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25∧8×4∧8＝（25x4）∧8=100∧8=10∧16谢谢，请采纳
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我们会通过消息、邮箱等方式尽快将举报结果通知您。已知F1（-c,0),F2（c,0)是椭圆x^2/a^2+y^2/b^2=1的左右焦点,过F1做倾斜角为60度的直线L叫
已知F1（-c,0),F2（c,0)是椭圆x^2/a^2+y^2/b^2=1的左右焦点,过F1做倾斜角为60度的直线L叫椭圆于A,B两点三角形ABF2的内切圆半径为7分之2乘根号3倍的c.求（1）椭圆离心率；（2）若lABl=8倍根号2,求椭圆标准方程.第一问都不会,俺非常受挫,
第一问,根据面积相等,rl/2=2a.|y1-y2|/2.这里r是内切圆半径,l是三角形ABF2周长,等于4a.那么由椭圆第二定义,设AF1=L1,BF1=L2,不妨设A在x轴上方.那么Lcos60°+ -c - -c∧2 /a∧2=L/e.得到L1=2b∧2/a(2-e).同理L2=2b∧2/a(2+e).所以（L1+L2）sin60°*2c/2=4ar=4a2√3c/7.即b∧2/a∧2(2-e) +b∧2/a∧2(2+e)=4/7.也就是(1-e∧2)/(2-e) + (1-e∧2)/(2+e)=4/7.整理得到e∧2=1/2 所以e=√2/2第二问,由第一问过程,|AB|=L1+L2=8√2.得到ab∧2/(4a∧2-c∧2)=√2 即a/7=√2所以标准方程是x∧2/98 + y∧2/49 =1
与《已知F1（-c,0),F2（c,0)是椭圆x^2/a^2+y^2/b^2=1的左右焦点,过F1做倾斜角为60度的直线L叫》相关的作业问题
已知直线L经过(0,-2)其倾斜角为60度.斜率为tan60=√3所以y=√3x-2圆心到直线的距离为|3-1-2|/√(1+3)=0圆心在L上……
设A(x1,y1),B(x2,y2),&M(x0,y0)x0=(x1+x2)/2,y0=(y1+y2)/2,直线斜率k=tan135°=-1,(y1-y2)/(x1-x2)=-1,代入椭圆方程,x1^2/a^2+y1^2/b^2=1,(1),x2^2/a^2+y2^2/b^2=1,(2),(1)-(2)式,b
求抛物线方程,只要确定p的值即可.本题之关键是如何深挖45°的作用.既然是45°,则三条线段在x轴上的射影也成等比数列.设B(x1,y1)、C(x2,y2),则|x1＋2|、|x2－x1|、|x2＋2|也成等比数列,即x1＋2、|x2－x1|、x2＋2成等比数列,∴|x2－x1|^2=(x1＋2)×(x2＋2)这个式子
在椭圆x^2/2+y^2=1中,a=√2,b=1,c=1,左焦点F1(-1,0) 过F1,倾斜角60°的直线方程是y=√3(x+1).代入椭圆方程,得到 x^2+2*3(x+1)^2=2 --->7x^2+12x+4=0 --->x1,x2=(-6+'-2√2)/7 y1,y2=3(x+1)=3(1+'2√2)/7=(
F1(-1,0),则直线AB的方程为：y=√3(x+1),设A(x1,y1),B(x2,y2),直线与椭圆联列方程组：y=√3(x+1),x^2/2+y^2=1；消去y得关于x的二次方程：7x^2/2+6x+2=0,显然x1,x2是该方程的两个根,则韦达定理得：x1+x2=-12/7,x1*x2=4/7；而由两点间距离
差的平方与和的平方之间的转换
F1F2是圆的直径,PF1F2为直角三角形,PF1+PF2=2a,PF1*PF2/2=26.F1F2=2c(2c)^2=PF1^2+PF2^2=(PF1+PF2)^2-2PF1*PF2=121c=11/2a=15/2b=√26a+b+c=13+√26你的串号我已经记下,采纳后我会帮你制作
设P点为（x,y）则向量PF1=（x-c,y）,向量PF2=（x+c,y）向量PF1*向量PF2=x^2-c^2+y^2=c^2得x^2+y^2=2c^2 又因为x^2/a^2+y^2/b^2=1得y^2=b^2-b^2*x^2/a^2代入前一式子,得（c^2/a^2）*x^2=3c^2-a^2 其中b^2=a^2-c
x²/a²+y²/b²=1一个焦点为F1(-√3,0)那么a²-b²=3①过点H(√3,1/2)所以3/a²+1/4b²=1②联立①②解方程组得a²=4,b²=1所以椭圆方程是x²/4+y²=1 再问
向量PF1·向量PF2=0,∠F1PF2=90°,设|PF1=m,|PF2|=n,则m²+n²=4c²,由椭圆定义m+n=2a…①,∵ (m+n)²-2mn=m²+n²,∴ mn=2(a²-c²)…②,由①,②知m,n是方程z²-
(1).点A(-3,1)关于 y=-2 的对称点为A'(-3,-5).又∵过A点,斜率为 -5/2的光线,经直线y=-2反射.∴过A'、F点的直线方程为：y+5= (5/2)(x+3).当y=0时,x=1.即：c=1.又∵椭圆 x²/a²+y²/b²=1,(a>b>0)的左准线的
F1F2=2C所以c=4又因为b^2=25所以b=5又因为a^2=b^2+c^2所以a=根号41
y²=2pxy=x-p/2(x-p/2)²=2pxx²-3px+p²/4=0xA+xB=3p|AB|=|AF|+|BF|=xA+p/2+xB+p/2=3p+p=4p=62p=3方程为y²=3x
y^2=1/2xF(1/8,0)|AB|=1
焦点为(3,0),则p=6,抛物线方程为y²=6x.直线被抛物线所截得的弦长为2p/sin²α,本题中α=45°.
那啥,虽然我还没学圆锥曲线,不过还是知道一点,就凑合着做吧.由题意知,c^2=a^2-b^2=a^2-4;在三角形PF1F2中运用余弦定理,得F1F2^2=PF1^2+PF2^2-2PF1*PF2*cos60整理得,4c^2=PF1^2+PF2^2-PF1*PF2=4a^2-16又有 (PF1+PF2)^2=PF1^2
过F1作F1P⊥MN,交MN与P,∵L的倾角为45°,在△F1PF2为等腰直角三角形,|F1F2|=√2|F1P|=2,c=1直线L的方程为y=x-1,椭圆的方程可设为x^2/a^2+y^2/(a^2-1)=1,将直线L方程带入,消去y,得(2a^2-1)x^2-2a^2x+a^4=0x1+x2=2a^2/(2a^2-
(1)根据题意 c=√3 a=2 a^2=4 b^2=a^2-c^2=4-3=1 ∴椭圆的方程为 x^2/4+y^2/1=1(2)设直线方程为 y=kx代人椭圆方程得：(1+4k^2)x^2=4 x=±2/√(1+4k^2) y==±2k/√(1+4k^2)A、B两点间距离为4√(1+k^2)/√(1+4k^2)三角形
设角AQB为k,Q（m,n）由对称性,只用考虑n大于等于0的情况 有m^2/a^2+n^2/b^2=1,m^2=a^2-a^2*n^2/b^2……* 对三角形AQB面积,有两种算法,以此建立等式：(1/2)*AQ*BQ*sink=(1/2)*AB*n 两边约去12,再平方代入m,n得到：[(m+a)^2+n^2]*[(30×10∧8×60×60×24×365等于多少_百度知道
30×10∧8×60×60×24×365等于多少
=3×10^9×=3×36x876×10^12=9=9.
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INTRODUCTION TO LINEAR ALGEBRA Third EditionMANUAL FOR INSTRUCTORSGilbert Stranggs@math.mit.eduMassachusetts Institute of Technologyhttp://web.mit.edu/18.06/www http://math.mit.edu/?gs http://www.wellesleycambridge.comWellesley-Cambridge Press Box 812060 Wellesley, Massachusetts 02482
Solutions to ExercisesProblem Set 1.1, page 61 Line through (1, 1, 1); same plane! 3 v = (2, 2) and w = (1, ?1). 4 3v + w = (7, 5) and v ? 3w = (?1, ?5) and cv + dw = (2c + d, c + 2d). 5 u + v = (?2, 3, 1) and u + v + w = (0, 0, 0) and 2u + 2v + w = (add ?rst answers) = (?2, 3, 1). 6 The components of every cv + dw add to zero. Choose c = 4 and d = 10 to get (4, 2, ?6). 8 The other diagonal is v ? w (or else w ? v ). Adding diagonals gives 2v (or 2w ). 9 The fourth corner can be (4, 4) or (4, 0) or (?2, 2). 10 i + j is the diagonal of the base.1 1 1 , 2 , 2 ). The 11 Five more corners (0, 0, 1), (1, 1, 0), (1, 0, 1), (0, 1, 1), (1, 1, 1). The center point is ( 2centers of the six faces are ( 1 , 1 , 0), ( 1 , 1 , 1) and (0, 1 , 1 ), (1, 1 , 1 ) and ( 1 , 0, 1 ), ( 1 , 1, 1 ). 2 2 2 2 2 2 2 2 2 2 2 2 12 A four-dimensional cube has 24 = 16 corners and 2 ? 4 = 8 three-dimensional sides and 24 two-dimensional faces and 32 one-dimensional edges. See Worked Example 2.4 A. 13 sum = sum = ?4:00 1:00 is 60? from horizontal = (cos π , sin π ) = (1 , 3 3 2 14 Sum = 12j since j = (0, 1) is added to every vector. 15 The point3 v 4 √ 3 ). 2+1 w is three-fourths of the way to v starting from w . The vector 41 v 4w is +1 41 halfway to u = 2 v+1 w , and the vector v + w is 2u (the far corner of the parallelogram). 216 All combinations with c + d = 1 are on the line through v and w . The point V = ?v + 2w is on that line beyond w . 17 The vectors cv + cw ?ll out the line passing through (0, 0) and u =1 v 2+1 w . It continues 2beyond v + w and (0, 0). With c ≥ 0, half this line is removed and the “ray” starts at (0, 0). 18 The combinations with 0 ≤ c ≤ 1 and 0 ≤ d ≤ 1 ?ll the parallelogram with sides v and w . 19 With c ≥ 0 and d ≥ 0 we get the “cone” or “wedge” between v and w . 20 (a)1 u 3+1 v+1 w is the center of the triangle between u , 3 31 u 2+1 w is the center 2of the edge between u and w c + d + e = 1.(b) To ?ll in the triangle keep c ≥ 0, d ≥ 0, e ≥ 0, and3421 The sum is (v ? u ) + (w ? v ) + (u ? w ) = zero vector. 22 The vector1 (u 2+ v + w ) is outside the pyramid because c + d + e =1 2+1 2+1 2& 1.23 All vectors are combinations of u , v , and w . 24 Vectors cv are in both planes. 25 (a) Choose u = v = w = any nonzero vector and w to be a combination like u + v . 26 The solution is c = 2 and d = 4. Then 2(1, 2) + 4(3, 1) = (14, 8). 27 The combinations of (1, 0, 0) and (0, 1, 0) ?ll the xy plane in xyz space. 28 An example is (a, b) = (3, 6) and (c, d) = (1, 2). The ratios a/c and b/d are equal. Then ad = bc. Then (divide by bd) the ratios a/b and c/d are equal! (b) Choose u and v in di?erent directions,Problem Set 1.2, page 171 u ? v = 1.4, u ? w = 0, v ? w = 24 = w ? v . 2 u = 1 and v = 5 = w . Then 1.4 & (1)(5) and 24 & (5)(5).3 Unit vectors v / v = ( 3 , 4 ) = (.6, .8) and w / w = ( 4 , 3 ) = (.8, .6). The cosine of θ is 5 5 5 5 ? ? ? v w 24 v ? w = 25 . The vectors w , u , ?w make 0 , 90 , 180 angles with w . 4 u1 = v/ v = could be√1 (3, 1) 10and u 2 = w / w = 1 (2, 1, 2). U 1 = 3√1 (1, ?3) 10or√1 (?1, 3). 10U21 √ (1, ?2, 0). 55 (a) v ? (?v ) = ?1 so θ = 90? 6 (a) cos θ = (c) cos θ =(b) (v + w ) ? (v ? w ) = v ? v + w ? v ? v ? w ? w ? w = 1+()?()?1 = 0(c) (v ? 2w ) ? (v + 2w ) = v ? v ? 4w ? w = ?31 (2)(1) ?1+3 (2)(2)so θ = 60? or =1 2π 3radiansπ 3so θ = 60? or(b) cos θ = 0 so θ = 90? or π radians 2 √ π (d) cos θ = ?1/ 2 so θ = 135? or 34 .7 All vectors w = (c, 2c); all vectors (x, y, z ) with x + y + z = 0 all vectors perpendicular to (1, 1, 1) and (1, 2, 3) lie on a line. 8 (a) False (b) True: u ? (cv + dw ) = cu ? v + du ? w = 0 (c) True9 If v2 w2 /v1 w1 = ?1 then v2 w2 = ?v1 w1 or v1 w1 + v2 w2 = 0. 10 Slopes2 1and ? 1 multiply to give ?1: perpendicular. 211 v ? w & 0 means angle & 90? ; this is half of the plane. 12 (1, 1) perpendicular to (1, 5) ? c(1, 1) if 6 ? 2c = 0 or c = 3; v ? (w ? cv ) = 0 if c = v ? w /v ? v . 13 v = (1, 0, ?1), w = (0, 1, 0). 14 u = (1, ?1, 0, 0), v = (0, 0, 1, ?1), w = (1, 1, ?1, ?1). √ √ √ 15 1 (x + y ) = 5; cos θ = 2 16/ 10 10 = .8. 2 = 9 so v = 3; u = 1 w = (1, ?1, 0, . . . , 0). 3 √ √ 2 2 2 17 cos α = 1/ 2, cos β = 0, cos γ = ?1/ 2, cos2 α + cos2 β + cos2 γ = (v1 + v2 + v3 )/ v 16 v22= 1.518 v2= 42 + 22 = 20, w2= (?1)2 + 22 = 5, (3, 4)2= 25 = 20 + 5.19 v ? w = (5, 0) also has (length)2 = 25. Choose v = (1, 1) and w = (0, 1) which a (length of v )2 + (length of w )2 = 12 + 12 + 12 but (length of v ? w )2 = 1. 20 (v + w ) ? (v + w ) = (v + w ) ? v +(v + w ) ? w = v ? (v + w )+ w ? (v + w ) = v ? v + v ? w + w ? v + w ? w = v ? v + 2v ? w + w ? w . Notice v ? w = w ? v ! 