Skip to content
Textbook Guides

Solution Manual for Linear Algebra with Applications, 5th Edition

Solution Manual for Linear Algebra with Applications, 5th Edition is your go-to resource for solving complex textbook problems with clear solutions and explanations.

Preview 16 of 443 pages

See the first 16 pages exactly as they appear in the document.

  1. Page 1
    Read page 1 as text

    S OLUTIONS M ANUAL L INEAR A LGEBRA WITH A PPLICATIONS F IF TH E DITION Otto Bretscher Colby College

  2. Page 2
  3. Page 3
    Read page 3 as text

    Section 1.1 Chapter 1 Section 1.1 1.1. 1 [ x + 2 y = 1 2 x + 3 y = 1 ] − 2 × 1st equation → [ x + 2 y = 1 − y = − 1 ] ÷ ( − 1) → [ x + 2 y = 1 y = 1 ] − 2 × 2nd equation → [ x = − 1 y = 1 ] , so that ( x, y ) = ( − 1 , 1). 1.1. 2 [ 4 x + 3 y = 2 7 x + 5 y = 3 ] ÷ 4 → [ x + 3 4 y = 1 2 7 x + 5 y = 3 ] − 7 × 1st equation → [ x + 3 4 y = 1 2 − 1 4 y = − 1 2 ] × ( − 4) → [ x + 3 4 y = 1 2 y = 2 ] − 3 4 × 2nd equation → [ x = − 1 y = 2 ] , so that ( x, y ) = ( − 1 , 2). 1.1. 3 [ 2 x + 4 y = 3 3 x + 6 y = 2 ] ÷ 2 → [ x + 2 y = 3 2 3 x + 6 y = 2 ] − 3 × 1st equation → [ x + 2 y = 3 2 0 = − 5 2 ] . So there is no solution. 1.1. 4 [ 2 x + 4 y = 2 3 x + 6 y = 3 ] ÷ 2 → [ x + 2 y = 1 3 x + 6 y = 3 ] − 3 × 1st equation → [ x + 2 y = 1 0 = 0 ] This system has infinitely many solutions: if we choose y = t , an arbitrary real number, then the equation x + 2 y = 1 gives us x = 1 − 2 y = 1 − 2 t . Therefore the general solution is ( x, y ) = (1 − 2 t, t ), where t is an arbitrary real number. 1.1. 5 [ 2 x + 3 y = 0 4 x + 5 y = 0 ] ÷ 2 → [ x + 3 2 y = 0 4 x + 5 y = 0 ] − 4 × 1st equation → [ x + 3 2 y = 0 − y = 0 ] ÷ ( − 1) → [ x + 3 2 y = 0 y = 0 ] − 3 2 × 2nd equation → [ x = 0 y = 0 ] , so that ( x, y ) = (0 , 0). 1.1. 6   x + 2 y + 3 z = 8 x + 3 y + 3 z = 10 x + 2 y + 4 z = 9   − I − I →   x + 2 y + 3 z = 8 y = 2 z = 1   − 2( II ) →   x + 3 z = 4 y = 2 z = 1   − 3( III ) →   x = 1 y = 2 z = 1   , so that ( x, y, z ) = (1 , 2 , 1). 1.1. 7   x + 2 y + 3 z = 1 x + 3 y + 4 z = 3 x + 4 y + 5 z = 4   − I − I →   x + 2 y + 3 z = 1 y + z = 2 2 y + 2 z = 3   − 2( II ) − 2( II ) →   x + z = − 3 y + z = 2 0 = − 1   This system has no solution. 1

