The requested URL was not found on this server. Additionally, a 404 Not Found error was encountered while trying to use an ErrorDocument to handle the request. Apache/2.4.41 (Ubuntu) Server at e2shi.jhu.edu Port 443 The requested URL was not found on this server. Additionally, a 404 Not Found error was encountered while trying to use an ErrorDocument to handle the request. Apache/2.4.41 (Ubuntu) Server at e2shi.jhu.edu Port 443 Instructor Solutions Manual for Introduction to Linear Algebra with Applications Jim DeFranza St. Lawrence University Contents
1 Systems of Linear Equations 1 Solutions to All Exercises 1 Systems of Linear Equations and MatricesExercise Set 1. In Section 1 of the text, Gaussian Elimination is used to solve a linear system. This procedure utilizes three operations that when applied to a linear system resultin a new system that is equivalent to the original. Equivalent means that the linear systems have the same solutions. The three operations are:
When used judiciously these three operations allow us to reduce a linear system to a triangular linear system, which can be solved. A linear system is consistent if there isat least one solution and is inconsistent if there are no solutions. Every linear system has either a unique solution, infinitely many solutions or no solutions. For example, the triangular linear systems x 1 −x 2 +x 3 = 2 x 2 − 2 x 3 =− 1 x 3 = 2 ,x 1 − 2 x 2 +x 3 = 2 −x 2 + 2x 3 =− 3 , 2 x 1 +x 3 = 1 x 2 −x 3 = 2 0 = 4 have a unique solution, infinitely many solutions, and no solutions, respectively. In the second linear system, the variablex 3 is a free variable, and once assigned any real number the values ofx 1 andx 2 are determined. In this way the linear system has infinitely many solutions. If a linear system has the same form as the second system, but also has the additional equation 0 = 0,then the linear system will still have free variables. The third system is inconsistent since the last equation 0 = 4 is impossible. In some cases, the conditions on the right hand side of a linear system are not specified. Considerfor example, the linear system −x 1 −x 2 =a 2 x 1 + 2x 2 +x 3 =b 2 x 3 =c which is equivalent to −−−−−−−−−−−−−−−−→ −x 1 −x 2 =a x 3 =b+ 2a 0 =c− 2 b− 4 a .This linear system is consistent only for valuesa, bandcsuch thatc− 2 b− 4 a= 0. Solutions to Exercises 1 the given operations we obtain the equivalent triangular system x 1 −x 2 − 2 x 3 = 3 −x 1 + 2x 2 + 3x 3 = 1 2 x 1 − 2 x 2 − 2 x 3 =− 2 E 1 +E 2 →E 2−−−−−−−−−−→x 1 −x 2 − 2 x 3 = 3 x 2 +x 3 = 4 2 x 1 − 2 x 2 − 2 x 3 =− 2 (−2)E 1 +E 3 →E 3−−−−−−−−−−−−−−→2 Chapter 1 Systems of Linear Equations and Matrices x 1 −x 2 − 2 x 3 = 3 x 2 +x 3 = 4 2 x 3 =− 8 .Using back substitution, the linear system has the unique solution x 1 = 3, x 2 = 8, x 3 =− 4. 