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Solution Manual for Elementary Linear Algebra, 2nd Edition

Get the textbook answers you need with Solution Manual for Elementary Linear Algebra, 2nd Edition, a solutions manual packed with clear and concise explanations.

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Solution Manual for Elementary Linear Algebra, 2nd Edition - Page 1 preview imageElementary Linear AlgebraA Matrix ApproachSpenceInselFriedbergSecond Edition
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Solution Manual for Elementary Linear Algebra, 2nd Edition - Page 3 preview imageElementary Linear AlgebraA Matrix ApproachSpenceInselFriedbergSecond EditionInstructor’s Manual
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Solution Manual for Elementary Linear Algebra, 2nd Edition - Page 10 preview imageChapter1Matrices, Vectors, andSystems of LinearEquations1.1MATRICES AND VECTORS1.»8−42012164–2.»−21−5−3−4−1–3.»6−424810−4–4.»8−311131811–5.242406−48356.2447−11019357.»3−13575–8.2447−11019359.2423−14513510.»1−1711−3–11.24−1−20−32−43512.24−1−20−32−43513.»−31−2−4−1−562–14.2664−120615−3−906377515.»−62−4−8−2−10124–16.»−84−2−0010−64–17.not possible18.»7−33410−3−4–19.266471−303−34−4377520.»11412325−24−2–21.not possible22.2664124−4208−2416−8377523.2664−7−130−33−44377524.2664−7−130−33−44377525.−226.027.24302π3528.24−21.653529.»22e–30.»0.40–31.[2−3 0.4]32.[2e12 0]33.24150150√31035mph34.(a)The swimmer’s velocity isu=»√2√2–mph.EastNorth...................................................45◦xyswimmer instill waterFigure for Exercise 34(a)1
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Solution Manual for Elementary Linear Algebra, 2nd Edition - Page 11 preview image2Chapter 1Matrices, Vectors, and Systems of Linear Equations(b)The water’s velocity isv=»01–mph.Sothe new velocity of the swimmer isu+v=»√2√2 + 1–mph. The correspond-ing speed isp5 + 2√2≈2.798 mph.67EastNorth...................................................45◦xywater currentcombinedvelocityFigure for Exercise 34(b)35.(a)»150√2 + 50150√2–mph(b)50p37 + 6√2≈337.21 mph36.The three components of the vector represent,respectively, the average blood pressure, averagepulse rate, and the average cholesterol readingof the 20 people.37.True38.True39.True40.False, a scalar multiple of the zero matrix is thezero matrix.41.False, the transpose of anm×nmatrix is ann×mmatrix.42.True43.False, the rows ofBare 1×4 vectors.44.False, the (3,4)-entry of a matrix lies in row 3and column 4.45.True46.False, anm×nmatrix hasmnentries.47.True48.True49.True50.False, matrices must have the same size to beequal.51.True52.True53.True54.True55.True56.True57.Suppose thatAandBarem×nmatrices.(a)Thejth column ofA+Bandaj+bjarem×1 vectors. Theith component of thejth column ofA+Bis the (i, j)-entry ofA+B, which isaij+bij. By definition, theith components ofajandbjareaijandbij, respectively. So theith component ofaj+bjis alsoaij+bij.Thus thejthcolumns ofA+Bandaj+bjare equal.