21 2v ? w ≤ 2 v ( v + w ) . 22 Compare v ? v + w ? w with (v ? w ) ? (v ? w ) to ?nd that ?2v ? w = 0. Divide by ?2. 23 cos β = w1 / w and sin β = w2 / w . Then cos(β ?a) = cos β cos α+sin β sin α = v1 w1 / v v2 w2 / v w = v ? w/ v w . w cos θ w +2w leads to v + w2= v ? v + 2v ? w + w ? w ≤ v2+2 vw + w2=24 We know that (v ? w ) ? (v ? w ) = v ? v ? 2v ? w + w ? w . The Law of Cosines writes v for v ? w . When θ & 90 this is positive and v ? v + w ? w is larger than v ? w? 2.2 2 2 2 2 2 2 2 2 2 2 2 25 (a) v1 w1 + 2v1 w1 v2 w2 + v2 w2 ≤ v1 w1 + v1 w2 + v2 w1 + v2 w2 is true because the di?erence is 2 2 2 2 v1 w2 + v2 w1 ? 2v1 w1 v2 w2 which is (v1 w2 ? v2 w1 )2 ≥ 0.26 Example 6 gives |u1 ||U1 | ≤1 (u2 1 22 + U1 ) and |u2 ||U2 | ≤1 (u2 2 22 + U2 ). The whole line becomes1 .96 ≤ (.6)(.8) + (.8)(.6) ≤ 2 (.62 + .82 ) + 1 (.82 + .62 ) = 1. 227 The cosine of θ is x/x2 + y 2 , near side over hypotenuse. Then | cos θ|2 = x2 /(x2 + y 2 ) ≤ 1.?7 1428 Try v = (1, 2, ?3) and w = (?3, 1, 2) with cos θ =and θ = 120? . Write v ? w = xz + yz + xy1 w = ?2 .1 as 2 (x + y + z )2 ? 1 (x2 + y 2 + z 2 ). If x + y + z = 0 this is ? 1 (x2 + y 2 + z 2 ), so v ? w / v 2 229 The length v ? w is between 2 and 8. The dot product v ? w is between ?15 and 15. 30 The vectors w = (x, y ) with v ? w = x + 2y = 5 lie on a line in the xy plane. The shortest w is (1, 2) in the direction of v . 31 Three vectors in the plane could make angles & 90? with each other: (1, 0), (?1, 4), (?1, ?4). Four vectors could not do this (360? total angle). How many can do this in R3 or Rn ?Problem Set 1.31 (x, y, z ) = (2, 0, 0) and (0, 6, 0); n = (3, 1, ?1); dot product (3, 1, ?1) ? (2, ?6, 0) = 0. 2 4x ? y ? 2z = 1 is parallel to every plane 4x ? y ? 2z = d and perpendicular to n = (4, ?1, ?2). 3 (a) True (assuming n = 0) 4 (a) x + 5y + 2z = 14 (b) False (c) True. (c) y = 0.(b) x + 5y + 2z = 305 The plane changes to the symmetric plane on the other side of the origin. 6 x ? y ? z = 0. 7 x + 4y = 0; x + 4y = 14. 8 u = (2, 0, 0), v = (0, 2, 0), w = (0, 0, 2). Need c + d + e = 1.69 x + 4y + z + 2t = 8. 10 x ? 4y + 2z = 0. 11 We choose v 0 = (6, 0, 0) and then in-plane vectors (3, 1, 0) and (1, 0, 1). The points on the plane are v 0 + y (3, 1, 0) + z (1, 0, 1). 12 v 0 = (0, 0, 0); all vectors in the plane are combinations y (?2, 1, 0) + z ( 1 , 0, 1). 2 13 v 0 = (0, 0, 0); all solutions are combinations y (?1, 1, 0) + z (?1, 0, 1). 14 Particular point (9, 0); solution (3, 1); points are (9, 0) + y (3, 1) = (3y + 9, y ). 15 v 0 = (24, 0, 0, 0); solutions (?2, 1, 0, 0) and (?3, 0, 1, 0) and (?4, 0, 0, 1). Combine to get (24 ? 2y ? 3z ? 4t, y, z, t). 16 Choose v 0 = (0, 6, 0) with two zero components. Then set components to 1 to choose (1, 0, 0)3 and (0, ?3/2, 1). Combinations are (x, 6 ? 2 z, z ). √ √ √ 17 Now |d|/ n = 12/ 56 = 12/2 14 = 6/ 14. Same answer because same plane.18 (a) |d|/ n = 18/3 = 6 and v = (4, 4, 2) (b) |d|/ n = 0 and v = 0 √ (c) |d|/ n = 6/ 2 and v = 3n = (3, 0, ?3). 19 (a) Shortest distance is along perpendicular to line √ (c) The distance to (5, ?10) is 125. 20 (a) n = (a, b) √ |c|/ a2 + b2 . (b) t = c/(a2 + b2 ) (b) Need t + 4t = 25 or t = 5(c) This distance to tn = (ca, cb)/(a2 + b2 ) is21 Substitute x = 1 + t, y = 2t, z = 5 ? 2t to ?nd (1 + t) + 2(2t) ? 2(5 ? 2t) = 27 or ?9 + 9t = 27 or t = 4. Then tn = 12. 22 Shortest distance in w + tn lies on the plane when n ? w + tn ? n = d or t = (d ? n ? w )/n ? n . The distance is |d ? n ? w |/ n (which is |d|/ n when w = 0). 23 The vectors (1, 2, 3) and (1, ?1, ?1) are perpendicular to the line. Set x = 0 to ?nd y = ?16 and z = 14. Set y = 0 to ?nd x = 9/2 and z = 5/2. These particular points are (0, ?16, 14) and (9/2, 0, 5/2). 24 (a) n = (1, 1, 1, ?1) (d) v 0 = (1, 0, 0, 0) (b) |d|/ n =1 2(c) dn /n ? n = ( 1 , 1, 1,?1 ) 4 4 4 4(e) (?1, 1, 0, 0), (?1, 0, 1, 0), (1, 0, 0, 1)(f) all points (1 ? y ? z + t, y, z, t). 25 n = (1, 1, 1) or any nonzero (c, c, c). √ √ 26 cos θ = (0, 1, 1) ? (1, 0, 1)/ 2 2 = 1 so θ = 60? . 2Problem Set 2.1, page 291 The planes x = 2 and y = 3 and z = 4 are perpendicular to the x, y, z axes. 2 The vectors are i = (1, 0, 0) and j = (0, 1, 0) and k = (0, 0, 1) and b = (2, 3, 4) = 2i + 3j + 4k .73 The planes are the same: 2y = 6 is y = 3, and 3z = 12 is z = 4. The solution is the same intersection point. The but same combination x = x . 4 The sol the second plane and row 2 of the matrix and all columns of the matrix are changed. 5 If z = 2 then x + y = 0 and x ? y = z give the point (1, ?1, 2). If z = 0 then x + y = 6 and x ? y = 4 give the point (5, 1, 0). Halfway between is (3, 0, 1). 6 If x, y, z satisfy the ?rst two equations they also satisfy the third equation. The line L of solutions contains v = (1, 1, 0) and w = ( 1 , 1, 1 ) and u = 2 2 cv + dw with c + d = 1. 7 Equation 1 + equation 2 ? equation 3 is now 0 = ?4. Solution impossible. 8 Column 3 = Column 1; solutions (x, y, z ) = (1, 1, 0) or (0, 1, 1) and you can add any multiple of (?1, 0, 1); b = (4, 6, c) needs c = 10 for solvability. 9 Four planes in 4-dimensional space normally meet at a point. The solution to Ax = (3, 3, 3, 2) is x = (0, 0, 1, 2) if A has columns (1, 0, 0, 0), (1, 1, 0, 0), (1, 1, 1, 0), (1, 1, 1, 1). 10 Ax = (18, 5, 0), Ax = (3, 4, 5, 5). 11 Nine multiplications for Ax = (18, 5, 0). 12 (14, 22) and (0, 0) and (9, 7). 13 (z, y, x) and (0, 0, 0) and (3, 3, 6). 14 (a) x has n components, Ax has m components columns are in m-dimensional space. 15 2x + 3y + z + 5t = 8 is Ax = b with the 1 by 4 matrix A = [ 2 3 1 5 ]. The solutions x ?ll a 3D “plane” in 4 dimensions. ? ? ? ? 1 0 0 1 ?, P = ? ?. I=? 0 1 1 0 ? ? ? ? 0 1 ? 1 0 ?, 180? rotation from R2 = ? ? = ?I . R=? ?1 0 0 ?1 ? ? ? ? 0 1 0 0 0 1 ? ? ? ? ? ? ? ? P = ? 0 0 1 ? produces (y, z, x) and Q = ? 1 0 0 ? recovers (x, y, z ). ? ? ? ? 1 0 0 0 1 0 ? ? ? ? 1 0 0 ? ? 1 0 ? ?, E = ? E=? ??1 1 0 ?. ? ? ?1 1 0 0 1 ? ? ? ? 1 0 0 1 0 0 ? ? ? ? ? ? ? ? E = ? 0 1 0 ?, E ?1 = ? 0 1 0 ?, E v = (3, 4, 8), E ?1 E v = (3, 4, 5). ? ? ? ? 1 0 1 ?1 0 1 ? ? ? ? ? ? ? ? 1 0 0 0 5 0 ?, P 2 = ? ?, P1 v = ? ?, P2 P1 v = ? ?. P1 = ? 0 0 0 1 0 0 (b) Planes in n-dimensional space, but the1 v 2+ 1 w and all combinations 21617181920218?√ √ ? 2 ? 2 ?√ √ ?. 2 222 R =1 2? ? x ? ? ? ? 23 The dot product [ 1 4 5 ] ? y ? = (1 by 3)(3 by 1) is zero for points (x, y, z ) on a plane in ? ? z three dimensions. The columns of A are one-dimensional vectors. 24 A = [ 1 2 ; 3 4 ] and x = [ 5 ?2 ] and b = [ 1 7 ] . r = b ? A ? x prints as zero. 25 A ? v = [ 3 4 5 ] and v ? v = 50; v ? A gives an error message. 26 ones(4, 4) ? ones(4, 1) = [ 4 4 4 4 ] ; B ? w = [ 10 10 10 10 ] . 27 The row picture has two lines meeting at (4, 2). The column picture has 4(1, 1) + 2(?2, 1) = 4(column 1) + 2(column 2) = right side (0, 6). 28 The row picture shows 2 planes in 3-dimensional space. The column picture is in 2-dimensional space. The solutions normally lie on a line. 29 The row picture shows four lines. The column picture is in four -dimensional space. No solution unless the right side is a combination of the two columns. ? ? ? ? .7 .65 ?. The components always add to 1. They are always positive. 30 u 2 = ? ?, u 3 = ? .3 .35 31 u 7 , v 7 , w 7 are all close to (.6, .4). Their components still add to 1. ? ?? ? ? ? ? ? .8 .3 .6 .6 .8 .3 ? ? ? = ? ? = steady state s . No change when multiplied by ? ?. 32 ? .2 .7 .4 .4 .2 .7 ? ? ? ? 8 3 4 5+u 5?u+v 5?v ? ? ? ? ? ? ? ? 34 M = ? 1 5 9 ? = ? 5 ? u ? v 5 5 + u + v ?; M3 (1, 1, 1) = (15, 15, 15); ? ? ? ? 6 7 2 5+v 5+u?v 5?u M4 (1, 1, 1, 1) = (34, 34, 34, 34) because the numbers 1 to 16 add to 136 which is 4(34).Problem Set 2.2, page 401 Multiply by l =10 2= 5 and subtract to ?nd 2x + 3y = 14 and ?6y = 6.2 y = ?1 and then x = 2. Multiplying the right side by 4 will multiply (x, y ) by 4 to give the solution (x, y ) = (8, ?4). 3 Subtract ? 1 times equation 1 (or add 21 2times equation 1). The new second equation is 3y = 3.Then y = 1 and x = 5. If the right side changes sign, so does the solution: (x, y ) = (?5, ?1). 