Linear algebra with applications otto bretscher 5th edition solutions pdf

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Solution Manual for Linear Algebra with Applications 5th Edition by Bretscher Section 1.1 Chapter Section 1.1 x + 2y = 2x + 3y = 1.1.1 x + 2y = y=1 4x + 3y 7x + 5y 1.1.2 x + 34 y − 14 y −2 × 1st equation −2 × 2nd equation =2 =3 = 21 = 12 ữ4 ì(4) x + 2y y → x y → = 7x + 5y =3 −7 × 1st equation = 12 =2 x + 43 y y ÷(−1) → = −1 , so that (x, y) = (−1, 1) =1 x + 34 y → =1 = −1 → − 43 × 2nd equation → x = −1 , y = so that (x, y) = (−1, 2) 2x + 4y 3x + 6y 1.1.3 =3 =2 ÷2 → x + 2y 3x + 6y = 23 =2 → x + 2y 3x + 6y =1 =3 −3 × 1st equation → x + 2y = 32 = − 52 → x + 2y =1 =0 So there is no solution 2x + 4y 3x + 6y 1.1.4 =2 =3 ÷2 −3 × 1st equation This system has infinitely many solutions: if we choose y = t, an arbitrary real number, then the equation x + 2y = gives us x = − 2y = − 2t Therefore the general solution is (x, y) = (1 − 2t, t), where t is an arbitrary real number 2x + 3y 4x + 5y 1.1.5 x + 23 y −y =0 =0 ÷2 =0 =0 ÷(−1) → x + 23 y 4x + 5y → x + 23 y y =0 =0 =0 =0 −4 × 1st equation − 23 × 2nd equation → → x =0 , y =0 so that (x, y) = (0, 0)   x + 2y + 3z x + 2y + 3z = y 1.1.6  x + 3y + 3z = 10  −I →  z x + 2y + 4z = −I     x + 3z = −3(III) x=1  y = 2 →  y = , so that z =1 z=1   x + 2y + 3z 1.1.7  x + 3y + 4z x + 4y + 5z   x + 2y + 3z =1 y+z =  −I →  2y + 2z = −I This system has no solution  =8 −2(II) → = 2 =1 (x, y, z) = (1, 2, 1)   x+z = −2(II) → y +z = 2 = −2(II)  = −3 = 2 = −1 Copyright c 2013 Pearson Education, Inc Full file at https://TestbankDirect.eu/ Solution Manual for Linear Algebra with Applications 5th Edition by Bretscher Chapter    x + 2y + 3z = x + 2y + 3z =0 =  ÷(−3) → 1.1.8  4x + 5y + 6z =  −4(I) →  −3y − 6z −6y − 11z = 7x + 8y + 10z = −7(I)       x + 2y + 3z = −2(II) x−z =0 +III x =0  y + 2z = 0 →  y + 2z =  −2(III) →  y = , −6y − 11z = +6(II) z =0 z =0  so that (x, y, z) = (0, 0, 0)     x + 2y + 3z = x + 2y + 3z = 1.1.9  3x + 2y + z =  −3(I) →  −4y − 8z = −2  ÷(−4) → 7x + 2y − 3z = −7(I) −12y − 24z = −6     −2(II) x−z =0 x + 2y + 3z =  y + 2z →  y + 2z = 12  = 12  −12y − 24z = −6 +12(II) =0 This system has infinitely many solutions: if we choose z = t, an arbitrary real number, then we get x = z = t and y = 21 − 2z = 12 − 2t Therefore, the general solution is (x, y, z) = t, 21 − 2t, t , where t is an arbitrary real number  x + 2y + 3z 1.1.10  2x + 4y + 7z 3x + 7y + 11z  x + 2y + 3z  y + 2z z   x + 2y + 3z =1 z =  −2(I) →  y + 2z = −3(I)   x−z = −2(II)   =5 → y + 2z =0 z so that (x, y, z) = (−9, 5, 0) 1.1.11 x − 2y 3x + 5y =2 = 17 −3(I) x − 2y 11y → = −9   =1 Swap : = 0 → II ↔ III =5   +III x = −9 =  −2(III) →  y = , z =0 =0 =2 = 11 ÷11 → x − 2y y =2 =1 +2(II) so that (x, y) = (4, 1) See Figure 1.1 Figure 1.1: for Problem 1.1.11 1.1.12 x − 2y 2x − 4y =3 =6 −2(I) → x − 2y =3 =0 Copyright c 2013 Pearson Education, Inc Full file at https://TestbankDirect.eu/ → x y =4 , =1 Solution Manual for Linear Algebra with Applications 5th Edition by Bretscher Section 1.1 This system has infinitely many solutions: If we choose y = t, an arbitrary real number, then the equation x − 2y = gives us x = + 2y = + 2t Therefore the general solution is (x, y) = (3 + 2t, t), where t is an arbitrary real number (See Figure 1.2.) Figure 1.2: for Problem 1.1.12 1.1.13 x − 2y 2x − 4y =3 =8 −2(I) → x − 2y =3 , which has no solutions (See Figure 1.3.) =2 Figure 1.3: for Problem 1.1.13  =0 = , so that there is no solution; no point in space belongs to all three =1  x + 5z 1.1.14 The system reduces to  y − z planes Compare with Figure 2b  x =0 1.1.15 The system reduces to  y =  so the unique solution is (x, y, z) = (0, 0, 0) The three planes intersect at z =0 the origin   x + 5z 1.1.16 The system reduces to  y − z arbitrary number The three planes 1.1.17 x + 2y 3x + 5y =a =b −3(I) →  =0 =  , so the solutions are of the form (x, y, z) = (−5t, t, t), where t is an =0 intersect in a line; compare with Figure 2a x + 2y −y =a = −3a + b ÷(−1) → x + 2y y =a = 3a − b Copyright c 2013 Pearson Education, Inc Full file at https://TestbankDirect.eu/ −2(II) Solution Manual for Linear Algebra with Applications 5th Edition by Bretscher Chapter x = −5a + 2b , so that (x, y) = (−5a + 2b, 3a − b) y = 3a − b  x + 2y + 3z 1.1.18  x + 3y + 8z x + 2y + 2z  x − 7z  y + 5z −z   =a x + 2y + 3z = b  −I →  y + 5z = c −I −z   = 3a − 2b x − 7z = −a + b  →  y + 5z = −a + c ÷(−1) z  =a −2(II) = −a + b  → = −a + c   = 3a − 2b +7(III) x = −a + b  −5(III) →  y =a−c z  = 10a − 2b − 7c = −6a + b + 5c , =a−c so that (x, y, z) = (10a − 2b − 7c, −6a + b + 5c, a − c) x+z 1.1.19 The system reduces to  y − 2z   =1 = −3  =k−7 a The system has solutions if k − = 0, or k = b If k = then the system has infinitely many solutions c If k = then we can choose z = t freely and obtain the solutions (x, y, z) = (1 − t, −3 + 2t, t) x − 3z 1.1.20 The system reduces to  y + 2z (k − 4)z  = = =  1  k−2 This system has a unique solution if k − = 0, that is, if k = ±2 If k = 2, then the last equation is = 0, and there will be infinitely many solutions If k = −2, then the last equation is = −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 −(I) → y+ z = 30 x +z = 28 y −z = −4 → → ÷2 2z = 34 x +y = 24 −y +z = y+ z = 30 x +z = 28 y −z = −4 z = 17 ÷(−1) → −(III) +(III) We see that x = 11, y = 13, and z = 17 Copyright c 2013 Pearson Education, Inc Full file at https://TestbankDirect.eu/ → x +y y y+ x y −z z z = 24 = −4 = 30 = 11 = 13 = 17 −(II) −(II) Solution Manual for Linear Algebra with Applications 5th Edition by Bretscher 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 − Solving the system −2x + y x−y = −2 = −1 gives x = and y = There are seven children in this family 1.1.23 a Note that the demand D1 for product increases with the increase of price P2 ; likewise the demand D2 for product increases with the increase of price P1 This indicates that the two products are competing; some people will switch if one of the products gets more expensive 70 − 2P1 + P2 105 + P1 − P2 which yields the unique solution P1 = 26 and P2 = 46 = −14 + 3P1 , or = −7 + 2P2 b Setting D1 = S1 and D2 = S2 we obtain the system −5P1 + P2 P1 − 3P2 = −84 , = 112 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 + 0.1b Setting up a similar equation for Industry B we obtain the system a b = 1000 + 0.1b a − 0.1b or = 780 + 0.2a −0.2a + b which yields the unique solution a = 1100 and b = 1000 = 1000 , = 780 1.1.25 The total demand for the products of Industry A is 310 (the consumer demand) plus 0.3b (the demand from Industry B) The output a must meet this demand: a = 310 + 0.3b Setting up a similar equation for Industry B we obtain the system a b = 310 + 0.3b = 100 + 0.5a 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 or a − 0.3b −0.5a + b = 310 , = 100 d2 x dt2 = −a sin(t) − b cos(t) Substituting these expressions into the equation ddt2x − dx dt − x = cos(t) and simplifying gives (b − 2a) sin(t) + (−a − 2b) cos(t) = cos(t) Comparing the coefficients of sin(t) and cos(t) on both sides of −2a + b = the equation then yields the system , so that a = − 51 and b = − 25 See Figure 1.4 −a − 2b = Figure 1.4: for Problem 1.1.26 Copyright c 2013 Pearson Education, Inc Full file at https://TestbankDirect.eu/ Solution Manual for Linear Algebra with Applications 5th Edition by Bretscher Chapter Full file at https://TestbankDirect.eu/ 1.1.27 a Substituting λ = yields the system 7x − y −6x + 8y = 5x = 5y or 2x − y −6x + 3y =0 =0 or 2x − y =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) = − 13 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 vb and vs , respectively Using the formula speed = distance time and measuring time in hours we get: vb − vs = vb + vs = We find vb = 54 and vs = 41 The time it takes to travel from Stein to Constance is time = or 48 minutes distance speed = vb = hours, 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 = (v + s) 31 s) 23 = (v − ← downstream ← upstream The solution is v = 18 and s = 1.1.30 The thermal equilibrium condition requires that T1 =  −4T1 + T2 We can rewrite this system as  T1 − 4T2 + T3 T2 − 4T3 T2 +200+0+0 , T2 = T1 +T3 +200+0 , and T3 = T2 +400+0+0  = −200 = −200  = −400 The solution is (T1 , T2 , T3 ) = (75, 100, 125) 1.1.31 To assure that the graph goes through the point (1, −1), we substitute t = and f (t) = −1 into the equation f (t) = a + bt + ct2 to give −1 = a + b + c   a+b+c = −1 Proceeding likewise for the two other points, we obtain the system  a + 2b + 4c =  a + 3b + 9c = 13 The solution is a = 1, b = −5, and c = 3, and the polynomial is f (t) = − 5t + 3t2 (See Figure 1.5.) Copyright c 2013 Pearson Education, Inc Full file at https://TestbankDirect.eu/ Solution Manual for Linear Algebra with Applications 5th Edition by Bretscher Section 1.1 Full file at https://TestbankDirect.eu/ Figure 1.5: for Problem 1.1.31  a+b+c 1.1.32 Proceeding as in the previous exercise, we obtain the system  a + 2b + 4c a + 3b + 9c   a = 3p − 3q + r The unique solution is  b = −2.5p + 4q − 1.5r  