(b)The proof is similar to the proof of (a).58.SinceAis anm×nmatrix, 0Ais also anm×nmatrix. Because the (i, j)-entry of 0Ais 0aij=0, we see that 0Aequals them×nzero matrix.59.SinceAis anm×nmatrix, 1Ais also anm×nmatrix. Because the (i, j)-entry of 1Ais 1aij=aij, we see that 1AequalsA.60.Because bothAandBarem×nmatrices, bothA+BandB+Aarem×nmatrices. The (i, j)-entry ofA+Bisaij+bij, and the (i, j)-entryofB+Aisbij+aij. Sinceaij+bij=bij+aijby the commutative property of addition of realnumbers, the (i, j)-entries ofA+BandB+Aareequal for alliandj.Thus, since the matricesA+BandB+Ahave the same size and allpairs of corresponding entries are equal,A+B=B+A.61.IfOis them×nzero matrix, then bothAandA+Oarem×nmatrices;so we needonly show they have equal corresponding en-tries. The (i, j)-entry ofA+Oisaij+ 0 =aij,which is the (i, j)-entry ofA.62.The proof is similar to the proof of Exercise 61.63.The matrices (st)A,tA, ands(tA) are allm×nmatrices; so we need only show that the corre-sponding entries of (st)Aands(tA) are equal.The (i, j)-entry ofs(tA) isstimes the (i, j)-entry oftA, and so it equalss(taij) =st(aij),which is the (i, j)-entry of (st)A.Therefore(st)A=s(tA).64.The matrices (s+t)A,sA, andtAarem×nma-trices. Hence the matrices (s+t)AandsA+tAarem×nmatrices; so we need only show theyhave equal corresponding entries.The (i, j)-entry ofsA+tAis the sum of the (i, j)-entriesofsAandtA, that is,saij+taij. And the (i, j)-entry of (s+t)Ais (s+t)aij=saij+taij.65.The matrices (sA)TandsATaren×mmatri-ces; so we need only show they have equal corre-sponding entries. The (i, j)-entry of (sA)Tis the(j, i)-entry ofsA, which issaji. The (i, j)-entryofsATis the product ofsand the (i, j)-entry ofAT, which is alsosaji.66.The matrixATis ann×mmatrix; so the ma-trix (AT)Tis anm×nmatrix.Thus we needonly show that (AT)TandAhave equal corre-sponding entries.The (i, j)-entry of (AT)Tis
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Solution Manual for Elementary Linear Algebra, 2nd Edition - Page 12 preview image1.2Linear Combinations, Matrix-Vector Products, and Special Matrices3the (j, i)-entry ofAT, which in turn is the (i, j)-entry ofA.67.Ifi6=j, then the (i, j)-entry of a square zeromatrix is 0. Because such a matrix is square, itis a diagonal matrix.68.IfBis a diagonal matrix, thenBis square.HencecBis square, and the (i, j)-entry ofcBiscbij=c·0 = 0 ifi6=j. ThuscBis a diagonalmatrix.69.IfBis a diagonal matrix, thenBis square. SinceBTis the same size asBin this case,BTissquare.Ifi6=j, then the (i, j)-entry ofBTisbji= 0. SoBTis a diagonal matrix.70.Suppose thatBandCaren×ndiagonal ma-trices.ThenB+Cis also ann×nmatrix.Moreover, ifi6=j, the (i, j)-entry ofB+Cisbij+cij= 0 + 0 = 0.SoB+Cis a diagonalmatrix.71.»2558–and242565786843572.LetAbe a symmetric matrix.ThenA=AT.So the (i, j)-entry ofAequals the (i, j)-entry ofAT, which is the (j, i)-entry ofA.73.LetObe a square zero matrix. The (i, j)-entryofOis zero, whereas the (i, j)-entry ofOTis the(j, i)-entry ofO, which is also zero. SoO=OT,and henceOis a symmetric matrix.74.By Theorem 1.2(b), (cB)T=cBT=cB.75.By Theorem 1.1(a) and Theorem 1.2(a) and (c),we have(B+BT)T=BT+ (BT)T=BT+B=B+BT.76.By Theorem 1.2(a), (B+C)T=BT+CT=B+C.77.No. Consider2425657868435and»2658–.78.LetAbe a diagonal matrix. Ifi6=j, thenaij=0 andaji= 0 by definition. Also,aij=ajiifi=j. So every entry ofAequals the correspondingentry ofAT. ThereforeA=AT.79.The (i, i)-entries must all equal zero. By equat-ing the (i, i)-entries ofATand−A, we obtainaii=−aii, and soaii= 0.80.TakeB=»01−10–.IfCis any 2×2 skew-symmetric matrix, thenCT=−C.Thereforec12=−c21. By Exercise 79,c11=c22= 0. SoC=»0−c21c210–=−c21»01−10–=−c21B.81.LetA1=12(A+AT) andA2=12(A−AT). Itis easy to show thatA=A1+A2. By Exercises75 and 74,A1is symmetric. Also, by Theorem1.2(b), (a), and (c), we haveAT2= 12 (A−AT)T= 12 [AT−(AT)T]= 12 (AT−A) =−12 (A−AT) =−A2.82.