4 Subtract l =c atimes equation 1. The new second pivot multiplying y is d?(cb/a) or (ad?bc)/a.Then y = (ag ? cf )/(ad ? bc). 5 6x + 4y is 2 times 3x + 2y . There is no solution unless the right side is 2 ? 10 = 20. Then all points on the line 3x + 2y = 10 are solutions, including (0, 5) and (4, ?1).96 Singular system if b = 4, because 4x + 8y is 2 times 2x + 4y . Then g = 2 ? 16 = 32 makes the system solvable. The lines become the same : in?nitely many solutions like (8, 0) and (0, 4). 7 If a = 2 elimination must fail. The equations have no solution. If a = 0 elimination stops for a row exchange. Then 3y = ?3 gives y = ?1 and 4x + 6y = 6 gives x = 3. 8 If k = 3 elimination must fail: no solution. If k = ?3, elimination gives 0 = 0 in equation 2: in?nitely many solutions. If k = 0 a row exchange is needed: one solution. 9 6x ? 4y is 2 times (3x ? 2y ). Therefore we need b2 = 2b1 . Then there will be in?nitely many solutions. 10 The equation y = 1 comes from elimination. Then x = 4 and 5x ? 4y = c = 16. 11 2x + 3y + z = 8 y + 3z = 4 8z = 8 12 2x ? 3y = 3 y + z = 1 gives 2y ? 3z = 2 gives x = 2 y = 1 z = 1 2x ? 3y = 3 If a zero is at the start of row 2 or 3, that avoids a row operation. x=3 Subtract 2 × row 1 from row 2 Subtract 1 × row 1 from row 3 Subtract 2 × row 2 from row 3y + z = 1 and y = 1 ? 5z = 0 z=013 Subtract 2 times row 1 from row 2 to reach (d ? 10)y ? z = 2. Equation (3) is y ? z = 3. If d = 10 exchange rows 2 and 3. If d = 11 th third pivot is missing. 14 The second pivot position will contain ?2 ? b. If b = ?2 we exchange with row 3. If b = ?1 (singular case) the second equation is ?y ? z = 0. A solution is (1, 1, ?1). 15 (a) 0x + 0y + 2z = 4 x + 2y + 2z = 5 0x + 3y + 4z = 6 (exchange 1 and 2, then 2 and 3) (b) 0x + 3y + 4z = 4 x + 2y + 2z = 5 0x + 3y + 4z = 6 (rows 1 and 3 are not consistent)16 If row 1 = row 2, then row 2 is zero after the ? exchange the zero row with row 3 and there is no third pivot. If column 1 = column 2 there is no second pivot. 17 x + 2y + 3z = 0, 4x + 8y + 12z = 0, 5x + 10y + 15z = 0 has in?nitely many solutions. 18 Row 2 becomes 3y ? 4z = 5, then row 3 becomes (q + 4)z = t ? 5. If q = ?4 the system is singular ― no third pivot. Then if t = 5 the third equation is 0 = 0. Choosing z = 1 the equation 3y ? 4z = 5 gives y = 3 and equation 1 gives x = ?9. 19 (a) Another solution is1 (x 2+ X, y + Y, z + Z ).(b) If 25 planes meet at two points, theymeet along the whole line through those two points. 20 Singular if row 3 is a combination of rows 1 and 2. From the end view, the three planes form a triangle. This happens if rows 1 + 2 = row 3 on the left side but not the right side: for example x + y + z = 0, x ? 2y ? z = 1, 2x ? y = 1. No parallel planes but still no solution. 21 Pivots 2,3 4 5 , , 2 3 4in the equations 2x + y = 0,3 y 2+ z = 0,4 z 3+ t = 0,5 t 4= 5. Solution t = 4,z = ?3, y = 2, x = ?1. 22 The solution is (1, 2, 3, 4) instead of (?1, 2, ?3, 4).1023 The ?fth pivot is 6 . The nth pivot is (n+1) . 5 n ? ? ? 1 1 1 1 ? ? ? ? ? ? 24 A = ? a a + 1 a + 1 ? for any a, b, c leads to U = ? 0 ? ? ? b b+c b+c+3 0 ? ? a 2 ? if a = 2 or a = 0. 25 Elimination fails on ? a a1 1 01?? ? 1 ?. ? 326 a = 2 (equal columns), a = 4 (equal rows), a = 0 (zero column). ? ? ? ? 1 3 0 4 ? and ? ?. A = [ 1 1 0 0; 1 0 1 0; 27 Solvable for s = 10 (add equations); ? 1 7 2 6 0 0 1 1; 0 1 0 1 ] and U = [ 1 1 0 0; 0 ?1 1 0; 0 0 1 1; 0 0 0 0 ]. 28 Elimination leaves the diagonal matrix diag(3, 2, 1). Then x = 1, y = 1, z = 4. 29 A(2, :) = A(2, :) ? 3 ? A(1, :) Subtracts 3 times row 1 from row 2. 30 The average pivots for rand(3) without row exchanges were1 , 5, 10 2in one experiment―butpivots 2 and 3 can be arbitrarily large. Their averages are actually in?nite! With row exchanges in MATLAB’s lu code, the averages .75 and .50 and .365 are much more stable (and should be predictable, also for randn with normal instead of uniform probability distribution).Problem Set 2.3, page 50? 1 0 1 0 0 ? ? 1 0 1 7 0 ? ? 1 0 0 1 0 ?? 0 1 0 1 0 ? ? 0 1 0 0 0 ? ? ? 1 E21 = ??5 ? 0 2 E32 E21 b ? 1 0 ? ? 3 ??4 1 ? 0 0 ? ? 0? , ? 1 ? ? E32 = ? 0 ? 0 ? ? 0? , ? 1 ? ? P = ?0 ? 0 ?? ?? 1? ?1 ?? 0 0 ? ? ? ? 0? = ?0 ? ? 0 1 ? ? 1 ?. ? 0= (1, ?5, ?35) but E21 E32 b = (1, ?5, 0). Then row 3 feels no e?ect from row 1. ? ? ? ? ?? ?? ? 0 1 0 0 1 0 0 1 0 0 ?? ? ? ? ? ? ?? ?? ? ? ? ? ? ?? 0? , ?0 1 0? , ?0 1 0? ? E21 , E31 , E32 ? ??4 1 0? . ? ? ? ? ?? ?? ? 1 2 0 1 0 ?2 1 M = E32 E31 E21 10 ?2 1 ? ? ? ? ? ? ? ? 1 1 1 1 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 4 Elimination on column 4: b = ? 0 ? → ??4 ? → ??4 ? → ??4 ?. Then back substitution in ? ? ? ? ? ? ? ? 0 0 2 101 U x = (1, ?4, 10) gives z = ?5, y = 2 ,x= 1 . This solves Ax = (1, 0, 0). 25 Changing a33 from 7 to 11 will change the third pivot from 5 to 9. Changing a33 from 7 to 2 will change the pivot from 5 to no pivot. 6 If all columns are multiples of column 1, there is no second pivot. ? 1 0 1 0 0 ? ? ? 7 To reverse E31 , add 7 times row 1 to row 3. The matrix is R31 = ? 0 ? 7 ? ? 0 ?. ? 18 The same R31 from Problem 7 is changed to I . Thus E31 R31 = R31 E31 = I .11? 1 0 0 ?? ? ? ? 9 M = ? 0 0 1 ?. After the exchange, E must act on the new row 3. ? ? ?1 1 0 ? ? ? ? ? ? 1 0 1 1 0 1 2 0 1 ? ? ? ? ? ? ? ? ? ? ? ? 10 E13 = ? 0 1 0 ? ; ? 0 1 0 ? ; ? 0 1 0 ? . ? ? ? ? ? ? 0 0 1 1 0 1 1 0 1 ? ? 1 2 2 ? ? ? ? 11 A = ? 1 1 2 ? . ? ? 1 2 1 ? ? ? ? 9 8 7 1 2 3 ? ? ? ? ? ? ? ? 12 ? 6 5 4 ? , ? 0 1 ? 2 ?. ? ? ? ? 3 2 1 0 2 ?3 13 (a) E times the third column of B is the third column of EB row 3 to give nonzeros. 14 E21 has l21 = ? 1 , E32 has l32 = ? 2 , E43 has l43 = ? 3 . Otherwise the E ’s match the identity 2 3 4 matrix. ? ? ? ? ? ?1 ? 4 ? 7 ? 1 ?4 ? 7 1 0 ? ? ? ? ? ? ? ? ? ? 15 A = ? 1 ?2 ?5 ? → ? 0 ?6 ?12 ?. E32 = ? 0 1 ? ? ? ? ? 3 0 ?3 0 ?12 ?24 0 ?2 16 (a) X ? 2Y = 0 and X + Y = 33; X=22, Y=11 c = 1. a+ b+ c= 4 17 a=2 0 ? (b) E could add row 2 to? ? 0 ?. ? 1(b) 2m + c = 5 and 3m + c = 7; m = 2,a + 2b + 4c = 8 gives b = 1 .a + 3b + 9c = 14 c=1 ? ? ? 1 0 0 1 ? ? ? ? ? ? 18 EF = ? a 1 0 ?, F E = ? a ? ? ? b c 1 b + ac ? ? ? 0 1 0 0 0 ? ? ? ? ? ? 19 P Q = ? 0 0 1 ?, QP = ? 1 0 ? ? ? 1 0 0 0 1 more).0 1 c 1 ?0??10 1 00??10 1 3c0?? ? ? ? 2 0 ?, E = ? 2a ? ? 1 2b? ? ? ? 3 0 ?, F = ? 0 ? ? 1 0? ? 0? . ? 1? ? 2 2 2 2 0 ?, P = I, (?P ) = I, I = I, (?I ) = I (and many ? 0 ? ? ? ? ? ? ? ? ? rows20 (a) Each column is E times a column of B are multiples of [ 1 2 4 ]. ? ? ? 1 0 1 ?, F = ? 21 No. E = ? 1 1 0 22 (a) a3j xj (b) a21 ? a11(b) ?1 10 11 12 2 ? ?.4 4?=?1 22 44 81 1???, EF = ?1 11 2???, F E = ?2 11 1(c) x2 ? x1(d) (Ax )1 =a1j xj .23 E (EA) subtracts 4 times row 1 from row 2. AE subtracts 2 times column 2 of A from column 1.12? 24 [ A b ] = ? 2 4 3 1 1 17 ? ? 2 3 1 15 ? ?: 2x1 + 3x2 = 1 ?5x2 = 15 x1 = 5?→?0 ?5x2 = ?3.25 The last equation becomes 0 = 3. Change the original 6 to 3. Then row 1 + row 2 = row 3. ? ? ? ? ? ?? ? 1 4 1 0 1 4 1 0 ?7 4 ?→? ? → ? ?? ?. 26 (a) A ? 2 7 0 1 0 ?1 ?2 1 2 ?1 27 (a) No solution if d = 0 and c = 0 e?ect from a and b. 28 A = AI = A(BC ) = (AB )C = IC = C . 29 Given positive integers with ad ? bc = 1. Certainly c & a and b & d would be impossible. Also c & a and b & d would be impossible ? ? with integers.?This leaves ? row 1 ? & row ? 2 OR row 2 & 3 4 1 ?1 1 1 ?. Multiply by ? ? to get ? ?, then multiply row 1. An example is M = ? 2 3 0 1 2 3 ? ? ? ? ? ?? ?? ?? ? 1 0 1 1 1 1 1 0 1 0 1 1 ? to get ? ?. This shows that M = ? ?? ?? ?? ?. twice by ? ?1 1 0 1 0 1 1 1 1 1 0 1 ? ? ? ? 1 0 0 0 1 0 0 0 ? ? ? ? ? ? ? ? ? ?1 ? ?1 1 0 0? 1 0 0? ? and eventually M = “inverse of Pascal” = ? ? 30 E = ? ? ? ? ? ? 0 ?1 ? 1 ?2 1 0? 1 0? ? ? ? ? 0 0 ?1 1 ?1 3 ?3 1 reduces Pascal to I . (b) In?nitely many solutions if d = 0 and c = 0. NoProblem Set 2.4, page 591 BA = 3I is 5 by 5 AB = 5I is 3 by 3 ABD = 5D is 3 by 1. ABD: No A(B + C ): No. 2 (a) A (column 3 of B ) (b) (Row 1 of A) B (c) (Row 3 of A)(column 4 of B )(d) (Row 1 of C )D(column 1 of E ). ? ? 