2 the given operations we obtain the equivalent triangular system 2 x 1 − 2 x 2 −x 3 =− 3 x 1 − 3 x 2 +x 3 =− 2 x 1 − 2 x 2 = 2 E 1 ↔E 2−−−−−−→x 1 − 3 x 2 +x 3 =− 2 2 x 1 − 2 x 2 −x 3 =− 3 x 1 − 2 x 2 = 2 (−2)E 1 +E 2 →E 2−−−−−−−−−−−−−−→x 1 − 3 x 2 +x 3 =− 2 4 x 2 − 3 x 3 = 1 x 1 − 2 x 2 = 2 (−1)E 1 +E 3 →E 3−−−−−−−−−−−−−−→x 1 − 3 x 2 +x 3 =− 2 4 x 2 − 3 x 3 = 1 x 2 −x 3 = 4 E 2 ↔E 3−−−−−−→x 1 − 3 x 2 +x 3 =− 2 x 2 −x 3 = 4 4 x 2 − 3 x 3 = 1 (−4)E 2 +E 3 →E 3−−−−−−−−−−−−−−→x 1 − 3 x 2 +x 3 =− 2 x 2 −x 3 = 4 x 3 =− 15 .Using back substitution, the linear system has the unique solutionx 1 =− 20 , x 2 =− 11 , x 3 =− 15. 3 the given operations we obtain the equivalent triangular system x 1 + 3x 4 = 2 x 1 +x 2 + 4x 4 = 3 2 x 1 +x 3 + 8x 4 = 3 x 1 +x 2 +x 3 + 6x 4 = 2 (−1)E 1 +E 2 →E 2−−−−−−−−−−−−−−→x 1 + 3x 4 = 2 x 2 +x 4 = 1 2 x 1 +x 3 + 8x 4 = 3 x 1 +x 2 +x 3 + 6x 4 = 2 (−2)E 1 +E 3 →E 3−−−−−−−−−−−−−−→x 1 + 3x 4 = 2 x 2 +x 4 = 1 +x 3 + 2x 4 =− 1 x 1 +x 2 +x 3 + 6x 4 = 2 (−1)E 1 +E 4 →E 4−−−−−−−−−−−−−−→x 1 + 3x 4 = 2 x 2 +x 4 = 1 +x 3 + 2x 4 =− 1 x 2 +x 3 + 3x 4 = 0 (−1)E 2 +E 4 →E 4−−−−−−−−−−−−−−→x 1 + 3x 4 = 2 x 2 +x 4 = 1 x 3 + 2x 4 =− 1 x 3 + 2x 4 =− 1 (−1)E 3 +E 4 →E 4−−−−−−−−−−−−−−→x 1 + 3x 4 = 2 x 2 +x 4 = 1 x 3 + 2x 4 =− 1 0 = 0 .The final triangular linear system has more variables than equations, that is, there is a free variable. As a result there are infinitely many solutions. Specifically, using back substitution, the solutions are given by x 1 = 2− 3 x 4 , x 2 = 1−x 4 , x 3 =− 1 − 2 x 4 , x 4 ∈R. 4 the given operations we obtain the equivalent triangular system x 1 +x 3 =− 2 x 1 +x 2 + 4x 3 =− 1 2 x 1 + 2x 3 +x 4 =− 1 (−1)E 1 +E 2 →E 2−−−−−−−−−−−−−−→x 1 +x 3 =− 2 x 2 + 3x 3 = 1 2 x 1 + 2x 3 +x 4 =− 1 (−2)E 1 +E 3 →E 3−−−−−−−−−−−−−−→4 Chapter 1 Systems of Linear Equations and Matrices {− 2 x 1 +x 2 = 2 3 x 1 −x 2 + 2x 3 = 1 reduces to−−−−−−−→ {− 2 x 1 +x 2 = 2 x 2 + 4x 3 = 8 .There are infinitely many solutions with solution setS={(− 2 t+ 3,− 4 t+ 8, t)| t∈R}. 17 operation 2E 1 +E 2 →E 2 gives the equation− 3 x 2 − 3 x 3 − 4 x 4 =− 9 .Hence, the linear system has two free variables,x 3 andx 4 .The two parameter set of solutions isS= {(3 − 53 t,−s− 43 t+ 3, s, t )∣∣ s, t∈R }.18 linear system is in reduced form. The solution set is a twoparameter family given by S= {( 12 s− 3 t+ 5 2 , 3 t− 2 , s, t )| s, t∈R }.19 operation− 2 E 1 +E 2 →E 2 givesx=b− 2 a=a+ 2x=a+ 2(b− 2 a) = 2b− 3 a,so that the unique solution isx=− 2 a+b, y=− 3 a+ 2b. 20 linear system { 2 x+ 3y =a x +y =b reduces to−−−−−−−→ {2 x+ 3y =a y =a− 2 b ,so the solution isx=−a+ 3b, y=a− 2 b. 21 linear system is equivalent to the triangular linear system −x −z =b y =a+ 3b z =c− 7 b− 2 a ,which