 18x1 − 2x4 = (H) ⇒  18 0 −2  s    −2 −1 0 2x2 − 2x3 − x4 = (O) Full file at https://TestbankDirect.eu/ linear system − 19  25  − 18  − 89 Solution Manual for Linear Algebra for Engineers and Scientists Using Matlab by Hardy Full file at https://TestbankDirect.eu/ 46 Chapter ISM Linear Systems Let x4 = t, where t is a real parameter Then x1 = t/9, x2 = 25t/18, x3 = 8t/9 Choosing t = 18 gives the smallest solution (2, 25, 16, 18) moles and the reaction is C OM BU ST ION 2C8 H18 + 25O2 23 =⇒ 16CO2 + 18H2 O Suppose x1 moles of cellulose and x2 moles of oxygen produces x3 moles of carbon dioxide, x4 moles of water, plus x5 moles of carbon The reaction is C OM BU ST ION x1 C6 H10 O5 + x2 O2 =⇒ x3 CO2 + x4 H2 O + x5 C Equating numbers of atoms before and after combustion gives a × homogeneous linear system (S) which we solve by reducing the coefficient matrix to its reduced echelon form   −1 −1   −2 0   10 6x − x − x = (C)   −2 −1 (S) 10x1 − 2x4 = (H) ⇒    0 −5  5x1 + 2x2 − 2x3 − x4 = (O)   s − 65  0 − 65 Let x5 = k and x4 = t, where k and t are real parameters Then x1 = 0.2t, x2 = 1.2t − k, x3 = 1.2t − k Setting t = and k = 0, 1, gives the solutions (x1 , x2 , x3 , x4 , x5 ) = (1, 6, 6, 5, 0), (1, 5, 5, 5, 1), (1, 4, 4, 5, 2) moles 24 Suppose x1 moles of toluene and x2 moles of nitric acid react to produce x3 moles of trinitrotoluene and x4 moles of water The reaction is REA C T ION x1 C7 H8 + x2 HNO3 =⇒ x3 C7 H5 O6 N3 + x4 H2 O Equating numbers of atoms before and after combustion gives a × homogeneous (S) which we solve by reducing the coefficient matrix to its reduced echelon form     7x1 − 7x3 = (C)  −1     8x1 + x2 − 5x3 − 2x4 = (H)  −5 −2      (S) ⇒  s     −3 0 x2 − 3x3 = (N)     −6 −1 0 3x2 − 6x3 − x4 = (O) linear system − 31  −1    − 13  0 Let x4 = t, where t is a real parameter Then x1 = t/3, x2 = t, x3 = t/3 Setting t = 3k and k = 0, 1, 2, gives all solutions (x1 , x2 , x3 , x4 ) = (k, 3k, k, 3k) moles 1.3.5 Fitting Curves Full file at https://TestbankDirect.eu/ Solution Manual for Linear Algebra for Engineers and Scientists Using Matlab by Hardy Full file at https://TestbankDirect.eu/ ISM 25 1.3 Applications 47 Substitute pairs of coordinates into the equation mx + c = y to form a × linear system (S), which is solved by Gauss elimination (x = −1, y = 1.1) (S) −m + c = 1.1 2m + c = 3.2 (x = 2, y = 3.2) 1 1.1 3.2 −1 − r1 → r1 r2 − 2r1 → r2 −1 −2 and back-substituting, we have 3c = 5.4 ⇒ c = 1.8 and m = −1.1 + c = −1.1 + 1.8 = 0.7 The equation is y = 0.7x + 1.8 26 0.7N+ 1.8 −1.1 5.4 −1 O= −1 3 Figure 1.3 [25] Substitute pairs of coordinates into the equation a2 x2 + a1 x + a0 = y to form a × linear system (S), which is solved by Gauss-Jordan elimination  a + a1 + a0 = (x = y = 1)   (S) a2 − a1 + a0 = (x = −1, y = 1)   4a2 + 2a1 + a0 = (x = y = 2)     1 0 1 1     1 s  0  −1 2 0 3 O= (N2+ 2) /3 −2 −1 Figure 1.3 [26] The equation is y = (x2 + 2)/3 27 Substitute pairs of coordinates into the equation a2 x2 + a1 x + a0 = y to form a × linear system (S), which is solved by Gauss-Jordan elimination (S)  a2 − a1 + a0 = − (x = y = −1)      a0 = (x = 0, y = 1) O=  a2 + a1 + a0 = (x = y = 1)     4a2 + 2a1 + a0 = − (x = 2, y = −1)  0  M = [A| b] =  1 −1 1 1   −1   1  s 1  −1 + N− N2 0 0 0 −1  1   1 −1 −2 −2 −1 Figure 1.3 [27] Referring to Theorem 1.7(c), we have n = and m = > = n + In this case rank A = n + (= rank M) and there is a unique parabola with equation y = + x − x2 passing through all four points Full file at https://TestbankDirect.eu/ Solution Manual for Linear Algebra for Engineers