3 8 ?. 3 AB + AC = A(B + C ) = ? 6 9 4 A(BC ) = (AB )C = zero matrix ? ? ? ? n n 1 bn 2 2 ? and An = ? ?. 5 An = ? 0 1 0 0 ? ? ? 10 4 16 2 2 2 2 2 ? = A + AB + BA + B . But A + 2AB + B = ? 6 (A + B ) = ? 6 6 3 7 (a) True (b) False (c) True (d) False.2 0? ?.8 Rows of DA are 3?(row 1 of A) and 5?(row 2 of A). Both rows of EA are row 2 of A. Columns of AD are 3?(column 1 of A) and 5?(column 2 of A). Columns of AE are zero and column 1 of A + column 2 of A. ? ? a a+b ? and E (AF ) equals (EA)F because matrix multiplication is associative. 9 AF = ? c c+d13? 10 F A = ? a+c b+d ? ? a+c b+d ?? and then E (F A) = ? ?. E (F A) is not F (EA) because c d a + 2c b + 2d multiplication is not commutative. ? ? 0 0 1 ? ? ? ? 11 (a) B = 4I (b) B = 0 (c) B = ? 0 1 0 ? (d) Every row of B is 1, 0, 0, . . . ? ? 1 0 0 ? ? ? ? a 0 a b ? = BA = ? ? gives b = c = 0. Then AC = CA gives a = d : A = aI . 12 AB = ? c 0 0 0 13 (A ? B )2 = (B ? A)2 = A(A ? B ) ? B (A ? B ) = A2 ? AB ? BA + B 2 . 14 (a) True (b) False (c) True (b) mnp (d) False (take B = 0). (c) n3 (this is n2 dot products).15 (a) mn (every entry)16 By linearity (AB )c agrees with A(B c ). Also for all other columns of C . 17 (a) Use only column 2 of B (b) Use only row 2 of A ? ? ? ? ? ? 1 1 1 1 ?1 1 1/1 1/2 1/3 ? ? ? ? ? ? ? ? ? ? ? ? 18 A = ? 1 2 2 ?, ??1 1 ?1 ?, ? 2/1 2/2 2/3 ?. ? ? ? ? ? ? 1 2 3 1 ?1 1 3/1 3/2 3/3 (c)C(d) Use row 2 of ?rst A.19 Diagonal matrix, lower triangular, symmetric, all rows equal. Zero matrix. 20 (a) a11 ? (b) l31 = a31 /a11 ? 0 4 0 ? ? 0 0 4? ?, A3 = ? 0 0 0? ? 0 0 0 ?a31 21 (c) a32 ? ( a )a12 (d) a22 ? ( a )a12 . a11 11 ? ? ? ? ? ? 0 0 0 8 2y 4z ? ? ? ? ? ? ? ? ? ? ? ? ?0 0 0 0? ? 2z ? ? 4t ? ? ?, A4 = 0; then Av = ? ?, A2 v = ? ?, ? ? ? ? ? ? ?0 0 0 0? ? 2t ? ? 0? ? ? ? ? ? ? 0 0 0 0 0 00 ? ? ?0 21 A2 = ? ? ?0 ? 0 ? 8t ? ? ? ? ? 0? 4 ? A3 v = ? ? ?, A v = 0. ? 0? ? ? 0? 22 A = A2 = A3 = ? ? ? but AB = ?? and (AB )2 = 0. .5 ?.5 ? ? ? ?? ? ? ? 0 1 1 ?1 1 1 0 0 2 ? has A = ?I ; BC = ? ?? ?=? ?; 23 A = ? ?1 0 1 ?1 1 1 0 0 ? ?? ? ? ? 0 1 0 1 ?1 0 ?? ?=? ? = ?ED. DE = ? 1 0 ?1 0 0 1 ? ? ? ? ? ? 0 1 0 0 0 1 ? ? ? ? 0 1 ? ? ? ? has A2 = 0; A = ? 24 A = ? ? 0 0 1 ? has A2 = ? 0 0 0 ? but A3 = 0. ? ? ? ? 0 0 0 0 0 0 0 0 ? ? ? ? ? ? n n n n?1 2 2 ? 1 1 1 a a b n?1 ? ? ?, An ?, An ? ?. 25 An 1 = 2 = 2 3 = 0 1 1 1 0 0.5 ?.5?14? ? ? ? ? 1 0 3 ? ? ? ? ? ? ? ? ? ? 26 ? 2 ? 3 3 0 + ? 4 ? 1 2 1 = ? 6 ? ? ? ? ? 2 1 6 27 (a) (Row ? ? 3 of A)?(column ? x 0 ? ? ? ? ? ? (b) ? x ? 0 x x = ? 0 ? ? ? 0 0 28 A 3 6 6 0 ? ? 0 0 8 2 0 ? ? 3 3 14 8 0 ?? ? ? ? 0? + ?4 ? ? 0 1? ? ? ? 4 ? = ? 10 ? ? 1 7? ? 4 ?. ? 11 of B? ) = (Row 3 of A)?(column ? ? ?2 x x x 0 ? ? ? ? ? ? ? ? and 0 0 x = ?x? ?0 x x? ? ? ? ? 0 0 x 0 ;of B ) = ?0 0 x ? ? 0 x ?. ? 0 x; [????]B ; [????] ? ????? ????29303132x ? 1? ? ? Ax = ? x2 ? = x1 (column 1) + x2 (column 2) + ? ? ? . ? ? x3 ? ? ? ? ? 1 0 0 1 0 0 1 0 ? ? ? ? ? ? ? ? ? ? E21 = ? 1 1 0 ?, E31 = ? 0 1 0 ?, E = E31 E21 = ? 1 1 ? ? ? ? ? 0 0 1 ?4 0 1 ?4 0 ? ? ? ? ? ? ?2 0 1 1 1 ?, D = ? ?, D ? cb /a = ? ?. In Problem 30, c = ? 8 5 3 1 3 ? ?? ? ? ? A ?B x Ax ? By real part ? ?? ? = ? ? B A y Bx + Ay imaginary part.0??21 1 10?? ? ? ? 0 ?, then EA = ? 0 ? ? 1 0? ? 1 ?. ? 333 A times X will be the identity matrix I . ? ? ? ? ? 3 3 1 ? ? ? ? ? ? ? ? ? ? 34 The solution for b = ? 5 ? is 3x 1 + 5x 2 + 8x 3 = ? 8 ? ; A = ? ?1 ? ? ? ? ? 8 16 0 x1 , x2 , x3 . 35 S = D ? CA?1 B is the Schur complement: block version of d ? (cb/a). ? ? ? ? a+b a+b a+c b+d ? agrees with ? ? when b = c and a = d. 36 ? c+d c+d a+c b+d0 1 ?10?? ? 0 ? to produce ? 137 If A is “northwest” and B is “southeast” then AB is upper triangular and BA is lower triangular. One reason: Row i of A can have n ? i + 1 nonzeros, with zeros after that. Column j of B has j nonzeros, with zeros above that. If i & j then (row i of A) ? (column j of B ) = 0. So AB is upper triangular. Similarly BA is lower triangular. Problem 2.7.40 asks about inverses and transposes and permutations of a northwest A and a southeast ? ? ? 0 1 0 0 1 2 0 1 1 ? ? ? ? ? ? ?1 0 1 0 0? ?0 2 0 1 ? ? ? ? ? ? 2 38 A = ? 0 1 0 1 0 ? , A = ? 1 0 2 0 ? ? ? ? ? ? ?0 0 1 0 1? ?1 1 0 2 ? ? ? 1 0 0 1 0 0 1 1 0 gives diameter 3. B. ? ? 0 0 ? ? ? ? ?3 1? ? ? ? ? 3 1? , A = ?1 ? ? ? ? ?1 0? ? ? 2 3 ?3 0 3 1 11 3 0 3 11 1 3 0 33? ? 1? ? ? 1 ?, ? ? 3? ? 0A3 with A215? 0 1 0 0 0 1 0 0 0 ? ? 0 0 0 0 0 0 1 0 0 0 1 0 1 0 0 0 0 ? ? 0 0 0 0 1 0 0 0 0 0 1 1 0 0 0 0 0 ?? ? ?0 0 1 ? ? 39 A = ? 0 0 0 ? ? ?0 0 0 ? 1 0 0 so diameter 4.? ? ? ? ?0 0? ? ? ? ? 2 0? , A = ?0 ? ? ? ? ?1 1? ? ? 0 0? ? ? ? ?0 0? ? ? ? ? 3 1? , A = ?1 ? ? ? ? ?0 0? ? ? 0 0? ? 1? ? ? 0? ? ? 0? ? 0need also A4Problem Set 2.5, Page 72? 1 A?1 = ? ?, B ?1 = ? ?, C ?1 = ? ?. ?1 1 ? 5 3 2 ? ? 0 0 1 ? ? ? ? ?1 ?1 2 P = P; P = ? 1 0 0 ?. Always P ?1 = ?transpose? of P . ? ? 0 1 0 ? ? ? ? ? ? ? ? ? ? ? ? ? x .5 5 ?2 ?1 0 t ?.2 ?1 ? 1 1 ?, ? ? = ? ? so A = ? ?. A = ? ? and ? 3 ? ?=? 10 y ?.2 z .1 ?2 1 0 ?1 0 ? ?? ?? ??1 x t ?1 0 x t ?? ?? ? . any ? y z 0 1 y z1 301 4??1 20??7 ?4?00 1? ? and4 x + 2y = 1, 3x + 6y = 0: impossible. ? ? 1 ?1 ?. 5 U =? 0 ?1 6 (a) Multiply AB = AC by A?1 to ? ?nd B = ? C x y ?. (b) B and C can be any matrices ? ? x ?y 7 (a) In Ax = (1, 0, 0), equation 1 + equation 2 ? equation 3 is 0 = 1 must satisfy b1 + b2 = b3 (b) The right sides(c) Row 3 becomes a row of zeros―no third pivot. (b) Elimination keeps columns 1+2 = column8 (a) The vector x = (1, 1, ?1) solves Ax = 03. When columns 1 and 2 end in zeros so does column 3: no third pivot. 9 If you exchange rows 1 and 2 of A, you exchange columns 1 and 2 of A?1 . ? ? ? ? 0 0 0 1/5 3 ?2 0 0 ? ? ? ? ? ? ? ? ? ? ? ? 0 0 1 / 4 0 ? 4 3 0 0 ?, B ?1 = ? ? (invert each block). 10 A?1 = ? ? ? ? ? ? 0 ? 0 1/3 0 0 ? 0 6 ?5 ? ? ? ? ? 1/2 0 0 0 0 0 ?7 6 ? ? ? ? 1 0 0 0 ?, B = ? ?. 11 (a) A = I , B = ?I (b) A = ? 0 0 0 1 12 C = AB gives C ?1 = B ?1 A?1 so A?1 = BC ?1 . 13 M ?1 = C ?1 B ?1 A?1 so B ?1 = CM ?1 A.16? 14 B ?1 = A?1 ? 1 1 0 1 ??1 ? ? 1 0 1 ? ?: subtract column 2 of A?1 from column 1.= A?1 ? ?115 If A has a column of zeros, so does BA. So BA = I is impossible. There is no A?1 . ? ?? ? ? ? a b d ?b ad ? bc 0 ?? ? = ? ? = (ad ? bc)I . The inverse of one matrix is the 16 ? c d ?c a 0 ad ? bc other divided by ad ? bc. ? ?? ?? ? ? ? ? ? 1 1 1 1 1 ? ?? ?? ? ? ? ? ? ? ?? ?? ? ? ? ? ? 17 ? ? ??1 1 ? = ??1 ? = E; ? 1 1 ? = L = E ?1 ?? 1 1 1 ?? ?? ? ? ? ? ? ? ?1 1 ?1 1 1 0 ?1 1 1 1 1 after reversing the order and changing ?1 to +1. 18 A2 B = I can be written as A(AB ) = I . Therefore A?1 is AB . 19 The (1, 1) entry requires 4a ? 3b = 1; the (1, 2) entry requires 2b ? a = 0. Then b = a=2 . 5 1 5andFor the 5 by 5 case 5a ? 4b = 1 and 2b ? a = 0 give b =1 6and a =2 . 620 A ? ones(4, 1) is the zero vector so A cannot be invertible. 21 6 of ? 1 22 ? 2 ? 1 ? 3 ? 2 ? ? 23 ? 1 ? 0 ? 2 ? ? ?0 ? 0 ? 2 ? ? ?0 ? 0 ? 1 ? ? 24 ? 0 ? 0 the 16 are invertible, including all four with three 1’s. ? ? ? ? ? 3 1 0 1 3 1 0 1 0 7 ?3 ?→? ?→? ? = I A?1 ; 7 0 1 0 1 ?2 1 0 1 ?2 1 ? ? ? 3 1 0 1 0 ?8 3 ?→? ? = I A?1 . 8 0 1 0 1 3 ?1 ? ? ? 1 0 1 0 0 2 1 0 1 0 0 ? ? ? ? ? ? 2 1 0 1 0? → ?0 3/2 1 ?1/2 1 0? → ? ? ? 1 2 0 0 1 0 1 2 0 0 1 ? ? ? 1 0 1 0 0 2 1 0 1 0 0 ? ? ? ? ? ? 3/2 1 ?1/2 1 0 ? → ? 0 3/2 0 ?3/4 3/2 ?3/4 ? → ? ? ? 0 0 4/3 1/3 ?2/3 1 0 4/3 1/3 ?2/3 1 ? ? ? 0 0 3/2 ?1 1/2 1 0 0 3/4 ?1/2 1/4 ? ? ? ? ? ? 3/2 0 ?3/4 3/2 z ? 3/4 ? → ? 0 1 0 ?1/2 1 ?1/2 ?. ? ? ? 0 4/3 1/3 ?2/3 1 0 0 1 1/4 ?1/2 3/4 ? ? ? ? ? a b 1 0 0 1 a 0 1 0 ?b 1 0 0 1 ?a ac ? b ? ? ? ? ? ? ? ? ? ? 1 c 0 1 0 ? → ? 0 1 0 0 1 ?c ? → ? 0 1 0 0 1 ? c ?. ? ? ? ? ? 0 1 0 0 1 0 0 1 0 0 1 0 0 1 0 0 1 ? ? ? ? ? ? 3 ?1 ?1 1 0 ? ? ? ? ? ? ? ? ? ? ? ?1 ?1 1 ? 25 A = 4 ??1 3 ?1 ?; B ? 1 ? = ? 0 ? so B does not exist. ? ? ? ? ? ? ?1 ? 1 3 1 0 ? ? ? ? ? ?? ? ? ? ? 1 0 1 2 1 ?1 1 0 1 0 1 ?A = ? ?. Then ? ?? ?A = ? ?. Multiply by D = ? 26 ? ?2 1 0 2 0 1 ?2 1 0 2 0 ? ? 3 ?1 ? = A?1 . to reach I . Here D?1 E12 E21 = ? ?1 1/20 1/2? ?17? 1 0 0 ? ? 2 ?1 0 ?? ? ? ? ? ? ? ? ?1 27 A?1 = ??2 1 ?3 ? (notice the pattern); A = ??1 2 ? 1 ?. ? ? ? ? 