  4. Page 4
    Read page 4 as text

    Chapter 1 1.1. 8   x + 2 y + 3 z = 0 4 x + 5 y + 6 z = 0 7 x + 8 y + 10 z = 0   − 4( I ) − 7( I ) →   x + 2 y + 3 z = 0 − 3 y − 6 z = 0 − 6 y − 11 z = 0   ÷ ( − 3) →   x + 2 y + 3 z = 0 y + 2 z = 0 − 6 y − 11 z = 0   − 2( II ) +6( II ) →   x − z = 0 y + 2 z = 0 z = 0   + III − 2( III ) →   x = 0 y = 0 z = 0   , so that ( x, y, z ) = (0 , 0 , 0). 1.1. 9   x + 2 y + 3 z = 1 3 x + 2 y + z = 1 7 x + 2 y − 3 z = 1   − 3( I ) − 7( I ) →   x + 2 y + 3 z = 1 − 4 y − 8 z = − 2 − 12 y − 24 z = − 6   ÷ ( − 4) →   x + 2 y + 3 z = 1 y + 2 z = 1 2 − 12 y − 24 z = − 6   − 2( II ) +12( II ) →   x − z = 0 y + 2 z = 1 2 0 = 0   This system has infinitely many solutions: if we choose z = t , an arbitrary real number, then we get x = z = t and y = 1 2 − 2 z = 1 2 − 2 t . Therefore, the general solution is ( x, y, z ) = ( t, 1 2 − 2 t, t ) , where t is an arbitrary real number. 1.1. 10   x + 2 y + 3 z = 1 2 x + 4 y + 7 z = 2 3 x + 7 y + 11 z = 8   − 2( I ) − 3( I ) →   x + 2 y + 3 z = 1 z = 0 y + 2 z = 5   Swap : II ↔ III →   x + 2 y + 3 z = 1 y + 2 z = 5 z = 0   − 2( II ) →   x − z = − 9 y + 2 z = 5 z = 0   + III − 2( III ) →   x = − 9 y = 5 z = 0   , so that ( x, y, z ) = ( − 9 , 5 , 0). 1.1. 11 [ x − 2 y = 2 3 x + 5 y = 17 ] − 3( I ) → [ x − 2 y = 2 11 y = 11 ] ÷ 11 → [ x − 2 y = 2 y = 1 ] +2( II ) → [ x = 4 y = 1 ] , so that ( x, y ) = (4 , 1). See Figure 1.1. Figure 1.1: for Problem 1.1.11. 1.1. 12 [ x − 2 y = 3 2 x − 4 y = 6 ] − 2( I ) → [ x − 2 y = 3 0 = 0 ] 2

  5. Page 5
    Read page 5 as text

    Section 1.1 This system has infinitely many solutions: If we choose y = t , an arbitrary real number, then the equation x − 2 y = 3 gives us x = 3 + 2 y = 3 + 2 t . Therefore the general solution is ( x, y ) = (3 + 2 t, t ), where t is an arbitrary real number. (See Figure 1.2.) Figure 1.2: for Problem 1.1.12. 1.1. 13 [ x − 2 y = 3 2 x − 4 y = 8 ] − 2( I ) → [ x − 2 y = 3 0 = 2 ] , which has no solutions. (See Figure 1.3.) Figure 1.3: for Problem 1.1.13. 1.1. 14 The system reduces to   x + 5 z = 0 y − z = 0 0 = 1   , so that there is no solution; no point in space belongs to all three planes. Compare with Figure 2b. 1.1. 15 The system reduces to   x = 0 y = 0 z = 0   so the unique solution is ( x, y, z ) = (0 , 0 , 0). The three planes intersect at the origin. 1.1. 16 The system reduces to   x + 5 z = 0 y − z = 0 0 = 0   , so the solutions are of the form ( x, y, z ) = ( − 5 t, t, t ), where t is an arbitrary number. The three planes intersect in a line; compare with Figure 2a. 1.1. 17 [ x + 2 y = a 3 x + 5 y = b ] − 3( I ) → [ x + 2 y = a − y = − 3 a + b ] ÷ ( − 1) → [ x + 2 y = a y = 3 a − b ] − 2( II ) 3