c = 0.5p − q + 0.5r =p =q =r Only one polynomial of degree goes through the three given points, namely, f (t) = 3p − 3q + r + (−2.5p + 4q − 1.5r)t + (0.5p − q + 0.5r)t2 1.1.33 f (t) is of the form at2 + bt + c So f (1) = a(12 ) + b(1) + c = 3, and f (2) = a(22 ) + b(2) + c = Also, f ′ (t) = 2at + b, meaning that f ′ (1) = 2a + b =   a+b+c=3   So we have a system of equations:  4a + 2b + c =  2a + b =   a=2   which reduces to  b = −3  c=4 Thus, f (t) = 2t2 − 3t + is the only solution 1.1.34 f (t) is of the form at2 + bt + c So, f (1) = a(12 ) + b(1) + c = and f (2) = 4a + 2b + c = Also, 2 f (t)dt = (at2 + bt + c)dt = a3 t3 + 2b t2 + ct|21 = 83 a + 2b + 2c − ( a3 + b + c) = 73 a + 23 b + c = −1   a+b+c=1 So we have a system of equations:  4a + 2b + c =  3 a + b + c = −1   a=9 which reduces to  b = −28  c = 20 Copyright c 2013 Pearson Education, Inc Full file at https://TestbankDirect.eu/ Solution Manual for Linear Algebra with Applications 5th Edition by Bretscher Chapter Full file at https://TestbankDirect.eu/ Thus, f (t) = 9t2 − 28t + 20 is the only solution 1.1.35 f (t) is of the form at2 + bt + c f (1) = a + b + c = 1, f (3) = 9a + 3b + c = 3, and f ′ (t) = 2at + b, so f ′ (2) = 4a + b =   a+b+c=1   Now we set up our system to be  9a + 3b + c =  4a + b = a − 3c =   This reduces to  b + 43 c =  0=0   We write everything in terms of a, revealing c = 3a and b = − 4a So, f (t) = at2 + (1 − 4a)t + 3a for an arbitrary a 1.1.36 f (t) = at2 + bt + c, so f (1) = a + b + c = 1, f (3) = 9a + 3b + c = Also, f ′ (2) = 3, so 2(2)a + b = 4a + b =   a+b+c=1 Thus, our system is  9a + 3b + c =  4a + b = When we reduce this, however, our last equation becomes = 2, meaning that this system is inconsistent 1.1.37 f (t) = ae3t + be2t , so f (0) = a + b = and f ′ (t) = 3ae3t + 2be2t , so f ′ (0) = 3a + 2b = Thus we obtain the system which reveals a+b=1 , 3a + 2b = a=2 b = −1 So f (t) = 2e3t − e2t 1.1.38 f (t) = a cos(2t) + b sin(2t) and 3f (t) + 2f ′ (t) + f ′′ (t) = 17 cos(2t) f ′ (t) = 2b cos(2t) − 2a sin(2t) and f ′′ (t) = −4b sin(2t) − 4a cos(2t) So, 17 cos(2t) = 3(a cos(2t) + b sin(2t)) + 2(2b cos(2t) − 2a sin(2t)) + (−4b sin(2t) − 4a cos(2t)) = (−4a + 4b + 3a) cos(2t) + (−4b − 4a + 3b) sin(2t) = (−a + 4b) cos(2t) + (−4a − b) sin(2t) So, our system is: This reduces to: −a + 4b = 17 −4a − b = a = −1 b=4 So our function is f (t) = − cos(2t) + sin(2t) Copyright c 2013 Pearson Education, Inc Full file at https://TestbankDirect.eu/ Solution Manual for Linear Algebra with Applications 5th Edition by Bretscher Section 1.1 Full file at https://TestbankDirect.eu/ 1.1.39 Plugging the three points (x, y) into the equation a + bx + cy + x2 + y = 0, leads to a system of linear equations for the three unknowns (a, b, c) a + 5b + 5c + 25 + 25 a + 4b + 6c + 16 + 36 = = a + 6b + 2c + 36 + = 0 The solution is a = −20, b = −2, c = −4 −20 − 2x − 4y + x2 + y = is a circle of radius centered at (1, 2) 1.1.40 Plug the three points into the equation ax2 + bxy + cy = We obtain a system of linear equations a + 2b + 4c = 4a + 4b + 4c 9a + 3b + c = = 1 The solution is a = 3/20, b = −9/40, c = 13/40 This is the ellipse (3/20)x2 − (9/40)xy + (13/40)y =  −5a+2b = 4a−b x−z  1.1.41 The given system reduces to  y + 2z = = a − 2b + c    This system has solutions (in fact infinitely many) if a − 2b + c = The points (a, b, c) with this property form a plane through the origin 1.1.42 a x1 = −3 x2 = 14 + 3x1 = 14 + 3(−3) = x3 = − x1 − 2x2 = + − 10 = x4 = 33 + x1 − 8x2 + 5x3 − x4 = 33 − − 40 + 10 = 0, so that (x1 , x2 , x3 , x4 ) = (−3, 5, 2, 0) b x4 = x3 = − 2x4 = x2 = − 3x3 − 7x4 = − = −1 x1 = −3 − 2x2 + x3 − 4x4 = −3 + + = 1, so that (x1 , x2 , x3 , x4 ) = (1, −1, 2, 0) 1.1.43 a The two lines intersect unless t = (in which case both lines have slope −1) To draw a rough sketch of x(t), note that limt→∞ x(t) = limt→−∞ x(t) = −1 the line x + 2t y = t becomes almost horizontal and limt→2− x(t) = ∞, limt→2+ x(t) = −∞ Copyright c 2013 Pearson Education, Inc Full file at https://TestbankDirect.eu/ Solution Manual for Linear Algebra with Applications 5th Edition by Bretscher Chapter Full file at https://TestbankDirect.eu/ 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 and 2, and negative otherwise Apply similar reasoning to y(t) (See Figures 1.6 and 1.7.) b x(t) = −t t−2 , 2t−2 t−2 and y(t) = 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 1a + 1b + 1c = d and 3a + 5b + 0c = d Now the system a b + 25 c − 32 c a 3a −2d +d +b +5b = = +c −d −d = = 0 reduces to We can choose arbitrary