0 0 1 0 ?1 1 ? ? ? ? ? ? 2 2 0 1 2 0 ?1 1 1 0 ?1/2 1/2 ?→? ?→? ? = I A?1 . 28 ? 0 2 1 0 0 2 1 0 0 1 1/2 0 29 (a) True (AB has a row of zeros) (b) False (matrix of all 1’s) (c) True (inverse of A?1is A) (d) True (inverse of A2 is (A?1 )2 ). 30 Not invertible for c = 7 (equal columns), c = 2 (equal rows), c = 0 (zero column). ? ? a 0 ?b ? ? 1 ? ? 31 Elimination produces the pivots a and a ? b and a ? b. A?1 = ? ?a a 0 ?. a(a ? b) ? ? 0 ?a a ? ? 1 1 0 0 ? ? ? ? ?0 1 1 0? ?1 ?. The 5 by 5 A?1 also has 1’s on the diagonal and superdiagonal. 32 A = ? ? ? ?0 0 1 1? ? ? 0 0 0 1 33 x = (2, 2, 2, 1). 34 x = (1, 1, . . . , 1) has P x = Qx so (P ? Q)x = 0. ? ? ? ? ? ? I 0 A?1 0 ?D I ? and ? ? and ? ?. 35 ? ?C I ?D?1 CA?1 D?1 I 0 36 If AC = CA, multiply left and right by A?1 to ?nd CA?1 = A?1 C . If also BC = CB , then (using the associative law!!), (AB )C = A(BC ) = A(CB ) = (AC )B = (CA)B = C (AB ).37 A can be invertible but B is always singular. Each row of B will add to zero, from 0 + 1 + 2 ? 3, so the vector x = (1, 1, 1, 1) will give B x = 0. I thought A would be invertible as long as you put the 3’s on its main diagonal, but that’s wrong: ? 3 0 3 2 2 1 1 3 0 2 ?? 1 ? ? 0 1 0 3 2 2 1 0 3 3 ?? ? ?0 Ax = ? ? ?1 ? 1?? ? ?? ? 2? ? 1 ? ?? ? = 0 ?? ? 0 ? ??1 ? ?? ? ?1 3but? ? ?3 A=? ? ?2 ? 1? ? 2? ? ? 1? ? 0is invertible38 AD = pascal(4, 1) is its own inverse. 39 hilb(6) is not the exact Hilbert matrix because fractions are rounded o?. 40 The three Pascal matrices have S = LU = LLT and then inv(S ) = inv(LT )inv(L). Note that the triangular L is abs(pascal(n, 1)) in MATLAB. 41 For Ax = b with A = ones(4, 4) = singular matrix and b = ones(4, 1) in its column space, MATLAB will pick the shortest solution x = (1, 1, 1, 1)/4. Any vector in the nullspace of A could be added to this particular solution. 42 If AC = I for square matrices then C = A?1 (it is proved in 2I that CA = I will also be true). The same will be true for C ? . But a square matrix has only one inverse so C = C ? .1843 M M ?1 = = = (In ? U V ) (In + U (Im ? V U )?1 V ) In ? U V + U (Im ? V U )?1 V ? U V U (Im ? V U )?1 V In ? U V + U (Im ? V U )(Im ? V U )?1 V = In (formulas 1, 2, 4 are similar)Problem Set 2.6, page 84? 1 221= 1; L = ? = 1 and1 10 1??? times U x = c is Ax = b : ?1 1?? ? ? ? x 5 ? ? ? = ? ?. 2 y 7 13132= 2 (and33= 1): reverse the steps to recover x + 3y + 6z = 11 from U x = c :1 times (x + y + z = 5) + 2 times (y + 2z = 2) + 1 times (z = 2) gives x + 3y + 6z = 11. ? ?? ? ? ? ? ? ? ?? ? ? ? ? ? 1 0 c1 5 5 1 1 x1 5 3 ? ? ? = ? ?; c = ? ?. U x = c is ? ? ? ? = ? ?; x = ? ? . 3 Lc = b is ? 1 1 c2 7 2 0 1 x2 2 2 ? 1 1 2 1 1 0 ? ? ? 5 1 ?? ? ? ? ? ? ? ?? ? ? ? ? ? ? ? ? c ? = ? 7 ? ; c = ? 2 ?. U x = ? ?? ? ? ? ? ? ? 1 11 2 5 ?? 2 1 4 3 ? 0 ? ? 2 1 4 0 1 ? 0 ? ?? ? ? ? 1 1 1 ? ? ? ? 5 5 ?? ? ? ? ? ? ?? ? ? ? ? ? 2 ? ? x ? = ? 2 ?; x = ??2 ?. ?? ? ? ? ? ? 1 2 2 ? 1 1 0 ? 1 ? ? ? ? U. ? ?? ?? ? 4 Lc = ? 1 ? 1 ?? ? 5 EA = ? 0 ? ?3 1 ? ? 6 ?0 1 ? 0 ?2 ??? ?? ? ?0 ?? 1 6? ? ? ? 2? = ?0 ? ? 5 0 ? 1? ? ? ? 2 ? = U ; A = LU = ? 0 ? ? 5 3 ? 0 1 2 0 2 4 1 01 1 ? ? ?? ? ? ?? ? ??2 1 ? A = ? 0 ?? ? ? 1 0 0 1 0 ? 1 1 ?2 ? ?? 1??1 ? ? ? ? 2 3 ? = U . Then A = ? 2 ? ? 0 ?6 0 ?? 1 1 ? 1 1 ?? 1? ? ?1 ?1 0 ? U = E21 E32 U = LU . ? 1 ?? ? 7 E32 E31 E21 A = ? ? 1 ? ? This is ? 0 ? 0 ? 0 2 0 1?? ?? ?? ?? 1 ?3?? ?? ? ??2 ?? 0 1 2 0 ??? ?? ? ?2 ?? 1 3? ? 2 ?. ? 51 ? ? ? ? 0 ? = U . Then A = ? 2 ? ? 2 3 ? 1 1 ?c ? ?? 1? ? 0 ? U = LU . ? 1 ?? 1 1 ? ? 1 ? ? ? ?. This is ?? ? 8 E = E32 E31 E21 = ? ? L?1?? ?? ?? ?? 1 ?b ?1 1 ??? ?? ? ??a ??? ? ? ? ? = ? ?a 1 ? ? 1 ac ? b ?c ?1=A?1. 1 1 1 2 0 1 1 n 1 ?? ?? ?? ?? ?? d e f g d = 1, e = 1, then l = 1 f = 0 is not allowed no pivot in row 2? ? 9 2 by 2: d = 0 ? 1 ? 1? ? ? ? 2? = ? l ? ? 1 m? ? h? ? i10 c = 2 leads to zero in the second pivot position: exchange rows and the matrix will be OK. c = 1 leads to zero in the third pivot position. In this case the matrix is singular.19? 2 4 3 8 ? ? 2 ?? ? 11 A = ? 0 ? 01213141516? ? ? ? ? ? ?; A = LU has U = A (pivots on the diagonal); 9 ? has L = I and D = ? 3 ? ? ? 0 7 7 ? ? 1 2 4 ? ? ? ? A = LDU has U = D?1 A = ? 0 1 3 ? with 1’s on the diagonal. ? ? 0 0 1 ? ? ? ?? ? ? ?? ?? ? 2 4 1 0 2 4 1 0 2 0 1 2 ?=? ?? ?=? ?? ?? ? = LDU ; notice U is LT A=? 4 11 2 1 0 3 2 1 0 3 0 1 ? ?? ? ? ?? ?? ? 1 1 4 0 1 1 1 4 0 ? ?? ? ? ?? ?? ? ? ?? ? ? ?? ?? ? T A = ?4 ? ? 0 ?4 ?? ? ?0 1 4? = ?4 1 ?4 1 ?1 ? = LDL . ? ?? ? ? ?? ?? ? 0 ?1 1 0 0 4 0 ?1 1 4 0 0 1 ? ? ? ?? ? a a a a 1 a a a a a=0 ? ? ? ?? ? ? ? ? ?? ? ?a b b b ? ?1 1 ?? b ? a b ? a b ? a? b=a ? ?=? ?? ?. Need ? ? ? ?? ? ?a b c c ? ?1 1 1 ?? c ? b c ? b? c=b ? ? ? ?? ? a b c d 1 1 1 1 d?c d=c ? ? ? ?? ? a r r r 1 a r r r a=0 ? ? ? ?? ? ? ? ? ?? ? ?a b s s? ?1 1 ?? b ? r s ? r s ? r? b=r ? ?=? ?? ?. Need ? ? ? ?? ? ?a b c t ? ?1 1 1 ?? c ? s t ? s? c=s ? ? ? ?? ? a b c d 1 1 1 1 d?t d=t ? ? ? ? ? ? ? ? ? ? ? ? 1 0 2 2 2 4 2 ?5 ? ? c = ? ? gives c = ? ?. Then ? ? x = ? ? gives x = ? ?. 4 1 11 3 0 1 3 3 ? ? ? ? 2 4 2 ? times x is b = ? ?. Check that A = LU = ? 8 17 11 ? ? ? ? ? ? ? ? ? ? ? ? 1 0 0 4 4 1 1 1 4 3 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 1 1 0 ? c = ? 5 ? gives c = ? 1 ?. Then ? 0 1 1 ? x = ? 1 ? gives x = ? 0 ?. ? ? ? ? ? ? ? ? ? ? ? ? 1 1 1 6 1 0 0 1 1 1 (b) I goes to L?1 (c) LU goes to U .17 (a) L goes to I1 ?1 18 (a) Multiply LDU = L1 D1 U1 by inverses to get L? . The left side is lower 1 LD = D1 U1 Utriangular, the right side is upper triangular ? both sides are diagonal. (b) Since L, U, L1 , U1 have diagonals of 1’s we get D = D1 . Then ? ?? ? ? ? 1 1 1 0 a a 0 ? ?? ? ? ? ? ?? ? ? ? 19 ? 1 1 ?? 1 1 ? = LIU ; ? a a + b b ? = (same L) ? ?? ? ? ? 0 1 1 1 0 b b+c1 L? L is I and U1 U ?1 is I . ?1 ? a ? ? ? ? ? ? (same U ). b ? ? c20 A tridiagonal T has 2 nonzeros in the pivot row and only one nonzero below the pivot (so 1 operation to ? 1 2 ? ? ?2 3 T =? ? ?0 1 ? 0 0 ?nd the the new pivot!). T = bidiagonal L ? multiplier ? and 1 to ?nd ? ? 0 0 1 2 0 0 1 0 ? ? ? ? ? ? ? ? ? 0 ?1 1 0 ? ?2 1 0? 1 ? ?→ U = ? ?. Reverse steps by L = ? ? ? ? ? ?0 ? 0 ?1 2 3? 0 3 3? ? ? ? ? 3 4 0 0 0 1 0 0 times?U : 0 0 ? ? 0 0? ?. ? 1 0? ? 1 12021 For A, L has the 3 lower zeros but U may not have the upper zero. For B , L has the bottom left zero and U has the upper right zero. One zero in A and two zeros in B are ?lled in. ? ? ? ?? ? x x x 1 0 0 ? ? ? ? ? ? ?? ? ? ? ? ?? ? 22 ? x x x ? = ? ? 1 0 ? ? 0 ? (?’s are all known after the ?rst pivot is used). ? ? ? ?? ? x x x ? 1 0 ? ? ? ? ? ? ? ? 5 3 1 4 2 0 2 0 0 1 1 1 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 23 ? 3 3 1 ? → ? 2 2 0 ? → ? 2 2 0 ? = L. Then A = U L with U = ? 0 1 1 ?. ? ? ? ? ? ? ? ? 1 1 1 1 1 1 1 1 1 0 0 1 ? ? ? ? 1 1 0 0 5 1 1 0 0 5 ? ? ? ? ? ?? ? ? ? ? ? ? ? ?2 1 1 0 8? ? 0 ?1 1 0 ?2 ? ?1 1 x2 ?2 ?→? ?. Solve ? ? ? ? = ? ? for x2 = 3 24 ? ? ? ? ? ?0 1 3 2 8? ?0 1 1 0 4? 1 1 x3 4 ? ? ? ? 0 0 1 1 2 0 0 1 1 2 and x3 = 1 in the middle. Then x1 = 2 backward and x4 = 1 forward. 25 The 2 by 2 upper submatrix B has the ?rst two pivots 2, 7. Reason: Elimination on A starts in the upper left corner with elimination on B . 26 The ?rst three pivots for M are still 2, 7, 6. To be sure that 9 is the fourth pivot, put zeros in the ? 1 ? ? ?1 ? ? 27 ? 1 ? ? ?1 ? 1 rest of row 4 and column ? ? 1 1 1 1 1 ? ? ? ? 2 3 4 5? ?1 ? ? ? ? 3 6 10 15 ? = ? 1 ? ? ? ? 4 10 20 35 ? ? 1 ? ? 5 15 35 70 1 4. ?? 1 2 3 4 1 3 6 1 4 1 ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? 1 1 1 1 2 1 1 3 3 1 1 ? Pascal’s triangle in L and U . MATLAB’s lu code will wreck the pattern. chol does no row exchanges for symmetric matrices with positive pivots. ? ? 4? ? ? 6 ?. ? ? 4? ? 128 c = 6 and also c = 7 will make LU impossible (c = 6 needs a row exchange). 32 inv(A) ? b should take 3 times as long as A\b (n3 for A?1 vs n3 /3 multiplications for LU ). 34 The upper triangular part triu(A) should be about three times faster to invert. 35 Each new right side costs only n2 steps compared to n3 /3 for full elimination A\b . 