  6. Page 6
    Read page 6 as text

    Chapter 1 [ x = − 5 a + 2 b y = 3 a − b ] , so that ( x, y ) = ( − 5 a + 2 b, 3 a − b ). 1.1. 18   x + 2 y + 3 z = a x + 3 y + 8 z = b x + 2 y + 2 z = c   − I − I →   x + 2 y + 3 z = a y + 5 z = − a + b − z = − a + c   − 2( II ) →   x − 7 z = 3 a − 2 b y + 5 z = − a + b − z = − a + c   ÷ ( − 1) →   x − 7 z = 3 a − 2 b y + 5 z = − a + b z = a − c   +7( III ) − 5( III ) →   x = 10 a − 2 b − 7 c y = − 6 a + b + 5 c z = a − c   , so that ( x, y, z ) = (10 a − 2 b − 7 c, − 6 a + b + 5 c, a − c ). 1.1. 19 The system reduces to   x + z = 1 y − 2 z = − 3 0 = k − 7   . a. The system has solutions if k − 7 = 0, or k = 7. b. If k = 7 then the system has infinitely many solutions. c. If k = 7 then we can choose z = t freely and obtain the solutions ( x, y, z ) = (1 − t, − 3 + 2 t, t ). 1.1. 20 The system reduces to   x − 3 z = 1 y + 2 z = 1 ( k 2 − 4) z = k − 2   This system has a unique solution if k 2 − 4 6 = 0, that is, if k 6 = ± 2. If k = 2, then the last equation is 0 = 0, and there will be infinitely many solutions. If k = − 2, then the last equation is 0 = − 4, and there will be no solutions. 1.1. 21 Let x , y , and z represent the three numbers we seek. We set up a system of equations and solve systematically (although there are short cuts): ∣ ∣ ∣ ∣ ∣ ∣ x + y = 24 x + z = 28 y + z = 30 ∣ ∣ ∣ ∣ ∣ ∣ − ( I ) → ∣ ∣ ∣ ∣ ∣ ∣ x + y = 24 − y + z = 4 y + z = 30 ∣ ∣ ∣ ∣ ∣ ∣ ÷ ( − 1) → ∣ ∣ ∣ ∣ ∣ ∣ x + y = 24 y − z = − 4 y + z = 30 ∣ ∣ ∣ ∣ ∣ ∣ − ( II ) − ( II ) → ∣ ∣ ∣ ∣ ∣ ∣ x + z = 28 y − z = − 4 2 z = 34 ∣ ∣ ∣ ∣ ∣ ∣ ÷ 2 → ∣ ∣ ∣ ∣ ∣ ∣ x + z = 28 y − z = − 4 z = 17 ∣ ∣ ∣ ∣ ∣ ∣ − ( III ) +( III ) → ∣ ∣ ∣ ∣ ∣ ∣ x = 11 y = 13 z = 17 ∣ ∣ ∣ ∣ ∣ ∣ We see that x = 11 , y = 13 , and z = 17 . 4

  7. Page 7
    Read page 7 as text

    Section 1.1 1.1. 22 Let x = the number of male children and y = the number of female children. Then the statement “Emile has twice as many sisters as brothers” translates into y = 2( x − 1) and “Gertrude has as many brothers as sisters” translates into x = y − 1. Solving the system [ − 2 x + y = − 2 x − y = − 1 ] gives x = 3 and y = 4. There are seven children in this family. 1.1. 23 a Note that the demand D 1 for product 1 increases with the increase of price P 2 ; likewise the demand D 2 for product 2 increases with the increase of price P 1 . This indicates that the two products are competing; some people will switch if one of the products gets more expensive. b Setting D 1 = S 1 and D 2 = S 2 we obtain the system [ 70 − 2 P 1 + P 2 = − 14 + 3 P 1 105 + P 1 − P 2 = − 7 + 2 P 2 ] , or [ − 5 P 1 + P 2 = − 84 P 1 − 3 P 2 = 112 ] , which yields the unique solution P 1 = 26 and P 2 = 46. 1.1. 24 The total demand for the product of Industry A is 1000 (the consumer demand) plus 0.1b (the demand from Industry B). The output a must meet this demand: a = 1000 + 0 . 1 b . Setting up a similar equation for Industry B we obtain the system [ a = 1000 + 0 . 1 b b = 780 + 0 . 2 a ] or [ a − 0 . 1 b = 1000 − 0 . 2 a + b = 780 ] , which yields the unique solution a = 1100 and b = 1000. 1.1. 25 The total demand for the products of Industry A is 310 (the consumer demand) plus 0 . 3 b (the demand from Industry B). The output a must meet this demand: a = 310 + 0 . 3 b . Setting up a similar equation for Industry B we obtain the system [ a = 310 + 0 . 3 b b = 100 + 0 . 5 a ] or [ a − 0 . 3 b = 310 − 0 . 5 a + b = 100 ] , which yields the solution a = 400 and b = 300. 1.1. 26 Since x ( t ) = a sin( t ) + b cos( t ) we can compute dx dt = a cos( t ) − b sin( t ) and d 2 x dt 2 = − a sin( t ) − b cos( t ). Substituting these expressions into the equation d 2 x dt 2 − dx dt − x = cos( t ) and simplifying gives ( b − 2 a ) sin( t ) + ( − a − 2 b ) cos( t ) = cos( t ). Comparing the coefficients of sin( t ) and cos( t ) on both sides of the equation then yields the system [ − 2 a + b = 0 − a − 2 b = 1 ] , so that a = − 1 5 and b = − 2 5 . See Figure 1.4. Figure 1.4: for Problem 1.1.26. 5