real numbers for c and d; then a = − 25 c + 2d and b = 23 c − d For example, if we choose c = and d = 0, then a = −5 and b = 3, leading to the equation −5x + 3y + 2z = If we choose c = and d = 1, then a = and b = −1, giving the equation 2x − y = 10 Copyright c 2013 Pearson Education, Inc Full file at https://TestbankDirect.eu/ Solution Manual for Linear Algebra with Applications 5th Edition by Bretscher Section 1.3 Full file at https://TestbankDirect.eu/ 1.2.80 For each of the three statements, we set up an equation of the form (initial amount of grass) + (grass growth) = (grass consumed by cows), or (#of f ields)x + (#of f ields)(#of days)y = (#of cows)(#of days)z For the first statement, this produces the equation x + 2y = 6z, or x + 2y − 6z = Similarly, we obtain the equations 4x + 16y − 28z = and 2x + 10y − 15z = for the other two statements From this information, we     −5 −6        write the matrix   16 −28  , which reduces to  −  Thus our solutions are of the 10 −15 0 0     x 5t form  y  =  21 t  , where t is an arbitrary positive real number t z Section 1.3 1.3.1 a No solution, since the last row indicates = b The unique solution is x = 5, y = c Infinitely many solutions; the first variable can be chosen freely 1.3.2 The rank is since each row contains a leading one   1 1.3.3 This matrix has rank since its rref is  0  0   −1 1.3.4 This matrix has rank since its rref is  2 0 1.3.5 a x +y = 11 b The solution of the system in part (a) is x = 3, y = (See Figure 1.13.) 1.3.6 No solution, since any linear combination xv1 + yv2 of v1 and v2 will be parallel to v1 and v2 1.3.7 A unique solution, since there is only one parallelogram with sides along v1 and v2 and one vertex at the tip of v3 1.3.8 Infinitely many solution There are at least two obvious solutions Write v4 as a linear combination of v1 and v2 alone or as a linear combination of v3 and v2 alone Therefore, this linear system has infinitely many solutions, by Theorem 1.3.1 35 Copyright c 2013 Pearson Education, Inc Full file at https://TestbankDirect.eu/ Solution Manual for Linear Algebra with Applications 5th Edition by Bretscher Chapter Full file at https://TestbankDirect.eu/ Figure 1.13: for Problem 1.3.5  1.3.9      x 6y  = 4 9 z     1 1.3.10   ·  −2  = · + · (−2) + · = 1.3.11 Undefined since the two vectors not have the same number of components   6 1.3.12 [1 4] ·   = · + · + · + · = 70 1.3.13 29 =7 + 11 = 11 65 or   −1  1.3.14  = −1 +2 +1 = 4 1 · (−1) + · + · = · (−1) + · + · · + · 11 29 = = 11 · + · 11 65 or   −1  2 =   6 1.3.15 [1 4]   = · + · + · + · = 70 either way 1.3.16 −3 · + · (−3) = = · + · (−3) −3 36 Copyright c 2013 Pearson Education, Inc Full file at https://TestbankDirect.eu/ Solution Manual for Linear Algebra with Applications 5th Edition by Bretscher Section 1.3 Full file at https://TestbankDirect.eu/ 1.3.17 Undefined, since the matrix has three columns, but the vector has only two components        1 =   +   =  11  1.3.18   17 5   1 1.3.19  −5 −5  1.3.20 a  b           −1 1 −1    =  −5  +   +   =   −5 3  6 −9 18 27 36 45  158  70  1.3.21   81 123   1.3.22 By Theorem 1.3.4, the rref is  0  0 1 0 1.3.23 All variables are leading, that is, there is a leading one in each column of the rref: 0  0 1.3.24 By Theorem 1.3.4, rref (A) =  0  0 0 0 0  0   0  1.3.25 In this case, rref(A) has a row of zeros, so that rank(A) < 4; there will be a free variable The system Ax = c could have infinitely many solutions (for example, when c = 0) or no solutions (for example, when c = b), but it cannot have a unique solution, by Theorem 1.3.4 0 1.3.26 From Example 4c we know that rank(A) = 3, so that rref(A) =  0  0  0  Since all variables are leading, the system Ax = c cannot have infinitely many solutions, but it could have a unique solution (for example, if c = b) or no solutions at all (compare with Example 4d) 37 Copyright c 2013 Pearson Education, Inc Full file at https://TestbankDirect.eu/ Solution Manual for Linear Algebra with Applications 5th Edition by Bretscher Chapter Full file at https://TestbankDirect.eu/ 0 1.3.27 By Theorem 1.3.4, rref (A) =  0  0 0  0  1 0  1.3.28 There must be a leading one in each column: rref (A) =   0  1.3.29 A is  a and  0  a 0 of the form  b  0 c       0 5a b    =  3b  =   c −9 −9c 2 0  0    0  So a = 25 , b = and c = − 19 , and A =  0  0 0  − 19      b c e f    =   Thus, 5a + 3b − 9c = 2, 5d + 3e − 9f = 0, and h i −9 matrix to have rank is to make all the entries  in the second and third 0 columns zero, meaning that a = 52 , d = 0, and