36 This L comes from the ?1, 2, ?1 tridiagonal A = LDLT . (Row i of L) ? (Column j of L?1 ) =1?i i j +(1) i?1 j i= 0 for i & j so LL?1 = I . Then L?1 leads to A?1 = (L?1 )T D?1 L?1 .1 The ?1, 2, ?1 matrix has inverse A? ij = j (n ? i + 1)/(n + 1) for i ≥ j (reverse for i ≤ j ).Problem Set 2.7, page 95? 1 A =? ?1 c2 T1 0 c9 3 ?? ?, A?1?=? ?310 1/3? ?, (A?1 T? ) = (A )T ?1=?1 0?3 1/3? ?; AT = A and A?1 =0?c ?1? = (A?1 )T .2 In case AB = BA, transpose both sides: AT commutes with B T . 3 (AB )?1T= (B ?1 A?1 )T = (A?1 )T (B ?1 )T .21? 4 A=? 0 0 1 0 ? ? has A2 = 0. But the diagonal entries of AT A are dot products of columns of Awith themselves. If AT A = 0, zero dot products ? zero columns ? A = zero matrix. ? ? 2 T T 5 (a) x Ay = a22 = 5 (b) x A = 4 5 6 (c) Ay = ? ?. 5 ? ? AT C T ?; M T = M needs AT = A, B T = C, DT = D. 6 MT = ? B T DT 7 (a) False (needs A = AT ) (b) False (c) True (d) False.8 The 1 in column 1 then the 1 in column 2 has n ? 1 . . . (n! choices overall). ? ?? ?01 0 0010 0 10? ? 9 P1 P2 = ? 0 ? 1?? ?? 1? ?0 ?? 0 0? ? 1 ? = P2 P1 . ? 010 (3, 1, 2, 4), (2, 3, 1, 4) keep 4 6 more keeping 1 or 2 or 3 (2, 1, 4, 3) and (3, 4, 1, 2) ? 0 ? ? 11 P = ? 0 ? 1 exchanging 2 pairs. ? 1 0 ? ? 0 1 ?; ? 0 0 ? 1 0 0 1 0 ? ? 0 0 1 0 1 ? ? ? ? ? 1 ?, P2 = ? 0 ? ? 0 1T? ? No AP is lower triangular (this is a column exchange); P1 = ? 0 ? 0T T T T T? ? 0 ?. ? 012 (P x ) (P y ) = x P P y = x y because P P = I ; In general P x ? y = x ? P y = x ? P y : ? 0 ? ? ?0 ? 1 ? ? 0 P ? ? for the same P had P 4 = P . 1 0 0 ?? ? ? ? ? ? ? 1 1 1 0 ?? ? ? ? ? ? ? ?? ? ? ? ? ? ? 1? ?2? ? ?1? = ?2? ? ?0 ?? ? ? ? ? ? ? 0 3 2 3 1 0 1 0 0 ?? ? 1 ?? ? ?? ? 1? ?1? . ?? ? 0 2 001 0 00? ? 13 P = ? 0 ? 1? ? 1 ? 1 ? P = ? ? 0 014 There are n! permutation matrices of order n. Eventually two powers of P must be the same: P r = P s and P r?s = I . Certainly r ? s ≤ n! ? ? ? ? ? 0 ? P2 0 1 ? ? ? ? ? P = is 5 by 5 with P2 = and P3 = ? 0 ? P3 1 0 1 ? ? E 0 ? = P T with E 15 (a) P T (row 4) = row 1 (b) P = ? 0 E ?1 00? ? 1 ?. ? 0 0 ? ? 0 1 ? moves all rows. =? 1 0 ? ?.16 A2 ? B 2 and ABA are symmetric if A and B are symmetric. ? ? ? ? ? ? ? 1 1 0 1 1 1 1 ? ? ? has D = ? 17 (a) A = ? (b) A = ? (c) A = ? 1 1 1 1 1 0 00 ?12218 (a) 5 + 4 + 3 + 2 + 1 = 15 independent entries if A = AT 15 in LDLT T(b) L has 10 and D has 5: total(c) Zero diagonal if A = ?A, leaving 4 + 3 + 2 + 1 = 10. (b) (RT R)jj = (column j19 (a) The transpose of RT AR is RT AT RTT = RT AR = n by nof R) ? (column j of R) = length squared of column j . ? ? ? ?? ?? ? ? ? ? ?? ?? ? 1 3 1 0 1 0 1 3 1 b 1 0 1 0 1 b ?=? ?? ?? ?; ? ?=? ?? ?? ?. 20 ? 3 2 3 1 0 ?7 0 1 b c b 1 0 c ? b2 0 1 ? ? ? ? ?5 ? 7 d ? b2 e ? bc ?, ? ?. 21 Lower right 2 by 2 matrix is ? ?7 ?32 e ? bc f ? c2 ? ? ? ? ?? ? ? ? ? ?? 1 0 1 1 1 1 2 0 0 1 1 ? ? ? ?? ? ? ? ? ?? ? ? ? ? ?? ? ? ? ? ?? ? 22 ? 1 0 ? A = ?0 1 ?? ?? ?1 1? 1 1 ?; ? 0 1? A = ?1 1 ? ? ? ? ?? ? ? ? ? ?? 1 2 3 1 ?1 1 0 2 0 1 1 ? ? 0 0 1 ? ? ? ? 23 A = ? 1 0 0 ? = P and L = U = I ; exchanges rows 1C2 then rows 2C3. ? ? 0 1 0 ? ?? ? ? ?? ? 1 0 1 2 1 2 1 1 ? ?? ? ? ?? ? ? ?? ? ? ?? ? 24 ? ? ?0 3 8? = ?0 ?? 1 1 3 8 ?. If we wait to exchange, then ? ?? ? ? ?? ? 1 2 1 1 0 1/3 1 ?2/3 ?? ? ? ?? 1 1 2 1 1 ? ?? ?? ? ? ?? ?? ? A = L 1 P 1 U1 = ? 3 1 ? ?1 ? ? 0 1 2 ?. ?? ? ? ?? 0 0 2 1 1 ? ? ? ? 2 3 0 1 ? and P → ? ?; no more elimination 25 abs(A(1, 1)) = 0 and abs(A(2, 1)) & A → ? 0 1 1 0 ? ? 2 3 4 ? ? ? ? so L = I and U = new A. abs(A(1, 1)) = 0 and abs(A(2, 1)) & A → ? 0 0 1 ? and ? ? 0 5 6 ? ? ? ? ? ? 0 1 0 2 3 4 0 1 0 ? ? ? ? ? ? abs(A(2, 2)) = 0 ? ? ? ? ? ? P → ? 1 0 0 ?; ; A → ? 0 5 6 ?, L = I , P → ? 0 0 1 ? . ? ? ? ? abs(A(3, 2)) & tol ? ? 0 0 1 1 0 0 0 0 1 ? ? 1 1 0 ? ? ? ? 26 abs(A(1, 1)) = 0 so ?nd abs(A(2, 1)) & exchange rows to A = ? 0 1 2 ? and P = ? ? 2 5 4 ? ? ? ? ? ? 0 1 0 1 1 0 1 0 0 ? ? ? ? ? ? ? ? ? ? ? ? ? 1 0 0 ?; eliminate to A = ? 0 1 2 ? and L = ? 0 1 0 ?, same P ; abs(A(2, 2)) & tol ? ? ? ? ? ? 0 0 1 0 3 4 2 0 1 ? ? ? ? 1 1 0 1 0 0 ? ? ? ? ? ? ? ? so eliminate to A = ? 0 1 2 ? = ?nal U and L = ? 0 1 0 ?. ? ? ? ? 2 3 1 0 0 ?2 27 No solution23? 1 1 0 1 ? ? ? ? shows the elimination steps as actually done (L is a?ected by P ). ?? ? 28 L1 = ? 1 ? 229 One way to decide even vs. odd is to count all pairs that P has in the wrong order. Then P is even or odd when that count is even or odd. Hard step: show that an exchange always reverses that count! Then 3 or 5 exchanges will leave that count odd. ? ? ? ? ? 1 1 0 0 1 ? ? ? ? ? ? ? ? ? ? T 30 E21 = ??3 1 ? and E21 AE21 = ? 0 2 4 ? E32 = ? 1 ? ? ? ? ? 1 0 4 9 ?4? ? ? ? ?1T T and E32 E21 AE21 E32 = D. Elimination from both sides gives the symmetric LDLT directly. ? ?? ? ? ? yBC yBC + yBS 1 0 1 ? ?? ? ? ? ? ?? ? ? ? 31 Total currents are AT y = ??1 1 0 ? ? yCS ? = ??yBC + yCS ?. ? ?? ? ? ? 0 ? 1 ?1 yBS ?yCS ? yBSEither way (Ax )T y = x T (AT y ) = xB yBC + xB yBS ? xC yBC + xC yCS ? xS yCS ? xS yBS . ? ? ? ? ? ? 700 ? ? 1 50 ? ? ? ? x1 ? ? 1 40 2 6820 1 truck ? ? ? ?? ? 32 Inputs ? 40 1000 ? ? ? = A AT y = ? ? 3 ?= ? ? ? x2 ? ? 50 000 1 plane 2 50 3000 33 Ax ? y is the cost of inputs while x ? AT y is the value of outputs. 34 P 3 = I so three rotations for 360? ; P rotates around (1, 1, 1) by 120? . ? ? ? ?? ? 1 2 1 0 1 2 ?=? ?? ? = EH 35 ? 4 9 2 1 2 5 36 L(U T )?1 = triangular times triangular. The transpose of U T DU is U T DT U TT = U T DU again. 37 These are groups: Lower triangular with diagonal 1’s, diagonal invertible D, permutations P , orthogonal ? 0 1 2 ? ? ?1 2 3 38 ? ? ?2 3 0 ? 3 0 1 matrices with QT = Q?1 . ? 3 ? ? 0? ? (I don’t know any rules for constructions like this) ? 1? ? 2a b c d39 Reordering the rows and/or columns ofwill move the entry a.1 1 ?1 1 040 Certainly B T is northwest. B 2 is a full matrix! B ?1 is southeast:=0 1 1 ?1. The?1rows of B are in reverse order from a lower triangular L, so B = P L. Then B?1=LP ?1has the columns in reverse order from L?1 . So B ?1 is southeast. Northwest times southeast is upper triangular! B = P L and C = P U give BC = (P LP )U = upper times upper. 41 The i, j entry of P AP is the n ? i + 1, n ? j + 1 entry of A. The main diagonal reverses order.24Problem Set 3.1, Page 1071 x + y = y + x and x + (y + z ) = (x + y ) + z and (c1 + c2 )x = c1 x + c2 x . 2 The only broken rule is 1 times x equals x . 3 (a) cx may not be in our set: not closed under scalar multiplication. Also no 0 and no ?x (b) c(x + y ) is the usual (xy )c , while cx + cy is the usual (xc )(y c ). Those are equal. With c = 3, x = 2, y = 1 they equal 8. This is 3(2 + 1)!! The zero vector is the number 1. ? ? ? ? ? ? 0 0 1 ?1 ?2 2 1 ?; A = ? ? and ?A = ? ?. The smallest 4 The zero vector in M is ? 2 0 0 1 ?1 ?2 2 subspace containing A consists of all matrices cA. 5 (a) One possibility: The matrices cA form a subspace not containing B subspace must contain A ? B = I 6 h(x) = 3f (x) ? 4g (x) = 3x2 ? 20x. 7 Rule 8 is broken: If cf (x) is de?ned to be the usual f (cx) then (c1 + c2 )f = f ((c1 + c2 )x) is di?erent from c1 f + c2 f = usual f (c1 x) + f (c2 x). 8 If (f + g )(x) is the usual f (g (x)) then (g + f )x is g (f (x)) which is di?erent. In Rule 2 both sides are f (g (h(x))). Rule 4 is broken because there might be no inverse function f ?1 (x) such that f (f ?1 (x)) = x. If the inverse function exists it will be the vector ?f . 9 (a) The vectors with integer components allow addition, but not multiplication by1 2(b) Yes: the(c) All matrices whose main diagonal is all zero.(b) Remove the x axis from the xy plane (but leave the origin). Multiplication by any c is allowed but not all vector additions. 10 Only (a) (d) (e) are subspaces. ? ? ? a b a ? 11 (a) All matrices ? (b) All matrices ? 0 0 0 12 The sum of (4, 0, 0) and (0, 4, 0) is not on the plane. 13 P0 has the equation x + y ? 2z = 0; (2, 0, 1) and (0, 2, 1) and their sum (2, 2, 2) are in P0 . 14 (a) The subspaces of R2 are R2 itself, lines through (0, 0), and (0, 0) itself4 4a 0? ? (c) All diagonal matrices.(b) The sub-spaces of R are R itself, three-dimensional planes n ? v = 0, two-dimensional subspaces (n 1 ? v = 0 and n 2 ? v = 0), one-dimensional lines through (0, 0, 0, 0), and (0, 0, 0, 0) alone. 