has the unique solutionx= 2a+ 6b−c, y=a+ 3b, z=− 2 a− 7 b+c. 22 linear system − 3 x+ 2y +z =a x −y −z =b x −y− 2 z =c −E 2 +E 3 →E 3 , E 1 + 3E 2 →E 1 reduces to −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−→ −y − 2 z =a+ 3b x −y −z =b −z =−b+c ,so the solution isx=−a− 3 b+c, y=−a− 5 b+ 2c, z=b−c. 23 the operation 2E 1 +E 2 →E 2 gives the equation 0 = 2a+ 2,then the linear system is consistent fora=− 1. 24 { −x+ 3y =a 2 x− 6 y = 3 reduces to−−−−−−−→ {−x+ 3y =a 0 = 3 + 2a ,the linear system is consistent ifa=− 32. 25 the operation 2E 1 +E 2 →E 2 gives the equation 0 =a+b,then the linear system is consistent for b=−a. 26 { 6 x− 3 y =a − 2 x +y =b reduces to−−−−−−−→ {6 x− 3 y =a 0 = 13 a+b ,the linear system is consistent ifb=− 13 a. 1 Systems of Linear Equations 5 27 linear system is equivalent to the triangular linear system x− 2 y+ 4z =a 5 y− 9 z =− 2 a+b 0 =c−a−b and hence, is consistent for alla, b,andcsuch thatc−a−b= 0. 28 x −y + 2z =a 2 x + 4y − 3 z =b 4 x + 2y +z =c reduces to−−−−−−−→ x −y + 2z =a 6 y − 7 z =b− 2 a 0 =c− 2 a−b ,the linear system is consistent ifc− 2 a−b= 0. 29 operation− 2 E 1 +E 2 →E 2 gives the equivalent linear system { x+y =− 2 (a−2)y = 7 .Hence, ifa= 2,the linear system is inconsistent. 30 { 2 x−y = 4 ax+ 3y = 2 reduces to−−−−−−−→ {2 x−y = 4 3 + 12 a = 2− 2 a ,the linear system is inconsistent if 3 + 12 a= 0,that isa=− 6 .Notice that ifa=− 6 ,then 2− 2 a 6 = 0. 31 operation− 3 E 1 +E 2 →E 2 gives the equivalent linear system { x−y = 2 0 =a− 6 .Hence, the linear system is inconsistent for alla 6 = 6. 32 { 2 x−y =a 6 x− 3 y =a reduces to−−−−−−−→ {2 x−y =a 0 =− 2 a the linear system is inconsistent fora 6 = 0. 33 find the parabolay=ax 2 +bx+cthat passes through the specified points we solve the linear system c = 0. 25 a+b+c =− 1. 75 a−b+c = 4. 25 .The unique solution isa= 1, b=− 3 ,andc= 14 ,so the parabola isy=x 2 − 3 x+ 14 = (x− 32 ) 2− 2 .The vertex of the parabola is the point ( 32 ,− 2).34 find the parabolay=ax 2 +bx+cthat passes through the specified points we solve the linear system c = 2 9 a− 3 b+c =− 1 0. 25 a+ 0. 5 b+c = 0. 75 .1 Matrices and Elementary Row Operations 7
43 1 →E 1 , 9 E 2 →E 2 ,and−E 1 +E 2 →E 2 gives the equivalent linear system { 9 kx+ k 2 y = 9k (9−k 2 )y =− 27 − 9 k .Whether the linear system is consistent or inconsistent cannow be determined by examining the second equation. a= 3,the second equation becomes 0 =− 54 ,so the linear system is inconsistent.b=− 3 ,then the second equation becomes 0 = 0,so the linear system has infinitely many solutions.c 6 =± 3 ,then the linear system has a unique solution. 