  8. Page 8
    Read page 8 as text

    Chapter 1 1.1. 27 a Substituting λ = 5 yields the system [ 7 x − y = 5 x − 6 x + 8 y = 5 y ] or [ 2 x − y = 0 − 6 x + 3 y = 0 ] or [ 2 x − y = 0 0 = 0 ] . There are infinitely many solutions, of the form ( x, y ) = ( t 2 , t ) , where t is an arbitrary real number. b Proceeding as in part (a), we find ( x, y ) = ( − 1 3 t, t ) . c Proceedings as in part (a), we find only the solution (0 , 0). 1.1. 28 Use the distance from Stein to Schaffhausen (and from Stein to Constance) as the unit of length. Let the speed of the boat and the speed of the river flow be v b and v s , respectively. Using the formula speed = distance time and measuring time in hours we get: v b − v s = 1 v b + v s = 3 2 We find v b = 5 4 and v s = 1 4 . The time it takes to travel from Stein to Constance is time = distance speed = 1 v b = 4 5 hours, or 48 minutes. 1.1. 29 Let v be the speed of the boat relative to the water, and s be the speed of the stream; then the speed of the boat relative to the land is v + s downstream and v − s upstream. Using the fact that (distance) = (speed)(time), we obtain the system [ 8 = ( v + s ) 1 3 8 = ( v − s ) 2 3 ] ← downstream ← upstream The solution is v = 18 and s = 6. 1.1. 30 The thermal equilibrium condition requires that T 1 = T 2 +200+0+0 4 , T 2 = T 1 + T 3 +200+0 4 , and T 3 = T 2 +400+0+0 4 . We can rewrite this system as   − 4 T 1 + T 2 = − 200 T 1 − 4 T 2 + T 3 = − 200 T 2 − 4 T 3 = − 400   The solution is ( T 1 , T 2 , T 3 ) = (75 , 100 , 125). 1.1. 31 To assure that the graph goes through the point (1 , − 1), we substitute t = 1 and f ( t ) = − 1 into the equation f ( t ) = a + bt + ct 2 to give − 1 = a + b + c . Proceeding likewise for the two other points, we obtain the system   a + b + c = − 1 a + 2 b + 4 c = 3 a + 3 b + 9 c = 13   . The solution is a = 1 , b = − 5, and c = 3, and the polynomial is f ( t ) = 1 − 5 t + 3 t 2 . (See Figure 1.5.) 6

  9. Page 9
    Read page 9 as text

    Section 1.1 Figure 1.5: for Problem 1.1.31. 1.1. 32 Proceeding as in the previous exercise, we obtain the system   a + b + c = p a + 2 b + 4 c = q a + 3 b + 9 c = r ∣ ∣ ∣ ∣ ∣ ∣ . The unique solution is   a = 3 p − 3 q + r b = − 2 . 5 p + 4 q − 1 . 5 r c = 0 . 5 p − q + 0 . 5 r   . Only one polynomial of degree 2 goes through the three given points, namely, f ( t ) = 3 p − 3 q + r + ( − 2 . 5 p + 4 q − 1 . 5 r ) t + (0 . 5 p − q + 0 . 5 r ) t 2 . 1.1. 33 f ( t ) is of the form at 2 + bt + c . So f (1) = a (1 2 ) + b (1) + c = 3, and f (2) = a (2 2 ) + b (2) + c = 6. Also, f ′ ( t ) = 2 at + b , meaning that f ′ (1) = 2 a + b = 1. So we have a system of equations:    a + b + c = 3 4 a + 2 b + c = 6 2 a + b = 1    which reduces to    a = 2 b = − 3 c = 4    . Thus, f ( t ) = 2 t 2 − 3 t + 4 is the only solution. 1.1. 34 f ( t ) is of the form at 2 + bt + c . So, f (1) = a (1 2 ) + b (1) + c = 1 and f (2) = 4 a + 2 b + c = 0. Also, ∫ 2 1 f ( t ) dt = ∫ 2 1 ( at 2 + bt + c ) dt = a 3 t 3 + b 2 t 2 + ct | 2 1 = 8 3 a + 2 b + 2 c − ( a 3 + b 2 + c ) = 7 3 a + 3 2 b + c = − 1. So we have a system of equations:   a + b + c = 1 4 a + 2 b + c = 0 7 3 a + 3 2 b + c = − 1   which reduces to   a = 9 b = − 28 c = 20   . 7