15 (a) Two planes through (0, 0, 0) probably intersect in a line through (0, 0, 0) and line probably intersect in the point (0, 0, 0) (b) The plane(c) Suppose x is in S ∩ T and y is inS ∩ T . Both vectors are in both subspaces, so x + y and cx are in both subspaces. 16 The smallest subspace containing P and L is either P or R3 . ? ? ? ? 1 0 0 0 ?+? ? is not singular. 17 (a) The zero matrix is not invertible (b) ? 0 0 0 1 18 (a) True (b) True (b) False.19 The column space of A is the x axis = all vectors (x, 0, 0). The column space of B is the xy plane = all vectors (x, y, 0). The column space of C is the line of vectors (x, 2x, 0).2520 (a) Solution only if b2 = 2b1 and b3 = ?b1 (b) Solution only if b3 = ?b1 .21 A combination of the columns of C is also a combination of the columns of A ( B has a di?erent column space). 22 (a) Every b (b) Solvable only if b3 = 0 (c) Solvable only if b3 = b2 .23 The extra? column b ? enlarges the column space unless in the column space of A: ? b is already ? 1 0 1 (larger column space) 1 0 1 (b already in column space) ? ? ? [A b ] = ? 0 0 1 (no solution to Ax = b ) 0 1 1 (Ax = b has a solution) 24 The column space of AB is contained in (possibly equal to) the column space of A. If B = 0 and A = 0 then AB = 0 has a smaller column space than A. 25 The solution to Az = b + b ? is z = x + y . If b and b ? are in the column space so is b + b ? . 26 The column space of any invertible 5 by 5 matrix is R5 . The equation Ax = b is always solvable (by x = A?1 b ) so every b is in the column space. 27 (a) False ? 1 1 ? ? 28 A = ? 1 0 ? 0 1 (b) True ? ? 0 1 ? ? ? ? 0 ? or ? 1 ? ? 0 0 (c) True (d) False. ? ? ? 1 2 1 2 0 ? ? ? ? ? ? 0 1 ?; A = ? 2 4 0 ? (columns on 1 line). ? ? ? 1 1 3 6 029 Every b is in the column space so that space is R9 .Problem Set 3.2, Page 118? 1 2 0 0 2 1 0 4 2 0 6 ? ? 2 4 4 0 2 ? ? ? 1 (a) U = ? 0 ? 0 ? Free variables x2 , x4 , x5 ? 3? ? Pivot variables x1 , x3 0 ? ? (b) U = ? 0 ? 0 ? Free x3 ? 4? ? Pivot x1 , x2 02 (a) Free variables x2 , x4 , x5 and solutions (?2, 1, 0, 0, 0), (0, 0, ?2, 1, 0), (0, 0, ?3, 0, 1) (b) Free variable x3 : solution (1, ?1, 1). 3 The complete solutions are (?2x2 , x2 , ?2x4 ? 3x5 , x4 , x5 ) and (2x3 , ?x3 , x3 ). The nullspace contains only 0 when there are no free variables. ? ? ? ? 1 2 0 0 0 1 0 ?1 ? ? ? ? ? ? ? ? 4 R = ? 0 0 1 2 3 ?, R = ? 0 1 1 ?, R has the same nullspace as U and A. ? ? ? ? 0 0 0 0 0 0 0 0 ? ? ? ?? ? ? ? ? ?? ? ?1 3 5 1 0 ?1 3 5 ?1 3 5 1 0 ?1 3 5 ?=? ?? ?; ? ?=? ?? ?. 5 ? ?2 6 10 2 1 0 0 0 ?2 6 7 2 1 0 0 ?3 6 (a) Special solutions (3, 1, 0) and (5, 0, 1) (b) (3, 1, 0). Total count of pivot and free is n.7 (a) Nullspace of A is the plane ?x + 3y + 5z = 0; it contains all vectors (3y + 5z, y, z ) (b) The line through (3, 1, 0) has equations ?x + 3y + 5z = 0 and ?2x + 6y + 7z = 0. ? ? ? ? ? ? 1 ?3 ?5 1 ?3 0 1 0 ? with I = [ 1 ]; R = ? ? with I = ? ?. 8 R=? 0 0 0 0 0 1 0 1 9 (a) False (b) True (c) True (only n columns) (d) True (only m rows).26? 10 (a) Impossible above diagonal 1 1 2 1 1 ? ? 1 1 1 1 1 ?? ? (b) A = invertible = ? 1 ? 1 ? ?? ? 1? ? 2? ? (c) A = ? 1 ? 1 ?? ? 1? ? 1(d) A = 2I, U = 2I, R = I . ? ? ? 0 1 1 1 1 1 1 1 1 ? ? ? ? ? ? ?0 0 0 1 1 1 1? ?0 0 ? ? 11 ? ? ? ? ?0 0 0 0 1 0 0? ?0 0 ? ? ? 0 0 0 0 0 0 0 0 0 ? ? ? 1 1 0 1 1 1 0 0 0 ? ? ? ? ? ? ?0 0 1 1 1 1 0 0? ?0 ?, ? 12 ? ? ? ? ?0 0 0 0 0 0 1 0? ?0 ? ? ? 0 0 0 0 0 0 0 1 01 1 0 0 1 0 0 01 1 0 0 1 0 0 01 1 0 0 0 1 0 01 1 1 0 0 0 1 01000 0 0 01 0 0 01 0 0 01 1 0 01? ? 1? ? ? 1? ? 1 1 1 1 0 1 1 1 0? ? ?0 0 ? ? ?0 0 ? 0 0 ? 1 ? ? 1? ?. ? 1? ? 0? ? 1? ? ? 0? ? 013 If column 4 is all zero then x4 is a free variable. Its special solution is (0, 0, 0, 1, 0). 14 If column 1 = column 5 then x5 is a free variable. Its special solution is (?1, 0, 0, 0, 1). 15 There are n ? r special solutions. The nullspace contains only x = 0 when r = n. The column space is Rm when r = m. 16 The nullspace contains only x = 0 when A has 5 pivots. Also the column space is R5 , because we can solve Ax = b and every b is in the column space. 17 A = [ 1 ?3 ?1 ]; special solutions (3, 1, 0) and (1, 0, 1). 18 Fill in 12 then 3 then 1. 19 If LU x = 0, multiply by L?1 to ?nd U x = 0. Then U and LU have the same nullspace. 20 Column 5 is sure to have no pivot since it is a combination of earlier columns. With 4 pivots in the other columns, the special solution is s = (1, 0, 1, 0, 1). The nullspace contains all multiples of s (a line in R5 ). 21 Free variables 1 ? ? 22 A = ? 0 ? 0 ? 1 ? ? 23 A = ? 1 ? 5 ? 0 1 0 0 3 1 ? ?1 0 x3 , x4 : A = ? 0 ?1 ? 0 ?4 ? ? 0 ? 3 ?. ? 1 ?2 ? ?1/2 ? ? ? 2 ?. ? ?3 2 2 3 1 ? ?.24 This construction is impossible: 2 pivot columns, 2 free variables, only 3 columns. ? ? 1 ?1 0 0 ? ? ? ? 25 A = ? 1 0 ?1 0 ?. ? ? 1 0 0 ?1 ? ? 0 1 ?. 26 A = ? 0 02727 If nullspace = column space (r pivots) then n ? r = r. If n = 3 then 3 = 2r is impossible. 28 If A times zero, the column space of B is contained in the nullspace of ? every ? column ? of B is ? 1 1 1 1 ?, B = ? ?. A: A = ? 1 1 ? 1 ?1 29 R is most likely to be I ; R is most likely to be I with fourth row of zeros. ? ? ? ? 0 1 1 0 T ? shows that (a)(b)(c) are all false. Notice rref(A ) = ? ?. 30 A = ? 0 0 0 0 ? ? 1 0 0 ?2 ? ? ? ? 31 Three pivots (4 columns and 1 special solution); R = ? 0 1 0 ?1 ? (add any zero rows). ? ? 0 0 1 0 ? ? 1 0 0 ?, R = I . 32 Any zero rows come after these rows: R = [ 1 ?2 ?3 ], R = ? 0 1 0 ? ? ? ? ? ? ? ? ? ? 1 0 1 0 1 1 0 1 0 0 ?,? ?, ? ?, ? ?, ? ? 33 (a) ? (b) All 8 matrices are R’s ! 0 1 0 0 0 0 0 0 0 0 34 One reason: A and ?A have the same nullspace (and also the same column space).Problem Set 3.3, page 1281 (a) and (c) (d) is false because R might happen to have 1’s in nonpivot columns. ? ? ? ? ? ? 1 1 1 1 1 0 ?1 ?2 1 ?1 1 ?1 ? ? ? ? ? ? ? ? ? ? ? ? 2 R = ? 0 0 0 0 ? r = 1; R = ? 0 1 2 3 ? r = 2; R = ? 0 0 0 0? r = 1 ? ? ? ? ? ? 0 0 0 0 0 0 0 0 0 0 0 0 ? ? ? ? 1 2 0 ? ? RA 0 ? ? ? ?→ Zero row in the upper 3 RA = ? 0 0 1 ? RB = RA RA RC ?→ ? ? ? 0 RA 0 0 0 R moves all the way to the bottom. ? ? ? ? 0 I I ?. The nullspace matrix is N = ? ?. 4 If all pivot variables come last then R = ? 0 0 0 5 I think this is true.T T 6 A and ? A have ?the same rank r. But pivcol (the column number) is 2 for A and 1 for A : 0 1 0 ? ? ? ? A = ?0 0 0? . ? ? 0 0 0 ? ? ? ? ? 2 ?3 ? ? 1 0 ? ? ? ? ? ? 4 ?5 ? ? ? ? 7 The special solutions are the columns of N = ? and N = ? ?. 0 ? 2 ? ? ? ? ? 1 0? ? ? 0 1 0 128? 1 2 4 8 4 ? ? 2 6 ? ? a ? 3 ?3/2 ?, M = ? ? c 6 ?3 ?3 ? b bc/a ? ?.? ? 8 A = ?2 ? 4? ? ? ? 8 ?, B = ? 1 ? ? 16 29 If A has rank 1, the column space is a line in Rm . The nullspace is a plane in Rn (given by one equation). The column space of AT is a line in Rn . 10 u = (3, 1, 4), v = (1, 2, 2); u = (2, ?1), v = (1, 1, 3, 2). 11 A rank one matrix has one pivot. The second row of U is zero. ? ? ? ? 1 3 1 0 ? ?. 12 S = ? and S= 1 and S=? 1 4 0 1 13 P has rank r (the same as A) because elimination produces the same pivot columns. 14 The rank of RT is also r, and the example matrix A has rank 2: ? 1 3 ? ? PT? ? P = ?2 ? 2? ? 6? ? 7=?1 32 62 7?T? S =?1 32 7? ?? S=?1 23 7? ?.?15 Rank(AB ) = 1; rank(AM ) = 1 except AM = 0 if c = ?1/2. 16 (uv T )(w z T ) = u (v T w )z T has rank one unless v T w = 0. 17 (a) By matrix multiplication, each column of AB is A times the corresponding column of B . So a combination of columns of B turns into a combination of columns of AB . (b) The rank of B is r = 1. Multiplying by A cannot increase this rank. The rank stays the same for A1 = I and it drops to zero for A2 = 0 or A2 = [ 1 1; ?1 ?1 ]. 18 If we know that rank(B T AT ) ≤ rank(AT ), then since rank stays the same for transposes, we have rank(AB ) ≤ rank(A). 19 We are given AB = I which has rank n. Then rank(AB ) ≤ rank(A) forces rank(A) = n. 20 Certainly A and B have at most rank 2. Then their product AB has at most rank 2. Since BA is 3 by 3, it cannot be I even if AB = I : ? ? A=? 1 0 0 1 0 0 ? ? 