and Scientists Using Matlab by Hardy Full file at https://TestbankDirect.eu/ 48 Chapter ISM 28 Substitute pairs of coordinates into the equation a2 x2 + a1 x + a0 = y to form a × linear system (S), which is solved by Gauss-Jordan elimination  a0 = (x = 0, y = 1)      a2 + a1 + a0 = (x = 1, y = 0) (S)  a2 + a1 + a0 = (x = y = 1)     4a2 + 2a1 + a0 = − (x = 2, y = −5)     0 0 1   1 1 0 0      M = [A| b] =  s  1 1 1  0   −5 0 Linear Systems Referring to Theorem 1.7(c), n = and m = < = n + However, rank A = < = rank M and so no parabola passes through the four points 29 Substitute pairs of coordinates into the equation a2 x2 + a1 x + a0 = y to form a × linear system (S), which is solved by Gauss-Jordan elimination J= J= 16 1 2 s 0 − 81 − 14 −30 J= −60 Let a0 = t, where t is a real parameter Then a2 = − 14 + 8t and a1 = 32 − 3t and the family of parabolas is given by t y = (− + )x + ( 32 − 3t )x + t Substituting x = and y = into the latter equation gives t = −30 and so the particular curve has equation y = −4x2 + 24x − 30 J= 2 J= 0 Figure 1.3 [29] USING MATLAB 30 Let x1 , x2 , x3 , x4 denote, respectively, the number portfolios of type A, B, C, D used to fill the order Note that each xi is a nonnegative integer Equating blocks of bonds (federal, regional, provincial, municipal) in each portfolio to the total required gives a × linear system with augmented matrix M = [A| b] and its reduced form rref(M)= M* as shown     −2 1 11 2  17  −3 5     M= s  = M* 3   28 0 0 0 10 50 0 0 Note that r = rank A = rank M = and there are n − r = − = parameters required Let x3 = s and x4 = t, where s and t are real parameters Then x1 = − 3s + 2t ≥ ⇒ 6≥ 3s − 2t (1.4) x2 = + 3s − 5t ≥ ⇒ ≥ −3s + 5t (1.5) Full file at https://TestbankDirect.eu/ Solution Manual for Linear Algebra for Engineers and Scientists Using Matlab by Hardy Full file at https://TestbankDirect.eu/ ISM 1.3 Applications 49 Adding (1.4) and (1.4) gives 11 ≥ t ≥ and so t = 0, 1, 2, because x4 = t is nonnegative The possible solutions are classified as follows: t=0 x1 = x2 = − 3s + 3s t=2 x1 = 10 − 3s x2 = −5 + 3s s=0 s=1 s=2 s=2 s=3 (6, 5, 0, 0) * (3, 8, 1, 0) (0, 11, 2, 0) * (4, 1, 2, 2) (1, 4, 3, 2) s=0 s=1 s=2 t=1 x1 = x2 = − 3s 3s t=3 x1 = 12 − 3s s=4 x2 = −10 + 3s (8, 0, 0, 1) (5, 3, 1, 1) (2, 6, 2, 1) * (0, 2, 4, 3) The are three ways (starred in the table above) of filling the order using exactly two portfolios 31 Let x1 , x2 , x3 , x4 denote, respectively, the number portfolios of type A, B, C, D used to fill the order Note that each xi is a nonnegative integer Equating blocks of bonds (oil, mining, gold, commodity) in each portfolio to the total required gives a × linear system with augmented matrix M = [A| b] and its reduced form rref(M)= M* as shown     0 69 11   32  −1 −2      M=  = M* s  26   0 −1 4 1 20 0 0 Note that r = rank A = rank M = and there are n − r = − = parameters required Let x4 = t, where t are real parameters Then x1 = − t ≥ implies ≥ t and x2 = −2 + t ≥ implies t ≥ so that t = 2, 3, 4, The possible solutions are classified as follows: t = (3, 0, 6, 2) t = (1, 2, 8, 4) * t = (2, 1, 7, 3) t = (0, 3, 9, 5) * The are two ways (starred in the table above) of filling the order using exactly two portfolios Matrix multiplication (Text, page 58) gives the gain on each solution:  Smallest gain     7200 800    1000   8600          =    600   10000  32 600 11400 Biggest gain The problem is modeled by an × linear system with augmented in M corresponds to the loop or node indicated   45 0 15 0 150  20 15 −15 30 250       0 10 −15 0 130     0 0 −30 70 400    M=   −1 0 −1 0 0    −1 −1 0 0      0 0 −1 −1 −1 0 Full file at https://TestbankDirect.eu/ 1 −1 −1 matrix M = [A| b] Each row Loop ABGA Loop BCHGB Loop CDHC Loop GHF G Node B Node C Node G Node H Solution Manual for Linear Algebra for Engineers and Scientists Using Matlab by Hardy Full file at https://TestbankDirect.eu/ 50 Chapter ISM Linear Systems The reduced form rref(M)= M* shows that rank A = rank M = giving the unique solution (I1 , I2 , I3 , I4 , I5 , I6 , I7 ) = (4.5837, 8.3348, 10.2009, −1.8661, −3.7511, 1.8343, 6.5004) The current flow in branches CH and BG is in the opposite direction to those indicated in the figure 33 Refer to the solution to Exercise 17 and this exercise in the M-file ex01s3.m The commands syms V M = [5, 5, 10; 0, 5, −1, V; 1, −1, −1, 0] enters the symbolic bolic solution N= [ 1, 0, [ 0, 1, [ 0, 0, matrix M into the work space The command N=rref(M) returns the sym- 0, 0, 1, 4/3 + 1/15 ∗ V ] 2/15 ∗ V + 2/3] 2/3 − 1/15 ∗ V ] shows that I1 = 4/3 + 1/15 ∗ V , I2 = 2/15 ∗ V + 2/3, I3 = 2/3 − 1/15 ∗ V The command A = [N(1,4),N(2,4),N(3,4)] defines a symbolic vector of current values (in terms of V ) and the commands subs(A,9), subs(A,16) return the numeric values of A when V = 9, 16, respectively The command solve(A(3)) returns V = 10, the voltage at C when I3 = 34 Equating numbers of atoms (S) with augmented matrix  0 −2 1 −3 −1    4 −12 −4 M= 0 0   0 0 −1 0 before and after the reaction gives a × homogeneous M = [A| b] and reduced form rref(M)= M*   Fe 0 0 0   S  0 0 −1 0     0 −4 −1  O  s  −2  0 0  H    0  Mn 0 0  −2 0 0 0 K linear system − 54   − 14   −1    = M*  −8   − 14  − 18 Let x7 = t, where t is a real parameter Then x1 = 5t/4, x2 = t/4, x3 = t, x4 = 5t/8, x5 = t/4, x6 = t/8 Setting t = 8, we have the smallest solution 10FeSO4 + 2KMnO4 + 8H2 SO4 35 =⇒ 5Fe2 (SO4 )3 + 2MnSO4 + K2 SO4 + 8H2 O Equating numbers of atoms before and after the reaction gives a × homogeneous linear system (S) with augmented matrix M = [A| b] and reduced form rref(M)= M*     C 0 0 − 34 0 −2 0    −4  0 0 − 12  −2 0   H       −2 0 0 −1 0 O −4      M= s    0 0 −1  0 0 − 0  Na       0 0 −1 0  Cr 0 −1   0 −3 −1 0 0 − 11 Cl Full file at https://TestbankDirect.eu/ Solution Manual for Linear Algebra for Engineers and Scientists Using Matlab by Hardy Full file at https://TestbankDirect.eu/ ISM Review 51 Let x7 = t, where t is a real parameter Then x1 = 3t/4, x2 = t/2, x3 = 4t, x4 = 3t/4, x5 = t, x6 = 11t/4 Setting t = 4, we have the smallest solution 3C2 H5 OH + 2Na2 Cr2 O7 + 16HCl =⇒ 3C2 H4 O2 + 4CrCl3 + 11H2 O + 4NaCl Note on Vandermonde matrices Let A be the m × (n + 1) Vandermonde matrix shown in equation (1.55) on page 44 The special pattern of entries in A forces values of rank A stated in Theorem 1.7 When m ≤ n + the matrix A looks like  n  x1 xn−1 xm−1 · · · x1 ··· 1  xn xn−1 · · · xm−1 · · · x2    2  A=   = [ B | C]   xnm xn−1 m ··· xm−1 m · · · xm The matrix C is a square Vandermonde matrix and its determinant is nonzero if and only if the x-values of all data points are distinct (see Exercises 5.2 [18]) The implications are that C is invertible and that the rows of C, and therefore of A, are linearly independent, which implies that rank A = m Note that when m = n + we have C = A (cases (a) and (b) Theorem 1.7) If m > (n + 1), then A reduces In+ and so rank A = n + Then either rank A < rank M, which implies that no curve passes to O through all the data points, or rank A = rank M, which implies that there is a unique curve passing through all