  10. Page 10
    Read page 10 as text

    Chapter 1 Thus, f ( t ) = 9 t 2 − 28 t + 20 is the only solution. 1.1. 35 f ( t ) is of the form at 2 + bt + c . f (1) = a + b + c = 1, f (3) = 9 a + 3 b + c = 3, and f ′ ( t ) = 2 at + b , so f ′ (2) = 4 a + b = 1. Now we set up our system to be    a + b + c = 1 9 a + 3 b + c = 3 4 a + b = 1    . This reduces to    a − c 3 = 0 b + 4 3 c = 1 0 = 0    . We write everything in terms of a , revealing c = 3 a and b = 1 − 4 a . So, f ( t ) = at 2 + (1 − 4 a ) t + 3 a for an arbitrary a . 1.1. 36 f ( t ) = at 2 + bt + c , so f (1) = a + b + c = 1 , f (3) = 9 a + 3 b + c = 3. Also, f ′ (2) = 3, so 2(2) a + b = 4 a + b = 3. Thus, our system is   a + b + c = 1 9 a + 3 b + c = 3 4 a + b = 3   . When we reduce this, however, our last equation becomes 0 = 2, meaning that this system is inconsistent. 1.1. 37 f ( t ) = ae 3 t + be 2 t , so f (0) = a + b = 1 and f ′ ( t ) = 3 ae 3 t + 2 be 2 t , so f ′ (0) = 3 a + 2 b = 4. Thus we obtain the system [ a + b = 1 3 a + 2 b = 4 ] , which reveals [ a = 2 b = − 1 ] . So f ( t ) = 2 e 3 t − e 2 t . 1.1. 38 f ( t ) = a cos(2 t ) + b sin(2 t ) and 3 f ( t ) + 2 f ′ ( t ) + f ′′ ( t ) = 17 cos(2 t ). f ′ ( t ) = 2 b cos(2 t ) − 2 a sin(2 t ) and f ′′ ( t ) = − 4 b sin(2 t ) − 4 a cos(2 t ). So, 17 cos(2 t ) = 3( a cos(2 t ) + b sin(2 t )) + 2(2 b cos(2 t ) − 2 a sin(2 t )) + ( − 4 b sin(2 t ) − 4 a cos(2 t )) = ( − 4 a + 4 b + 3 a ) cos(2 t ) + ( − 4 b − 4 a + 3 b ) sin(2 t ) = ( − a + 4 b ) cos(2 t ) + ( − 4 a − b ) sin(2 t ). So, our system is: [ − a + 4 b = 17 − 4 a − b = 0 ] . This reduces to: [ a = − 1 b = 4 ] . So our function is f ( t ) = − cos(2 t ) + 4 sin(2 t ). 8