44 linear system kx+y+z = 0 x+ky+z = 0 x+y+kz = 0 reduces to−−−−−−−→ x+ ky+ z = 0 (1−k)y+ (k−1)z = 0 (2 +k)(1−k)z = 0 .a. The linear system has a unique solution ifk 6 = 1 andk 6 =− 2 .b. Ifk=− 2 ,the solution set is a one parameter family.c= 1,the solution set is a two parameter family. Exercise Set 1. Matrices are used to provide an alternative way to representa linear system. Reducing a linear system to triangular form is then equivalent to row reducing the augmented matrix corresponding to the linear system to a triangular matrix. For example, the augmented matrix for the linear system −x 1 −x 2 −x 3 − 2 x 4 = 1 2 x 1 + 2x 2 +x 3 − 2 x 4 = 2 x 1 − 2 x 2 +x 3 + 2x 4 =− 2 is − 1 − 1 − 1 − 2 12 2 1 − 2 21 −2 1 2 − 2.The coefficient matrix is the 3×4 matrix consisting of the coefficients of each variable, thatis, the augmented matrix with the augmented column 12− 2deleted. The first four columns of the augmented matrix cor- respond to the variablesx 1 , x 2 , x 4 ,andx 4 ,respectively and the augmented column to the constants on the right of each equation. Reducing the linear system using thethree valid operations is equivalent to reducing the augmented matrix to a triangular matrix using the row operations:
In the above example, the augmented matrix can be reduced to either -1 − 1 − 1 − 2 10 -3 0 0 − 10 0 -1 − 6 4 or 1 0 0 − 4 8 / 30 1 0 0 1 / 30 0 1 6 − 4The left matrix is in row echelon form and the right is in reduced row echelon form. The framed terms are the pivots of the matrix. The pivot entries correspond todependent variables and the non-pivot entries correspond to free variables. In this example, the free variable isx 4 andx 1 , x 2 ,andx 3 depend onx 4 .So the linear system has infinitely many solutions given byx 1 = 83 + 4x 4 , x 2 = 13 , x 3 =− 4 − 6 x 4 ,andx 4 is an 8 Chapter 1 Systems of Linear Equations and Matrices arbitrary real number. For a linear system with the same number of equations as variables, there will be a unique solution if and only if the coefficient matrix can be rowreduced to the matrix with each diagonal entry 1 and all others 0. Solutions to Exercises 1.[2 − 3 5−1 1 − 3]2.[2 − 2 13 0 1]3.2 0 − 1 41 4 1 24 1 − 1 1 4.−3 1 1 20 0 − 4 0−4 2 − 3 15.[2 0 − 1 41 4 1 2]6.[4 1 − 1 14 −4 2 − 2]7.2 4 2 2 − 24 − 2 − 3 − 2 21 3 3 − 3 − 4 8.3 0 −3 4 − 3−4 2 − 2 − 4 40 4 −3 2 − 39 linear system has the unique solutionx= − 1 , y= 12 , z= 0.
11 linear system is consistent with free vari- ablez are infinitely many solutions given byx=− 3 − 2 z, y= 2 +z, z∈R. 12 linear system is consistent with free vari- ablez are infinitely many solutions given byx= 4 + 13 z, y= 43 − 3 z, z∈R.
21 matrix is in reduced row echelon form. 22 matrix is in reduced row echelon form.
24 the pivot in row two is not a one, the matrix is not in reduced row echelon form. 25 matrix is in reduced row echelon form. 26 matrix is in reduced row echelon form.