the data points (case (c) in Theorem 1.7) 36 Refer to the M-file ch01s3ex36.m The polynomial is (to two decimal places) y ≃ 1.17x3 − 1.17x2 − 4.66x + 6.66 A value of y(x) is interpolated if x lies inside the range of x-values and is extrapolated if x lies outside the range of x-values Interpolating the value y when x = −1.37, we have y ≃ 7.7804 and extrapolating the value of y when x = −3.36, we have y ≃ −35.0798 37 10 7.8704 −3 −2 −1 Figure 1.3 [36] Project See the M-file ch01s3ex37.m CHAPTER REVIEW False The expression may possibly be viewed as a combination of two EEOs, namely, scaling equation 2: E2 → E2 and then replacing equation 2: E2 − 5E1 → E2 False The inverse of r3 − r2 → r3 is r3 + r2 → r3 Full file at https://TestbankDirect.eu/ Solution Manual for Linear Algebra for Engineers and Scientists Using Matlab by Hardy Full file at https://TestbankDirect.eu/ 52 Chapter ISM Linear Systems True Swapping the same pair of rows twice returns the original matrix False For example, the equation x1 + 2x2 = defines a × linear system False Forward elimination on a linear system will determine the pivot variables and free variables Further analysis (back-substitution or backward elimination) is normally required to say whether or not a variable is uniquely determined False If a variable is not free, then it is a pivot variable, which may or may not be uniquely determined False Keep in mind that square linear systems can have zero, one or infinitely many solutions True Any EEO performed on a linear system can be viewed as an ERO performed on the corresponding augmented matrix, and vice versa True There is a sequence of EROs that reduces the coefficient matrix A to its reduced form A* Applying the sequence of inverse EROs (in reverse order) to the matrix A* y , where y has components (0, 0, , 1), we obtain M = [A| b] and a required column vector b 10 True See Theorem 1.4 11 False For example, the homogeneous linear system 12 False For example, removing any single equation from the inconsistent linear system  x + x2 + x3 =   x1 + x2 + x3 =   x1 + x2 + x3 = x1 + x2 = 2x1 + 2x2 = has infinitely many solutions results in an inconsistent × linear system 13 14 True Their ranks are equal   False For example, A =     has rank A = 0 15 False The rank is equal to the number of nonzero rows in any echelon form for A 16 False The rank of an m × n matrix cannot exceed min{m, n} Full file at https://TestbankDirect.eu/ Solution Manual for Linear Algebra for Engineers and Scientists Using Matlab by Hardy Full file at https://TestbankDirect.eu/ ISM Review 53  17  True For example,  18 True We have A s only if c =    −2 Hence rank A = if and only if c = and rank A = if and 5−c 19 True We have r = rank A ≤ m < n and there are n − r > parameters in the solution set 20 True See Theorem 1.4 We have rank A = = rank M and there will be − = parameter in the solution set Full file at https://TestbankDirect.eu/ ... consistent for some b, and have infinitely many solutions for other b Full file at https://TestbankDirect.eu/ Solution Manual for Linear Algebra for Engineers and Scientists Using Matlab by Hardy. .. https://TestbankDirect.eu/ Solution Manual for Linear Algebra for Engineers and Scientists Using Matlab by Hardy Full file at https://TestbankDirect.eu/ ISM Chapter Linear Systems and a linear system (S)′... Figure 1.1 [36] Solution Manual for Linear Algebra for Engineers and Scientists Using Matlab by Hardy Full file at https://TestbankDirect.eu/ ISM 37 1.1 Solving Linear Systems 11 Perform forward elimination