(a)Because the (i, i)-entry ofA+Bisaii+bii,we havetrace(A+B)= (a11+b11) +· · ·+ (ann+bnn)= (a11+· · ·+ann) + (b11+· · ·+bnn)= trace(A) + trace(B).(b)The proof is similar to the proof of (a).(c)The proof is similar to the proof of (a).83.Theith component ofap+bqisapi+bqi, whichis nonnegative. Also, the sum of the componentsofap+bqis(ap1+bq1) +· · ·+ (apn+bqn)=a(p1+· · ·+pn) +b(q1+· · ·+qn)=a(1) +b(1) =a+b= 1.84.(a)2666646.5−0.5−1.9−2.89.6−2.91.5−3.017.40.4−15.55.2−1.0−3.7−7.317.55.21.43.516.8377775(b)266664−1.33.4−4.010.43.04.9−2.46.6−3.9−4.19.4−8.61.7−0.1−14.5−0.2−4.74.1−0.7−1.8377775(c)26643.97.410.3−0.11.90.8−0.3−1.1−2.52.3−2.60.2−7.2−9.72.11.60.20.611.610.637751.2LINEAR COMBINATIONS,MATRIX-VECTOR PRODUCTS,AND SPECIAL MATRICES1.»1214–2.24−547353.249010354.»2232–5.»ab–6.[18]7.»225–8.24abc35
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Solution Manual for Elementary Linear Algebra, 2nd Edition - Page 13 preview image4Chapter 1Matrices, Vectors, and Systems of Linear Equations9.24satbuc3510.[6]11.242−6103512.24−3423513.»−16–14.243−123515.»2113–16.»269–17.12»√2−√2√2√2–,12»−√2√2–18.»1001–,e119.12»1−√3√31–, 12»3− √33√3 + 1–20.12»√3−11√3–, 12»√3−21 + 2√3–21.12»−√31−1−√3–,12»√3−33√3 + 1–22.−1√2»11−11–,−1√2»1−3–23.»32–24.12»4√3 + 1√3−4–25.12»3− √33√3 + 1–26.12»2−5√32√3 + 5–27.12»3−3√3–28.»√31–29.»11–= (1)»10–+ (1)»01–30.»1−1–= 14»4−4–31.not possible32.»11–= (1)»10–+ (1)»01–33.not possible34.»11–= (1)»10–+ (−1)»0−1–+ 0»00–35.»−111–= 3»13–+ (−2)»2−1–36.»11–= 0»10–+ 0»0−1–+ (1)»11–37.»38–= 7»12–+ (−2)»23–+ 0»−2−5–38.»ab–=„a+ 2b3« »11–+„a−b3« »2−1–39.not possible40. u= 42401235+ (−2)24−1303541.u= 0242−1235+ 1243−2135+ 024−4133542.u= 52410035+ 62401035+ 7240013543.u= (−4)2410035+ (−5)2401035+ (−6)240013544.u= 0241−1135+ 0240−2335+ 124−1323545.True46.False.If the coefficients of the linear combina-tion 3»22–+ (−6)»11–=»00–were positive, thesum could not equal the zero vector.47.True48.True49.True50.False, the matrix-vector product of a 2×3 ma-trix and a 3×1 vector is a 2×1 vector.51.False, the matrix-vector product is a linear com-bination of thecolumnsof the matrix.52.False, the product of a matrix and a standardvector is a column of the matrix.53.True54.False, the matrix-vector product of anm×nmatrix and a vector inRnyields a vector inRm.55.False, every vector inR2is a linear combinationof twononparallelvectors.56.True57.False, a standard vector is a vector with a singlecomponent equal to 1 and the others equal to 0.58.True59.False, considerA=»1−1−11–andu=»11–.60.True61.False,Aθuis the vector obtained by rotatinguby acounterclockwiserotation of the angleθ.62.False, considerA=»1−1−11–,u=»11–, andv=»22–.63.True64.True65.Ifθ= 0, thenAθ=I2. SoAθv=I2v=vbyTheorem 1.3(h).66.We haveA180◦v=»−100−1–v=−I2v=−v.