1 0 ? AB = I and BA = I.? ? B = ?0 ? 0? ? 1? ? 021 (a) A and B will both have the same nullspace and row space as R (same R for both matrices). (b) A equals an invertible matrix times B , when they share the same R. A key fact! ? ? ? ? ? 1 3 0 2 ?1 1 0 ? ? ? ? ? 1 3 0 2 ? 1 ? ? ? ? ? (nonzero rows of R). 22 A = ? 0 0 1 4 ?3 ? = ? 0 1 ? ? ? ? ? ? 0 0 1 4 ?3 1 3 1 6 ?4 1 129? 1 ? 0 ? ? 1 ? 4? ? ? 0 8 1 1 1 0 0 0 1 1 1 ? 1 0 0 ? 1 1 1 1 0 0 0 0 ? ? 0 0 0 0 0 0 ?? ? 23 A = (pivot columns)(nonzero rows of R) = ? 1 ? 1 ? 1 ? 0 ? ? 1 ? 4? ? ? 0 8 ? 1 0 0 1 1 0 1 0 0 ? 1? ?=? ?1 ? 1 1 0 ? ? 0? ? ? ? 0 ?+? 0 ? ? 0 0 0 4 8 0 0 0 0 0 0? ? 4 ?. ? 8 ?1 1 1? ? B = ?1 ? 1? ?=? ?1 ? 1 1? ? ? ? 0? + ?0 ? ? 0 0? ? 4? . ? 824 The m by n matrix Z has r ones at the start of its main diagonal. Otherwise Z is all zeros. 25 Y = Z because the form ? 1 1 2 ? ? 26 If c = 1, R = ? 0 0 0 ? 0 0 0 is decided by the rank which is the same for A and AT . ? ? ? 2 1 0 2 2 ? ? ? ? ? ? 0 ? has x2 , x3 , x4 free. If c = 1, R = ? 0 1 0 0 ? has x3 , x4 free. ? ? ? 0 0 0 0 0 ? ? ? ? ? 1 ? 2 ?2 ?2 ? 2 ? ? ? ? ? ? ? ? ? 1 ? 0 0 0? 0? ? (c = 1) and N = ? ? (c = 1) in N = ? ? ? ? ? ? 0 ? 1 1 0? 0? ? ? ? ? 0 0 1 0 1 ? ? ? 1 1 ?2 ? and x1 if c = 2, R = ? ? and x2 R = I if c = 1, 2 0 0 0 ? ? ? ? 2 1 in N = ? ? (c = 1) or N = ? ? (c = 2) or N = 2 by 0 empty matrix. 1 0 ? ? I = ? ? ; N = empty. ?ISpecial solutions? If c = 1, R = ?0 0Special solutions ? I ? N27 N = ? ? ; ?IProblem Set 3.4, page 136? 2 4 5 3 6 7 5 4 6 2 b1 ? ? 2 4 1 ?1 6 1 ?1 4 2 ?2 b1 ? ? 2 4 1 0 6 1 0 4 2 0 b1 b2 ? b1 b3 + b2 ? 2b1 ? ? ? ? ? ? ? 1 ?2 ? 2 ? ? ? ? b2 ? → ? 0 ? ? b3 0 ? ? ? ? b2 ? b1 ? → ? 0 ? ? b3 ? b1 0Ax = b has a solution when b3 + b2 ? 2b1 = 0; the column space contains all combinations of (2, 2, 2) and (4, 5, 3) which is the plane b3 + b2 ? 2b1 = 0 (!); the nullspace contains all combinations of s 1 = (?1, ?1, 1, 0) and s 2 = (2, ?2, 0, 1); x complete = x p + c1 s 1 + c2 s 2 ; ? 1 0 1 0 ? 1 1 0 ?2 2 0 4 ?? ? [ R d ] = ?0 ? 0 ? ?? ? ?1 ? gives the particular solution x p = (4, ?1, 0, 0). ? 0 ? ? ?213b1213b111/23/25? ? ? ? ? ? ? ? ? ? ? ? 2 ? 6 3 9 b2 ? → ? 0 0 0 b2 ? 3b1 ? Then [ R d ] = ? 0 0 0 0 ? Ax = b ? ? ? ? ? ? 4 2 6 b3 0 0 0 b3 ? 2b1 0 0 0 0 has a solution when b2 ? 3b1 = 0 and b3 ? 2b1 = 0; the column space is the line through (2, 6, 4)30which is the intersection of the planes b2 ? 3b1 = 0 and b3 ? 2b1 = 0; the nullspace contains all combinations of s 1 = (?1/2, 1, 0) and s 2 = (?3/2, 0, 1); particular solution x p = (5, 0, 0) and complete solution x p + c1 s 1 + c2 s 2 . ? ? ? ? ?2 ?3 ? ? ? ? ? ? ? ? 3 x = ? ? + x2 ? 1 ?. complete ? 0 ? ? ? 1 0 ? ? ? ? ? ? 1/2 ?3 0 ? ? ? ? ? ? ? ? ? ? ? ? ? 0 ? ? 1? ? 0? ? + x2 ? ? + x4 ? ?. 4 x =? ? ? ? ? ? complete ? ? 1/2 ? ? 0? ??2 ? ? ? ? ? ? ? 0 0 1 ? ? ? ? 5b1 ? 2b2 2 ? ? ? ? ? ? ? ? 5 Solvable if 2b1 + b2 = b3 . Then x = ? b2 ? 2b1 ? + x3 ? 0 ?. ? ? ? ? 0 1 ? 6 (a) Solvable if b2 = 2b1 and 3b1 ? 3b3 + b4 = 0. Then x = ? ? (no free variables) b3 ? 2b1 ? ? ? ? 5b1 ? 2b3 ?1 ? ? ? ? ? ? ? ? 3b1 ? 3b3 + b4 = 0. Then x = ? b3 ? 2b1 ? + x3 ??1 ?. ? ? ? ? 0 1 ? 1 b2 row 3 ? 2 (row 2) + 4 (row 1) ? ? ?1 b2 ? 3b1 ? → is the zero row ? ?2 b3 ? 2b1 [ 0 0 0 b3 ? 2b2 + 4b1 ] independent rows. (b) Need b3 = 2b2 . Row 3 ? 2 row 2 = 0. ? 3 5 b1 ? ? ? = [ A b ]; 2 2 b2 ? 2b1 ? 0 0 b3 + b2 ? 5b1 5b1 ? 2b3 ?(b) Solvable if b2 = 2b1 and?13 8 41 2 0b1? ? 7 ?3 ? 21 3 ? ? ? ? b2 ? → ? 0 ?1 ? ? b3 0 ?2??8 (a) Every b is in the column space: ? ?? 1 0 0 1 2 ? ?? ? ?? 9 L[ U c ] = ? 2 1 0? ?0 0 ? ?? 3 ?1 1 0 0x p = (?9, 0, 3, 0) so ?9(1, 2, 3) + 3(3, 8, 7) = (0, 6, ?6) is exactly Ax p = b . ? ? ? ? 1 0 ?1 2 ? x = ? ?. 10 ? 0 1 ?1 4 11 A 1 by 3 system has at least two free variables. 12 (a) x 1 ? x 2 and 0 solve Ax = 0 (b) 2x 1 ? 2x 2 solves Ax = 0; 2x 1 ? x 2 solves Ax = b .13 (a) The particular solution is multiplied by (b) Any solution can be the par? x p? ?always ? ? ? ? 1? ? ? √ 3 3 x 6 1 2 ? ? ? = ? ?. Then ? ? is shorter (length 2) than ? ? ticular solution (c) ? 3 3 y 6 1 0 (d) The “homogeneous” solution in the nullspace is x n = 0 when A is invertible. 14 If column 5 has no pivot, x5 is a free variable. The zero vector is not the only solution to Ax = 0. If Ax = b has a solution, it has in?nitely many solutions. 15 If row 3 of U has no pivot, that is a zero row. U x = c is solvable only if c3 = 0. Ax = b might not be solvable, because U may have other zero rows.3116 The largest rank is 3. Then there is a pivot in every row. The solution always exists. The column space is R3 . An example is A = [ I F ] for any 3 by 2 matrix F . 17 The largest rank is 4. There is a pivot in every column. The solution is unique. The nullspace contains only the zero vector. An example is A = [ I ; G ] for any 4 by 2 matrix G. 18 Rank = 3; rank = 3 unless q = 2 (then rank = 2). 19 All ranks = 2. ? ? 20 A = ? 1 0 ?? ?? 3 4 1 0 ? 1 0 0 ?? 1 0 1 0 ? ? ?? ? ?? ? ?; A = ? ?2 1 0? ?0 2 ?2 3 ?. ? ? ? ? 2 1 0 ?3 0 1 0 3 1 0 0 11 ?5 ? ? ? ? ? ? ? ? ? ? ? ? ? ? x 4 ?1 ?1 x 4 ?1 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 21 (a) ? y ? = ? 0 ? + y ? 1 ? + z ? 0 ? (b) ? y ? = ? 0 ? + z ? 0 ?. ? ? ? ? ? ? ? ? ? ? ? ? ? ? z 0 0 1 z 0 1 22 If Ax 1 = b and Ax 2 = b then we can add x 1 ? x 2 to any solution of Ax = B . But there will be no solution to Ax = B if B is not in the column space. 23 For A, q = 3 gives rank 1, every other q gives rank 2. For B , q = 6 gives rank 1, every other q gives rank 2. ? ? 1 24 (a) ? ? (b) [ 1 1 ] 1(c) [ 0 ] or any r & m, r & n(d) Invertible. (d) r = m = n.25 (a) r & m, always r ≤ n (b) r = m, r & n ? ? 1 0 ?2 ? ? ? ? 26 R = ? 0 1 2 ?, R = I . ? ? 0 0 0(c) r & m, r = n27 R has n pivots equal to 1. Zeros above and below pivots make R = I . ? ? ? ? ? ? ? ? ? ? ? ? ?2 ?1 ? ? ? ? 35 1 2 0 ?1 ? ? ?→? ?; x n = ? ?→? ? xp = ? 28 ? ? 1 ?; ? ? 0 ?. ? ? ? ? 48 001 2 0 2 The pivot columns contain I so ?1 and 2 go into x p . ? ? ? ? ? ? 0 1 0 0 ?1 1 0 0 0 ? ? ? ? ? ? ? ? ? ? ? ? 29 R = ? 0 0 1 0 ? and x n = ? 1 ?; ? 0 0 1 2 ?: no solution because of row 3. ? ? ? ? ? ? 0 0 0 0 0 0 0 0 5 ? ? ? ? ? ? ? ? ? ? 4 ?2 ? ? ? ? 1 0 2 3 2 1 0 2 3 2 1 0 2 0 ?4 ? ? ? ? ? ? ? ? ? ? ??3 ? ? 0? ? ? ? ? ? ? ? ? ? 30 ? 1 3 2 0 5 ? → ? 0 3 0 ?3 3 ? → ? 0 1 0 0 3 ?; x p = ? ? ? and x n = x3 ? ?. ? ? ? ? ? ? ? 0? ? 1? ? ? ? ? 2 0 4 9 10 0 0 0 3 6 0 0 0 1 2 ?2 0 ? ? 1 1 ? ? ? ? 31 A = ? 0 2 ?; B cannot exist since 2 equations in 3 unknowns cannot have a unique solution. ? ? 0 332? 1 0 1 2 2 0 0 1 0 0 ?? 1 3 ? ? ? ? ? 7 ?7 ? ? ? ? ? 2? ? ? ? ? and x = ? ??2 ? + x3 ? 2 ? and then no solution. ? ? ? ? ? 0? ? 0 1 0 1 ?? ? ?1 32 A = LU = ? ? ?2 ? 1 ? ? 1 0 ?. 36 A = ? 3 0?? ?? 0 ? ? 0 ?1 ?? ?? 0? ?0 0 ?? 1 0 0Problem Set 3.5, page 150? 1 1 1 0 1 ?? c1 ? ? ? 1 ?0 ? 0 ?? ? ?? ? 1 ? ? c2 ? = 0 gives c3 = c2 = c1 = 0. But v 1 + v 2 ? 4v 3 + v 4 = 0 (dependent). ?? ? 1 c32 v 1 , v 2 , v 3 are independent. All six vectors are on the plane (1, 1, 1, 1) ? v = 0 so no four of these six vectors can be independent. 3 If a = 0 then column 1 = 0; if d = 0 then b(column 1) ? a(column 2) = 0; if f = 0 then all columns end in zero (all are perpendicular to (0, 0, 1), all in the xy plane, must be dependent). ? ?? ? ? ? a b c x 0 ? ?? ? ? ? ? ?? ? ? ? 4 U x = ? 0 d e ? ? y ? = ? 0 ? gives z = 0 then y = 0 then x = 0. ? ?? ? ? ? 0 0 f z 0 ? ? ? ? ? ? 1 2 3 1 2 3 1 2 3 ? ? ? ? ? ? ? ? ? ? ? ? 5 (a) ? 3 1 2 ? → ? 0 ?5 ?7 ? → ? 0 ?5 ?7 ?: invertible ? independent columns ? ? ? ? ? ? 2 3 1 0 ?1 ? 5 0 0 ?18/5 ? ? ? ? ? ? ? ? ? ? 1 2 ?3 1 2 ?3 1 2 ?3 1 0 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? (b) ??3 1 2? → ?0 7 ?7 ? → ? 0 7 ?7 ?; A ? 1 ? = ? 0 ?, columns add to 0. ? ? ? ? ? ? ? ? ? ? 2 ?3 1 0 ?7 7 0 0 0 1 0 6 Columns 1, 2, 4 are independent. Also 1, 3, 4 and 2, 3, 4 and others (but not 1, 2, 3). Same column numbers (not same columns!) for A. 7 The sum v 1 ? v 2 + v 3 = 0 because (w 2 ? w 3 ) ? (w 1 ? w 3 ) + (w 1 ? w 2 ) = 0. 8 If c1 (w 2 + w 3 )+ c2 (w 1 + w 3 )+ c3 (w 1 + w 2 ) = 0 then (c2 + c3 )w 1 +(c1 + c3 )w 2 +(c1 +