]R 1 +R 2 →R 2−−−−−−−−−−→[2 30 4].The next step is to make the pivots 1, and eliminate the term above the pivot in row two. This gives [ 2 3 0 4 ]14R 2 →R 2−−−−−−−→[2 30 1](−3)R 2 +R 1 →R 1−−−−−−−−−−−−−−→[2 00 1]12R 1 →R 1−−−−−−−→[1 00 1].30 matrix [−3 23 3]reduces to−−−−−−−→ [1 00 1].10 Chapter 1 Systems of Linear Equations and Matrices The linear system is inconsistent. 42 augmented matrix 3 0 − 2 − 3−2 0 1 − 20 0 − 1 2reduces to −−−−−−−→ 1 0 0 00 1 0 00 0 0 1.The linear system is inconsistent. 43 augmented matrix for the linear system and the reduced row echelon form are [ 3 2 3 − 3 1 2 − 1 − 2 ]−→[1 0 2 − 120 1 − 32 − 34].As a result, the variablex 3 is free and there are infinitely many solutions to the linear system given by x 1 =− 12 − 2 x 3 , x 2 =− 34 + 32 x 3 , x 3 ∈R. 44 augmented matrix [ 0 − 3 − 1 2 1 0 1 − 2 ]reduces to−−−−−−−→ [1 0 1 − 20 1 13 − 23].As a result, the variablex 3 is free and there are infinitely many solutions to the linear system given by x 1 =− 2 −x 3 , x 2 =− 23 − 13 x 3 , x 3 ∈R. 45 augmented matrix for the linear system and the reduced row echelon form are −1 0 3 1 22 3 −3 1 22 − 2 − 2 − 1 − 2−→1 0 0 1210 1 0 1210 0 1 121.As a result, the variablex 4 is free and there are infinitely many solutions to the linear system given by x 1 = 1− 12 x 4 , x 2 = 1− 12 x 4 , x 3 = 1− 12 x 4 , x 4 ∈R. 46 augmented matrix − 3 −1 3 3 − 31 −1 1 1 3−3 3 −1 2 1reduces to −−−−−−−→ 1 0 0 3440 1 0 9460 0 1 525.As a result, the variablex 4 is free and there are infinitely many solutions to the linear system given by x 1 = 4− 34 x 4 , x 2 = 6− 94 x 4 , x 3 = 5− 52 x 4 , x 4 ∈R. 47 augmented matrix for the linear system and the reduced row echelon form are 3 −3 1 3 − 31 1 − 1 − 2 34 −2 0 1 0−→1 0 − 13 − 1210 1 − 23 − 3220 0 0 0 0.As a result, the variablesx 3 andx 4 are free and there are infinitely many solutions to the linearsystem given byx 1 =− 8 −x 3 − 8 x 4 , x 2 =− 11 −x 3 − 11 x 4 , x 3 ∈R, x 4 ∈R. 48 augmented matrix −3 2 − 1 − 2 21 −1 0 − 3 34 −3 1 − 1 1reduces to −−−−−−−→ 1 0 1 8 80 1 1 11 − 110 0 0 0 0.As a result, the variablesx 3 andx 4 are free and there are infinitely many solutions to the linearsystem given byx 1 =− 8 −x 3 − 8 x 4 , x 2 =− 11 −x 3 − 11 x 4 , x 3 ∈R, x 4 ∈R. 49 augmented matrix for the linear system and the row echelon form are 1 Matrix Algebra 11 1 2 − 1 a 2 3 − 2 b − 1 −1 1 c −→1 2 − 1 a 0 −1 0 − 2 a+b 0 0 0 −a+b+c .a linear system is consistent precisely when the last equation, from the row echelon form, is consistent. That is, whenc−a+b= 0.b, the linear system is inconsistent whenc−a+b 6 = 0.c those values ofa, b,andcfor which the linear system is consistent, there is a free variable, so that there are infinitely many solutions.d linear system is consistent ifa= 1, b= 0, c= 1 the variables are denoted byx, y andz,then one solution is obtained by settingz= 1,that is,x=− 2 , y= 2, z= 1. 50 the augmented matrix gives [ a 1 1 2 a− 1 1 ]→[2 a− 1 1 a 1 1 ]→[1 a− 21 12 a 1 1 ]→[1 a− 21 12 0 −a(a 2 −1)+ 1 1 − 12 a ].a. Ifa 6 =− 1 ,the linear system is consistent. b. If the linear system is consistent anda 6 = 2,then the solution is unique. If the linear system is consistent anda= 2 there are infinitely many solutions. c. Let a= 1 unique solution isx= 12 , y= 12. 