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Solution Manual for Elementary Linear Algebra, 2nd Edition - Page 14 preview image1.2Linear Combinations, Matrix-Vector Products, and Special Matrices567.Letv=»ab–. ThenAθ(Aβv)=»cosθ−sinθsinθcosθ– „»cosβ−sinβsinβcosβ– »ab–«=»cosθ−sinθsinθcosθ– »acosβ−bsinβasinβ+bcosβ–=»acosθcosβ−bcosθsinβasinθcosβ−bsinθsinβ–+»−asinθsinβ−bsinθcosβacosθsinβ+bcosθcosβ–=»acos (θ+β)−bsin (θ+β)asin (θ+β) +bcos (θ+β)–=Aθ+βv.68.Letu=»ab–. ThenATθ(Aθu)=»cosθsinθ−sinθcosθ– „»cosθ−sinθsinθcosθ– »ab–«=»cosθsinθ−sinθcosθ– »acosθ−bsinθasinθ+bcosθ–=»acos2θ−bsinθcosθ−asinθcosθ+bsin2θ–+»asin2θ+bsinθcosθasinθcosθ+bcos2θ–=»a(sin2θ+ cos2θ)b(sin2θ+ cos2θ)–=»ab–=u.Similarly,Aθ(ATθu) =u.69.(a)As in Example 3, the populations are givenby the entries ofA»400300–=»349351–;sothere will be 349,000 people in the city and351,000 in the suburbs.(b)ComputingA»349351–=»307.180392.820–, we seethat there will be 307,180 people in thecity and 392,820 in the suburbs.70.Au=a2414735+b2425835+c243693571.Au=»−1001– »ab–=»−ab–, the reflection ofuabout they-axis72.We haveA(Au) =A„»−1001– »ab–«=»−1001– »−ab–=»ab–=u.73.B=»100−1–74.(a)C=A180◦=»−100−1–(b)We haveA(Cu) =»−1001– „»−100−1– »ab–«=»−1001– »−a−b–=»a−b–.In a similar fashion, we haveC(Au)=»a−b–=BuandB(Cu) =C(Bu) =Au.(c)The first equation shows that reflectingabout thex-axis can be accomplished byeither first rotating by 180◦and then re-flecting about they-axis, or first reflect-ing about they-axis and then rotating by180◦.The second equation shows that reflectingabout they-axis may be accomplished ei-ther by first rotating by 180◦and then re-flecting about thex-axis, or first reflect-ing about thex-axis and then rotating by180◦.75.Au=»a0–, the projection ofuon thex-axis76.This exercise is similar to Exercise 72.77.Ifv=»a0–, thenAv=»1000– »a0–=»a0–=v.78.B=»0001–79.(a)We haveA(Cu) =»1000– „»−100−1– »ab–«=»1000– »−a−b–=»−a0–,andC(Au) =»−100−1– „»1000– »ab–«
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Solution Manual for Elementary Linear Algebra, 2nd Edition - Page 15 preview image6Chapter 1Matrices, Vectors, and Systems of Linear Equations=»−100−1– »a0–=»−a0–.(b)Rotating a vector by 180◦and then pro-jecting the result on thex-axis is equiva-lent to projecting a vector on thex-axisand then rotating the result by 180◦.80.The sum of the two linear combinationsau1+bu2andcu1+du2is(au1+bu2)+(cu1+du2) = (a+c)u1+(b+d)u2,which is also a linear combination ofu1andu2.81.Writev=a1u1+a2u2andw=b1u1+b2u2,wherea1,a2,b1, andb2are scalars.A linearcombination ofvandwhas the formcv+dw=c(a1u1+a2u2) +d(b1u1+b2u2)= (ca1+db1)u1+ (ca2+db2)u2,which is also a linear combination ofu1andu2.82.The proof is similar to that of Exercise 81.83.We haveA(cu) = (cu1)a1+ (cu2)a2+· · ·+ (cun)an=c(u1a1+u2a2+· · ·+unan) =c(Au).Similarly, (cA)u=c(Au).84.We