  11. Page 11
    Read page 11 as text

    Section 1.1 1.1. 39 Plugging the three points ( x, y ) into the equation a + bx + cy + x 2 + y 2 = 0, leads to a system of linear equations for the three unknowns ( a, b, c ). a + 5 b + 5 c + 25 + 25 = 0 a + 4 b + 6 c + 16 + 36 = 0 a + 6 b + 2 c + 36 + 4 = 0 . The solution is a = − 20 , b = − 2 , c = − 4. − 20 − 2 x − 4 y + x 2 + y 2 = 0 is a circle of radius 5 centered at (1 , 2). 1.1. 40 Plug the three points into the equation ax 2 + bxy + cy 2 = 1. We obtain a system of linear equations a + 2 b + 4 c = 1 4 a + 4 b + 4 c = 1 9 a + 3 b + c = 1 . The solution is a = 3 / 20 , b = − 9 / 40 , c = 13 / 40. This is the ellipse (3 / 20) x 2 − (9 / 40) xy + (13 / 40) y 2 = 1. 1.1. 41 The given system reduces to    x − z = − 5 a +2 b 3 y + 2 z = 4 a − b 3 0 = a − 2 b + c    . This system has solutions (in fact infinitely many) if a − 2 b + c = 0. The points ( a, b, c ) with this property form a plane through the origin. 1.1. 42 a x 1 = − 3 x 2 = 14 + 3 x 1 = 14 + 3( − 3) = 5 x 3 = 9 − x 1 − 2 x 2 = 9 + 3 − 10 = 2 x 4 = 33 + x 1 − 8 x 2 + 5 x 3 − x 4 = 33 − 3 − 40 + 10 = 0 , so that ( x 1 , x 2 , x 3 , x 4 ) = ( − 3 , 5 , 2 , 0). b x 4 = 0 x 3 = 2 − 2 x 4 = 2 x 2 = 5 − 3 x 3 − 7 x 4 = 5 − 6 = − 1 x 1 = − 3 − 2 x 2 + x 3 − 4 x 4 = − 3 + 2 + 2 = 1 , so that ( x 1 , x 2 , x 3 , x 4 ) = (1 , − 1 , 2 , 0). 1.1. 43 a The two lines intersect unless t = 2 (in which case both lines have slope − 1). To draw a rough sketch of x ( t ), note that lim t →∞ x ( t ) = lim t →−∞ x ( t ) = − 1 ( the line x + t 2 y = t becomes almost horizontal ) and lim t → 2 − x ( t ) = ∞ , lim t → 2 + x ( t ) = −∞ . 9

  12. Page 12
    Read page 12 as text

    Chapter 1 Figure 1.6: for Problem 1.1.43a. Figure 1.7: for Problem 1.1.43a. Also note that x ( t ) is positive if t is between 0 and 2, and negative otherwise. Apply similar reasoning to y ( t ). (See Figures 1.6 and 1.7.) b x ( t ) = − t t − 2 , and y ( t ) = 2 t − 2 t − 2 . 1.1. 44 We can think of the line through the points (1 , 1 , 1) and (3 , 5 , 0) as the intersection of any two planes through these two points; each of these planes will be defined by an equation of the form ax + by + cz = d. It is required that 1 a + 1 b + 1 c = d and 3 a + 5 b + 0 c = d. Now the system [ a + b + c − d = 0 3 a +5 b − d = 0 ] reduces to [ a + 5 2 c − 2 d = 0 b − 3 2 c + d = 0 ] . We can choose arbitrary real numbers for c and d ; then a = − 5 2 c + 2 d and b = 3 2 c − d. For example, if we choose c = 2 and d = 0 , then a = − 5 and b = 3, leading to the equation − 5 x + 3 y + 2 z = 0 . If we choose c = 0 and d = 1 , then a = 2 and b = − 1 , giving the equation 2 x − y = 1. 10

  13. Page 13
    Read page 13 as text

    Section 1.1 We have found one possible answer: [ − 5 x +3 y +2 z = 0 2 x − y = 1 ] . 1.1. 45 To eliminate the arbitrary constant t , we can solve the last equation for t to give t = z − 2, and substitute z − 2 for t in the first two equations, obtaining [ x = 6 + 5( z − 2) y = 4 + 3( z − 2) ] or [ x − 5 z = − 4 y − 3 z = − 2 ] . This system does the job. 1.1. 46 Let b = Boris’ money, m = Marina’s money, and c = cost of a chocolate bar. We are told that [ b + 1 2 m = c 1 2 b + m = 2 c ] , with solution ( b, m ) = (0 , 2 c ). Boris has no money. 1.1. 47 Let us start by reducing the system:   x + 2 y + 3 z = 39 x + 3 y + 2 z = 34 3 x + 2 y + z = 26   − I − 3( I ) →   x + 2 y + 3 z = 39 y − z = − 5 − 4 y − 8 z = − 91   Note that the last two equations are exactly those we get when we substitute x = 39 − 2 y − 3 z : either way, we end up with the system [ y − z = − 5 − 4 y − 8 z = − 91 ] . 1.1. 48 a We set up two equations here, with our variables: x 1 = servings of rice, x 2 = servings of yogurt. So our system is: [ 3 x 1 +12 x 2 = 60 30 x 1 +20 x 2 = 300 ] . Solving this system reveals that x 1 = 8 , x 2 = 3. b Again, we set up our equations: [ 3 x 1 +12 x 2 = P 30 x 1 +20 x 2 = C ] , and reduce them to find that x 1 = − P 15 + C 25 , while x 2 = P 10 − C 100 . 1.1. 49 Let x 1 = number of one-dollar bills, x 2 = the number of five-dollar bills, and x 3 = the number of ten-dollar bills. Then our system looks like: [ x 1 + x 2 + x 3 = 32 x 1 + 5 x 2 + 10 x 3 = 100 ] , which reduces to give us solutions that fit: x 1 = 15 + 5 4 x 3 , x 2 = 17 − 9 4 x 3 , where x 3 can be chosen freely. Now let’s keep in mind that x 1 , x 2 , and x 3 must be positive integers and see what conditions this imposes on the variable x 3 . We see that since x 1 and x 2 must be integers, x 3 must be a multiple of 4. Furthermore, x 3 must be positive, and x 2 = 17 − 9 4 x 3 must be positive as well, meaning that x 3 < 68 9 . These constraints leave us with only one possibility, x 3 = 4, and we can compute the corresponding values x 1 = 15 + 5 4 x 3 = 20 and x 2 = 17 − 9 4 x 3 = 8. Thus, we have 20 one-dollar bills, 8 five-dollar bills, and 4 ten-dollar bills. 1.1. 50 Let x 1 , x 2 , x 3 be the number of 20 cent, 50 cent, and 2 Euro coins, respectively. Then we need solutions to the system: [ x 1 + x 2 + x 3 = 1000 . 2 x 1 + . 5 x 2 +2 x 3 = 1000 ] 11