g = 15 Thus, one possible matrix is  0  0 a 1.3.30 We must satisfy the equation  d g 5g + 3h − 9i = One way to force our 1.3.31 A is  a and  0   a b c of the form  d e  0 f       5a + 3b − 9c b c d e    =  3d − 9e  =   −9 −9f f Clearly, f must equal − 91 Then, since 3d = 9e, we can choose any non-zero value for the free variable e, and d will be 3e So, if we choose for e, then d = 3e = Lastly, we must resolve 5a + 3b − 9c = Here, b and c are If we let b = c = Then, a = 2−3(1)+9(1) = 58 the free variables, and a = 2−3b+9c 5 8  1 So, in our example, A =  1 0 − 91 1.3.32 For this problem, we set up the same three equations as in Exercise 30 However, here, we  must enforce that  −2 −2 −2 our matrix, A, contains no zero entries One possible solution to this problem is the matrix   −1 −1 −1 38 Copyright c 2013 Pearson Education, Inc Full file at https://TestbankDirect.eu/ Solution Manual for Linear Algebra with Applications 5th Edition by Bretscher Section 1.3 Full file at https://TestbankDirect.eu/  x1  x2      1.3.33 The ith component of Ax is [0 0]   = xi (The is in the ith position.)  xi    xn  Therefore, Ax = x       c b a 1.3.34 a Ae1 =  d  , Ae2 =  e , and Ae3 =  f  k h g   b Be1 = [v1 v2 v3 ]   = 1v1 + 0v2 + 0v3 = v1 Likewise, Be2 = v2 and Be3 = v3 1.3.35 Write A = [v1 v2 vi vm ], then         Aei = [v1 v2 vi vm ]   = 0v1 + 0v2 + · · · + 1vi + · · · + 0vm = vi = ith column of A      1.3.36 By Exercise 35, the ith column of A is Aei , for i = 1, 2, Therefore, A =  x1 = − 2x2 or x3 =1     − 2t x1 Let x2 = t Then the solutions are of the form  x2  =  t , where t is an arbitrary real number x3 1.3.37 We have to solve the system x1 + 2x2   x3 = = 1.3.38 We will illustrate our reasoning with an example We generate the “random” × matrix   0.141 0.592 0.653 A =  0.589 0.793 0.238  0.462 0.643 0.383 Since the entries of this matrix are chosen from a large pool of numbers (in our case 1000, from 0.000 to 0.999), it is unlikely that any of the entries will be zero (and even less likely that the whole first column will consist of zeros) means that we will usually be able to to turn the first   apply the Gauss-Jordan  column  This  algorithm  4.199 4.631 0.141 0.592 0.653 into  ; this is indeed possible in our example:  0.589 0.793 0.238  −→  −1.680 −2.490  −1.297 −1.757 0.462 0.643 0.383 39 Copyright c 2013 Pearson Education, Inc Full file at https://TestbankDirect.eu/ Solution Manual for Linear Algebra with Applications 5th Edition by Bretscher Chapter Full file at https://TestbankDirect.eu/ Again, it is unlikely thatany entries in the second column of the new matrix will be zero Therefore, we can turn the second column into     0 Likewise, we will be able to clear up the third column, so that rref(A) =   0 We summarize: As we apply Gauss-Jordan elimination to a random matrix A (of any size), it is unlikely that we will ever encounter a zero on the diagonal Therefore, rref(A) is likely to have all ones along the diagonal   a b , where a, b, and c are arbitrary c 1.3.39 We will usually get rref(A) =  0 0 1.3.40 We will usually have rref(A) =  0  0  0  (Compare with the summary to Exercise 38.)  1.3.41 If Ax = b is a “random” system, then rref(A) will usually be  0 solution  0 , so that we will have a unique 1.3.42 If Ax = b is a “random” system of three equations with four unknowns, then rref(A) will usually be   0 a  b  (by Exercise 39), so that the system will have infinitely many solutions (x4 is a free variable) 0 c 1.3.43 If Ax = b is a “random” system of equations with three unknowns, then rref[A b] will usually be   0    0    , so that the system is inconsistent   0  0 1.3.44 Let E = rref(A), and note that all the entries in the last row of E must be zero, by the definition of rref If c is any vector in Rn whose last component isn’t zero, then the system Ex = c will be inconsistent Now consider the elementary row operations that transform A into E, and apply the opposite operations, in reversed order, to the augmented matrix E c You end up with an augmented matrix A b that represents an inconsistent system Ax = b, as required 40 Copyright c 2013 Pearson Education, Inc Full file at https://TestbankDirect.eu/ Solution Manual for Linear Algebra with Applications 5th Edition by Bretscher Section 1.3 Full file at https://TestbankDirect.eu/     x1 kx1 1.3.45 Write A = [v1 v2 vm ] and x =   Then A(kx) = [v1 vm ]   = kx1 v1 + · · · + kxm vm and xm kxm k(Ax) = k(x1 v1 + · · · + xm vm ) = kx1 v1 + · · · + kxm vm The two results agree, as claimed 