have(A+B)u=u1(a1+b1) +· · ·+un(an+bn)=u1a1+u1b1+· · ·+unan+unbn= (u1a1+· · ·+unan)+ (u1b1+· · ·+unbn)=Au+Bu.85.We haveAej=0a1+· · ·+ 0aj−1+ 1aj+ 0aj+1+· · ·+ 0an=aj.86.SupposeBw=Awfor allw. Letw=ej. ThenBej=Aej.From Theorem 1.3(e), it followsthatbj=ajfor allj. SoB=A.87.The vectorA0is anm×1 vector. By definitionA0= 0a1+ 0a2+· · ·+ 0an=0.88.Every column ofOis them×1 zero vector. SoOv=v10+v20+· · ·+vn0=0.89.Thejth column ofInisej. SoInv=v1e1+v2e2+· · ·+vnen=v.90.Usingp=»400300–, we computeAp, A(Ap), . . .until we have ten vectors. From the final vector,we see that there will be 155,610 people living inthe city and 544,389 people living in the suburbsafter ten years.91.(a)266424.645.026.0−41.43775(b)2664134.144.47.6104.83775(c)2664128.480.663.525.83775(d)2664653.09399.77528.23−394.5237751.3SYSTEMS OF LINEAR EQUATIONS1.(a)»0−12130–(b)»0−120130−1–2.(a)ˆ2−13˜(b)ˆ2−134˜3.(a)2412−13−3435(b)24123−132−341354.(a)»102−12−101–(b)»102−132−1010–5.(a)2402−3−11220135(b)2402−34−112−62010356.(a)241−2171−20102−44835(b)241−21751−201032−4487357.2402−442−263−111−102−3358.24−330−69−263−1102−44235
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Solution Manual for Elementary Linear Algebra, 2nd Edition - Page 16 preview image1.3Systems of Linear Equations79.241−102−30433−502−4423510.24−263−111−102−302−4423511.241−102−3−263−1101−2213512.241−102−3−2015−13−502−4423513.241−102−3−263−11−8268063514.241−102−3−263−1120−48−43515.2664−240−11−12−46−321377516.266641−20−1212−122−46−3213777517.26641−20−11−1006−321377518.26641−20−11−12−460−41377519.26641−202−46−11−1−321377520.26641−20−3212−46−11−1377521.26641−20−11−12−46−103377522.2664−10−2−11−12−46−321377523.Yes, because 1(1)−4(−2) + 3(−1)=6 and1(−5)−2(−1) =−3. Alternatively,»1−403001−2–26641−2−5−13775=»6−3–.24.No, because 1(2)−4(0) + 3(1) = 56= 6. Alter-natively, ifAis the coefficient matrix, and thegiven vector isv, thenAv=»5−3–6=»6−3–.25.No, because the left side of the second equationyields 1(2)−2(1) = 06=−3. Alternatively,»1−403001−2–266430213775=»60–6=»6−3–.26.Yes, the components of the vector satisfy bothequations.Alternatively, if the given vector isv, thenAv=»6−3–.27.no28.yes29.yes30.yes31.yes32.no33.yes34.yes35.no36.yes37.no38.no39.x1= 2 +x2x2free40.x1=−4x2=541.x1= 6 + 2x2x2free42.x1= 5 + 4x2x2free43.not consistent44.x1=−6x2=345.x1= 4 + 2x2x2freex3= 346.not consistent47.x1=3x4x2=4x4x3=−5x4x4freeand2664x1x2x3x43775=x4266434−51377548.x1= 9 +x3−3x4x2= 8−2x3+ 5x4x3freex4freeand2664x1x2x3x43775=x326641−2103775+x42664−35013775+26649800377549.x1freex2=−3x3=−4x4=5and2664x1x2x3x43775=x1266410003775+26640−3−45377550.x1=−3 + 2x2x2freex3=−4x4=5and2664x1x2x3x43775=x2266421003775+2664−30−45377551.x1= 6−3x2+ 2x4x2freex3= 7−4x4x4freeand2664x1x2x3x43775=x22664−31003775+x4266420−413775+266460703775
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