  14. Page 14
    Read page 14 as text

    Chapter 1 this system reduces to: [ x 1 − 5 x 3 = − 5000 3 x 2 +6 x 3 = 8000 3 ] . Our solutions are then of the form   x 1 x 2 x 3   =   5 x 3 − 5000 3 − 6 x 3 + 8000 3 x 3   . Unfortunately for the meter maids, there are no integer solutions to this problem. If x 3 is an integer, then neither x 1 nor x 2 will be an integer, and no one will ever claim the Ferrari. Section 1.2 1.2. 1   1 1 − 2 . . . 5 2 3 4 . . . 2   − 2( I ) →   1 1 − 2 . . . 5 0 1 8 . . . − 8   − II →   1 0 − 10 . . . 13 0 1 8 . . . − 8   [ x − 10 z = 13 y + 8 z = − 8 ] −→ [ x = 13 + 10 z y = − 8 − 8 z ]   x y z   =   13 + 10 t − 8 − 8 t t   , where t is an arbitrary real number. 1.2. 2   3 4 − 1 . . . 8 6 8 − 2 . . . 3   ÷ 3 →   1 4 3 − 1 3 . . . 8 3 6 8 − 2 . . . 3   − 6( I ) →   1 4 3 − 1 3 . . . 8 3 0 0 0 . . . − 13   This system has no solutions, since the last row represents the equation 0 = − 13. 1.2. 3 x = 4 − 2 y − 3 z y and z are free variables; let y = s and z = t .   x y z   =   4 − 2 s − 3 t s t   , where s and t are arbitrary real numbers. 1.2. 4     1 1 . . . 1 2 − 1 . . . 5 3 4 . . . 2     − 2( I ) − 3( I ) →     1 1 . . . 1 0 − 3 . . . 3 0 1 . . . − 1     ÷ ( − 3) →     1 1 . . . 1 0 1 . . . − 1 0 1 . . . − 1     − II − II →     1 0 . . . 2 0 1 . . . − 1 0 0 . . . 0     , so that x = 2 y = − 1 . 12