1.3.46 Since a, d, and f are all nonzero, we can divide the first row by a, the second row by d, and the third row by f to obtain   ab ac  e  d 0 It follows that the rank of the matrix is 1.3.47 a x = is a solution b This holds by part (a) and Theorem 1.3.3 c If x1 and x2 are solutions, then Ax1 = and Ax2 = Therefore, A(x1 + x2 ) = Ax1 + Ax2 = + = 0, so that x1 + x2 is a solution as well Note that we have used Theorem 1.3.10a d A(kx) = k(Ax) = k = We have used Theorem 1.3.10b 1.3.48 The fact that x1 is a solution of Ax = b means that Ax1 = b a A(x1 + xh ) = Ax1 + Axh = b + = b b A(x2 − x1 ) = Ax2 − Ax1 = b − b = c Parts (a) and (b) show that the solutions of Ax = b are exactly the vectors of the form x1 + xh , where xh is a solution of Ax = 0; indeed if x2 is a solution of Ax = b, we can write x2 = x1 + (x2 − x1 ), and x2 − x1 will be a solution of Ax = 0, by part (b) Geometrically, the vectors of the form x1 + xh are those whose tips are on the line L in Figure 1.14; the line L runs through the tip of x1 and is parallel to the given line consisting of the solutions of Ax = Figure 1.14: for Problem 1.3.48c 41 Copyright c 2013 Pearson Education, Inc Full file at https://TestbankDirect.eu/ Solution Manual for Linear Algebra with Applications 5th Edition by Bretscher Chapter Full file at https://TestbankDirect.eu/ 1.3.49 a This system has either infinitely many solutions (if the right-most column of rref[A b] does not contain a leading one), or no solutions (if the right-most column does contain a leading one) b This system has either a unique solution (if rank[A b] = 3), or no solution (if rank[A b] = 4) c The right-most column of rref[A b] must contain a leading one, so that the system has no solutions d This system has infinitely many solutions, since there is one free variable 1.3.50 The right-most column of rref[A b] must contain a leading one, so that the system has no solutions 1.3.51 For Bx to be defined, the number of columns of B, which is s, must equal the number of components of x, which is p, so that we must have s = p Then Bx will be a vector in Rr ; for A(Bx) to be defined we must have m = r Summary: We must have s = p and m = r 1.3.52 A(Bx) = A so that C = −1 x1 x2 = 1 −x2 x1 = −x2 2x1 − x2 = −1 −1 x1 , x2 −1 −1   x1 1.3.53 Yes; write A = [v1 vm ], B = [w1 wm ], and x =   xm   x1 Then (A + B)x = [v1 + w1 vm + wm ]   = x1 (v1 + w1 ) + · · · + xm (vm + wm ) and xm     x1 x1 Ax + Bx = [v1 vm ]   + [w1 wm ]   = x1 v1 + · · · + xm vm + x1 w1 + · · · + xm wm xm xm The two results agree, as claimed 1.3.54 The vectors of the form c1 v1 + c2 v2 form a plane through the origin containing v1 and v2 ; in Figure 1.15 we draw a typical vector in this plane       1.3.55 We are looking for constants a and b such that a   + b   =       a + 4b = The resulting system  2a + 5b =  has the unique solution a = −1, b = 2, so that   is indeed a linear 3a + 6b =     combination of the vector   and   42 Copyright c 2013 Pearson Education, Inc Full file at https://TestbankDirect.eu/ Solution Manual for Linear Algebra with Applications 5th Edition by Bretscher Section 1.3 Full file at https://TestbankDirect.eu/ Figure 1.15: for Problem 1.3.54          −2 30  −5  2 6 7  −1            1.3.56 We can use technology to determine that the system  38  = x1   + x2   + x3   + x4   is           56 62   30  −1    inconsistent; therefore, the vector  38  fails to be a linear combination of the other four vectors   56 62  1.3.57 Pick a vector on each line, say Then write 11 on y = as a linear combination of x 2 and and = +b : a 11 3 on y = 3x The unique solution is a = 2, b = 3, so that the desired representation is is on the line y = x2 ; + = 11 is on line y = 3x         −1 1.3.58 We want  b  = k1   + k2   + k3  −3  , for some k1 , k2 and k3 −2 c       1 Note that we can rewrite this right-hand side as k1   + 2k2   − k3   2   = (k1 + 2k2 − k3 )   It follows that k1 + 2k2 − k3 = 3, so that b = and c = 43 Copyright c 2013 Pearson Education, Inc Full file at https://TestbankDirect.eu/ Solution Manual for Linear Algebra with Applications 5th Edition by Bretscher Chapter Full file at https://TestbankDirect.eu/         a+b 1    a + 2b  1 7 1.3.59   = a   + b   =   a + 3b c a + 4b d a +b = , which we quickly solve to find a = and b = Then, c = a +2b = a + 3b = + = and d = a + 4b = + = 11 So we have a small system:           a k2 + 2k3 b 0 0 0   1.3.60 We need   = k1   + k2   + k3   =   From this we see that a, c and d can c 3k1 + 4k2 + 5k3 d 0 6k3 be any value, while b