  15. Page 15
    Read page 15 as text

    Section 1.2 1.2. 5       0 0 1 1 . . . 0 0 1 1 0 . . . 0 1 1 0 0 . . . 0 1 0 0 1 . . . 0       swap : I ↔ III →       1 1 0 0 . . . 0 0 1 1 0 . . . 0 0 0 1 1 . . . 0 1 0 0 1 . . . 0       − I →       1 1 0 0 . . . 0 0 1 1 0 . . . 0 0 0 1 1 . . . 0 0 − 1 0 1 . . . 0       − II + II →        1 0 − 1 0 . . . 0 0 1 1 0 . . . 0 0 0 1 1 . . . 0 0 0 1 1 . . . 0        + III − III − III →        1 0 0 1 . . . 0 0 1 0 − 1 . . . 0 0 0 1 1 . . . 0 0 0 0 0 . . . 0          x 1 + x 4 = 0 x 2 − x 4 = 0 x 3 + x 4 = 0   −→   x 1 = − x 4 x 2 = x 4 x 3 = − x 4      x 1 x 2 x 3 x 4    =    − t t − t t    , where t is an arbitrary real number. 1.2. 6 The system is in rref already.   x 1 = 3 + 7 x 2 − x 5 x 3 = 2 + 2 x 5 x 4 = 1 − x 5   Let x 2 = t and x 5 = r .      x 1 x 2 x 3 x 4 x 5      =      3 + 7 t − r t 2 + 2 r 1 − r r      1.2. 7        1 2 0 2 3 . . . 0 0 0 1 3 2 . . . 0 0 0 1 4 − 1 . . . 0 0 0 0 0 1 . . . 0        − II →        1 2 0 2 3 . . . 0 0 0 1 3 2 . . . 0 0 0 0 1 − 3 . . . 0 0 0 0 0 1 . . . 0        − 2( III ) − 3( III ) →        1 2 0 0 9 . . . 0 0 0 1 0 11 . . . 0 0 0 0 1 − 3 . . . 0 0 0 0 0 1 . . . 0        − 9( IV ) − 11( IV ) +3( IV ) →       1 2 0 0 0 . . . 0 0 0 1 0 0 . . . 0 0 0 0 1 0 . . . 0 0 0 0 0 1 . . . 0          x 1 + 2 x 2 = 0 x 3 = 0 x 4 = 0 x 5 = 0    −→    x 1 = − 2 x 2 x 3 = 0 x 4 = 0 x 5 = 0    13

  16. Page 16
    Read page 16 as text

    Chapter 1 Let x 2 = t .      x 1 x 2 x 3 x 4 x 5      =      − 2 t t 0 0 0      , where t is an arbitrary real number. 1.2. 8 [ 0 1 0 2 3 . . . 0 0 0 0 4 8 . . . 0 ] ÷ 4 → [ 0 1 0 2 3 . . . 0 0 0 0 1 2 . . . 0 ] − 2( II )   0 1 0 0 − 1 . . . 0 0 0 0 1 2 . . . 0   [ x 2 − x 5 = 0 x 4 + 2 x 5 = 0 ] −→ [ x 2 = x 5 x 4 = − 2 x 5 ] Let x 1 = r, x 3 = s, x 5 = t .      x 1 x 2 x 3 x 4 x 5      =      r t s − 2 t t      , where r, t and s are arbitrary real numbers. 1.2. 9     0 0 0 1 2 − 1 . . . 2 1 2 0 0 1 − 1 . . . 0 1 2 2 0 − 1 1 . . . 2     swap : I ↔ II →     1 2 0 0 1 − 1 . . . 0 0 0 0 1 2 − 1 . . . 2 1 2 2 0 − 2 1 . . . 2     − I →     1 2 0 0 1 − 1 . . . 0 0 0 0 1 2 − 1 . . . 2 0 0 2 0 − 2 2 . . . 2     swap : II ↔ III →     1 2 0 0 1 − 1 . . . 0 0 0 2 0 − 2 2 . . . 2 0 0 0 1 2 − 1 . . . 2     ÷ 2 →     1 2 0 0 1 − 1 . . . 0 0 0 1 0 − 1 1 . . . 1 0 0 0 1 2 − 1 . . . 2       x 1 + 2 x 2 + x 5 − x 6 = 0 x 3 − x 5 + x 6 = 1 x 4 + 2 x 5 − x 6 = 2   −→   x 1 = − 2 x 2 − x 5 + x 6 x 3 = 1 + x 5 − x 6 x 4 = 2 − 2 x 5 + x 6   Let x 2 = r, x 5 = s , and x 6 = t .        x 1 x 2 x 3 x 4 x 5 x 6        =        − 2 r − s + t r 1 + s − t 2 − 2 s + t s t        , where r, s and t are arbitrary real numbers. 1.2. 10 The system reduces to   x 1 + x 4 = 1 x 2 − 3 x 4 = 2 x 3 + 2 x 4 = − 3   −→   x 1 = 1 − x 4 x 2 = 2 + 3 x 4 x 3 = − 3 − 2 x 4   14

427 more pages in the full document

Unlock it once and it stays in your library, ready to chat with or turn into flashcards.

Unlock the full document

More Mathematics study materials

See all

Study this document with CramX

Once it is in your library, every tool can work from it.