must equal zero 1.3.61 We need to solve the system      1  c  = x  + y  c2  with augmented matrix      The matrix reduces to         c   c2  c−2 c2 − 5c +     This system is consistent if and only if c = or c = Thus the vector is a linear combination if c = or c = 1.3.62 We need to solve the system       1  c  = x a  + y b  c2 a2 b2 with augmented matrix     a  a2 The matrix reduces to    b − a  0  b b2    c   c2 c−a (c − a)(c − b)      This system is consistent if and only if c = a or c = b Thus the vector is a linear combination if c = a or c = b 44 Copyright c 2013 Pearson Education, Inc Full file at https://TestbankDirect.eu/ Solution Manual for Linear Algebra with Applications 5th Edition by Bretscher True or False Full file at https://TestbankDirect.eu/ 1.3.63 This is the line parallel to w which goes through the end point of the vector v 1.3.64 This is the line segment connecting the head of the vector v to the head of the vector v + w 1.3.65 This is the full parallelogram spanned by the two vectors v and w 1.3.66 Write b = − a and av + bw = av + (1 − a)w = w + a(v − w) to see that this is the line segment connecting the head of the vector v to the head of the vector w 1.3.67 This is the full triangle with its vertices at the origin and at the heads of the vectors v and w 1.3.68 Writing u · v = u · w as u · (v − w) = 0, we see that this is the line perpendicular to the vector v − w  0 1.3.69 We write out the augmented matrix:  1 1 So x = −a+b+c , y= a−b+c and z = a+b−c a  0   b  and reduce it to   c 0   −a+b+c a−b+c a+b−c      1.3.70 We find it useful to let s = x1 + x2 + · · · + xn Adding up all n equations of the system, and realizing that n Now the ith the term xi is missing from the ith equation, we see that (n − 1)s = b1 + · · · + bn , or, s = b1 +···+b n−1 b1 +···+bn equation of the system can be written as s − xi = bi , so that xi = s − bi = n−1 − bi True or False Ch 1.TF.1 T, by Theorem 1.3.8 Ch 1.TF.2 T, by Definition 1.3.9 Ch 1.TF.3 T, by Definition Ch 1.TF.4 F; Consider the equation x + y + z = 0, repeated four times Ch 1.TF.5 F, by Example 3a of Section 1.3 Ch 1.TF.6 T, by Definition 1.3.7 Ch 1.TF.7 T, by Theorem 1.3.4 Ch 1.TF.8 F, by Theorem 1.3.1 Ch 1.TF.9 F, by Theorem 1.3.4 Ch 1.TF.10 F; As a counter-example, consider the zero matrix 45 Copyright c 2013 Pearson Education, Inc Full file at https://TestbankDirect.eu/ Solution Manual for Linear Algebra with Applications 5th Edition by Bretscher Chapter Full file at https://TestbankDirect.eu/ Ch 1.TF.11 T; The last component of the left-hand side is zero for all vectors x Ch 1.TF.12 T; A = , for example Ch 1.TF.13 T; Find rref Ch 1.TF.14 T; Find rref Ch 1.TF.15 F; Consider the × matrix A that contains all zeroes, except for a in the lower left corner = 2A Ch 1.TF.16 F; Note that A for all × matrices A Ch 1.TF.17 F; The rank is Ch 1.TF.18 F; The product on the left-hand side has two components  −3 Ch 1.TF.19 T; Let A =  −5 −7  0 , for example       Ch 1.TF.20 T; We have   =   −    Ch 1.TF.21 F; Let A =      and b =  , for example Ch 1.TF.22 T, by Exercise 1.3.44 Ch 1.TF.23 F; Find rref to see that the rank is always Ch 1.TF.24 T; Note that v = 1v + 0w Ch 1.TF.25 F; Let u = , for example ,w = ,v = 0 Ch 1.TF.26 T; Note that = 0v + 0w  Ch 1.TF.27 F; Let A =  0 to A all we want, we will   1 0  and B =  0 0 always end up with a  0 , for example We can apply elementary row operations 0 matrix that has all zeros in the first column Ch 1.TF.28 T; If u = av + bw and v = cp + dq + er, then u = acp + adq + aer + bw 46 Copyright c 2013 Pearson Education, Inc Full file at https://TestbankDirect.eu/ Solution Manual for Linear Algebra with Applications 5th Edition by Bretscher True or False Full file at https://TestbankDirect.eu/ Ch 1.TF.29 F; The system x = 2, y = 3, x + y = has a unique solution Ch 1.TF.30 F; Let A = Ch 1.TF.31 0  0  0 , for example 0     1 F; If A   = 0, then x =   is a solution to A However, since rank(A) = 3, rref A = 3  0 0  , meaning that only is a solution to Ax = 0 Ch 1.TF.32 F; If b = 0, then having a row of zeroes in rref(A) does not force the system to be inconsistent Ch 1.TF.33 T; By Example 4d of Section 1.3, the equation Ax = has the unique solution x = Now note that A(v − w) = 0, so that v − w = and v = w Ch 1.TF.34 T; Note that rank(A) = 4, by Theorem 1.3.4 Ch 1.TF.35 F; Let u = , v= , w= , for example 0   −2t Ch 1.TF.36 T; We use rref to solve the system Ax = and find x =  −3t , where t is an arbitrary constant t   −2 Letting t = 1, we find [u v w]  −3  = −2u − 3v + w = 0, so that w = 2u + 3v Ch 1.TF.37 F; Let A = B = , for example 1  Ch 1.TF.38 T; Matrices A and B can both be transformed into I =  0 operations backwards, we can transform I into B Thus we can first transform A   0  Running the elementary into I and then I into B Ch 1.TF.39 T; If v = au + bw, then Av = A(au + bw) = A(au) + A(bw) = aAv + bAw Ch 1.TF.40 T; check that the three defining properties of a matrix in rref still hold F; If b = 0, then having a row of zeroes in rref(A) does not force the system to be inconsistent Ch 1.TF.41 T; Ax = b is inconsistent if and only if rank A b = rank(A)+1, since there will be an extra leading one in the last column of the augmented matrix: (See Figure 1.16.) 47 Copyright c 2013 Pearson Education, Inc Full file at https://TestbankDirect.eu/ Solution Manual for Linear Algebra with Applications 5th Edition by Bretscher Chapter Full file at https://TestbankDirect.eu/ Figure 1.16: for Problem T/F 41 Ch 1.TF.42 T; The system Ax = b is consistent, by Example 4b, and there are, in fact, infinitely many solutions, by Example 4c Note that Ax = b is a system of three equations with four unknowns Ch 1.TF.43 T; Recall that we use rref A to solve the system Ax = Now, rref A = rref(A) = rref(B) = rref B Then, since rref(A) = rref(B) , they must have the same solutions Ch 1.TF.44 F; Consider If we remove the first column, then the remaining matrix fails to be in rref 0 Ch 1.TF.45 T; First we list all possible matrices rref(M ), where M is a × matrix, and show the corresponding solutions for M x = 0: rref(M ) 0 1 a 0 0 0 0 solutions of M x = {0} −at , for an arbitrary t t t , for an arbitrary t R2 Now, we see that if rref(A) = rref(B), then the systems Ax = and Bx = must have different solutions Thus, it must be that if the two systems have the same solutions, then rref(A) = rref(B) Ch 1.TF.46 T First note that the product of the diagonal entries is nonzero if (and only if) all three diagonal entries are nonzero     a 0 ÷a 0 If all the diagonal entries are nonzero, then A =  b c  ÷c →  b′  d e f ÷f d ′ e′   0 →  , showing that rank A = 0 Conversely, if a = or c = or f = 0, rank A ≤ For example, if a and c are    ÷a a 0 →  b′ A =  b c  ÷c d e d e then it is easy to verify that rref A will contain a row of zeros, so that nonzero but f = 0, then    0 0  →  , with rank A = 0 0 48 Copyright c 2013 Pearson Education, Inc Full file at https://TestbankDirect.eu/ Solution Manual for Linear Algebra with Applications 5th Edition by Bretscher True or False Full file at https://TestbankDirect.eu/ a c Ch 1.TF.47 T If a = 0, then → b/a → 0 b d ÷a → b/a c d , showing that rank If a = 0, then b and c are both nonzero, so that b c d a c b d −c(I) → b/a (ad − bc)/a a/(ad − bc) = reduces to 0 as claimed Ch 1.TF.48 T If w = au + bv, then u + v + w = (a + 1)u + (b + 1)v = (a − b)u + (b + 1)(u + v) Ch 1.TF.49 T If Av = b and Aw = c, then A(v + w) = b + c, showing that the system Ax = b + c is consistent Suppose A is an n × m matrix Since Ax = b has a unique solution, rank A must be m (by Example 1.3.3c), implying that the system Ax = b + c has a unique solution as well (by Example 1.3.4d) Ch 1.TF.50 F Think about constructing a 0-1 matrix A of size × with rank A = row by row The rows must be chosen so that rref A will not contain a row of zeros, which implies that no two rows of A can be equal For the first row we have = 23 − choices: anything except [ 0 ] For the second row we have six choices left: anything except the row of zeros and the first row For the third row we have at most five choices, since we cannot chose the row of zeros, the first row, or the second row Thus, at most × × = 210 of the 0-1-matrices of size × have rank 3, out of a total of 29 = 512 matrices 49 Copyright c 2013 Pearson Education, Inc Full file at https://TestbankDirect.eu/ ... https://TestbankDirect.eu/ Solution Manual for Linear Algebra with Applications 5th Edition by Bretscher Chapter Full file at https://TestbankDirect.eu/ Figure 1.6: for Problem 1.1.43a Figure 1.7: for Problem... https://TestbankDirect.eu/ Solution Manual for Linear Algebra with Applications 5th Edition by Bretscher Section 1.1 Full file at https://TestbankDirect.eu/ Figure 1.5: for Problem 1.1.31  a+b+c... https://TestbankDirect.eu/ Solution Manual for Linear Algebra with Applications 5th Edition by Bretscher Chapter Full file at https://TestbankDirect.eu/ Thus, f (t) = 9t2 − 28t + 20 is the only solution 1.1.35

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