51 augmented matrix for the linear system and the reduced row echelon form are −2 3 1 a 1 1 − 1 b 0 5 − 1 c −→1 0 − 45 − 12 a+ 103 b 0 1 − 1515 c 0 0 0 a+ 2b−c .a. The linear system is consistent precisely when the last equation, from the reduced row echelon form, is consistent. That is, whena+2b−c= 0.b, the linear system is inconsistent whena+2b−c= 0 6 = 0. c. For those values ofa, b,andcfor which the linear system is consistent, there is a free variable, so that there are infinitely many solutions.d linear system is consistent ifa= 0, b= 0, c= 0 the variables are denoted byx, yandz,then one solution is obtained by settingz= 1,that is,x= 45 , y= 15 , z= 1. [1 00 1],[1 00 0],[1 10 0],[0 10 0].Exercise Set 1. Addition and scalar multiplication are defined componentwise allowing algebra to be performed on ex- pressions involving matrices. Many of the properties enjoyed by the real numbers also hold for matrices. For example, addition is commutative and associative, the matrix of all zeros plays the same role as 0 in the real numbers since the zero matrix added to any matrixAisA each component of a matrixAis negated, denoted by−A,thenA+ (−A) is the zero matrix. Matrix multiplication is also defined. The matrixAB that is the product ofAwithB,is obtained by taking the dot product of each row vector ofAwith each column vector ofB order of multiplication is important since it is not always the case thatABandBA are the same matrix. When simplifying expressions with matrices, care is then needed and the multiplication of matrices can be reversed only when it is assumed or known that the matrices commute. The distributive property does hold for matrices, so thatA(B+C) =AB+AC this case however, it is also necessary to note that (B+C)A=BA+CAagain since matrix multiplication is not commutative. The transpose of a matrixA,denoted byAt,is obtained by interchanging the rows and columns of a matrix. There are important properties of the transpose operation you shouldalso be familiar with before solving the exercises. Of particular importance is (AB)t=BtAt properties are (A+B)t=At+Bt,(cA)t=cAt,and (At)t=A class of matrices that is introduced in Section 1 and considered throughout the text are the symmetric matrices. A matrixAis symmetric it is equal to its transpose, that is,At=A example, in the case of 2×2 matrices, A= [a b c d ]=At= [a c b d ]⇔b=c. 1 Matrix Algebra 13 14.(A+B)C=[0 − 31 0][2 0− 1 − 1]=[3 32 0]15. 2 A(B− 3 C) =[10 − 18− 24 0]16.(A+ 2B)(3C) =[21 9−6 0]
is 2At−Bt= 7 5−1 3− 3 − 2.18 3×2 and 2Ais 2× 3 ,the expression Bt− 2 Ais not defined.
productABtis defined withABt= [− 7 − 4−5 1] 20t= [− 7 − 5−4 1]21.(At+Bt)C= −1 76 84 12 22. SinceCis 2×2 andAt+Btis 3× 2 ,the expression is not defined. 23.(AtC)B= 0 20 150 0 0− 18 − 22 − 15 24 3×2 andBtis 3× 2 ,the expression is not defined. 25=AC= [− 5 − 15 1]
[0 20 5]andB= [1 10 0],then AB= [0 00 0].27 product A 2 =AA= [a b 0 c ][a b 0 c ]=[a 2 ab+bc 0 c 2 ]=[1 00 1]if and only ifa 2 = 1, c 2 = 1,andab+bc=b(a+c) = 0 is,a=± 1 , c=± 1 ,andb(a+c) = 0,so thatA has one of the forms [1 00 1],[1 b 0 − 1 ],[− 1 b 0 1 ],or [−1 00 − 1].28= [2 a+c 2 b+d a+c b+d ]andM A= [2 a+b a+b 2 c+d c+d ],thenAM=M Aif and only if 2a+c= 2 [a+b, 2 b+d=a+b, a+c= 2c+d, bed=c+d is,b=candd=a−b,so the matrices have the form a b b a−b ].
[1 10 0]andB= [− 1 − 11 1].ThenAB= [0 00 0],and neither of the matrices are the zero matrix. Notice that, this can not happen with real numbers. That is, if the product of two real numbers is zero, then at least one of the numbers must be zero. 30= [a b c d ]andB= [e f g h ],so thatAB−BA= [1 00 1]if and only if [ bg−cf (af+bh−(be+f d) (ce+dg)−(ag+ch) cf−bg ]=[1 00 1].Sobg−cfandcf−bgmust both be 1, which is not possible. 31 product [ 1 2 a 0 ][3 b −4 1 ]=[− 5 b+ 2 3 a ab ]14 Chapter 1 Systems of Linear Equations and Matrices will equal [−5 612 16]if and onlyb+ 2 = 6, 3 a= 12,andab= 16 is,a=b= 4. 32= [a b c d ]andB= [e f g h ].Since AB−BA=[bg−cf (af+bh−(be+f d) (ce+dg)−(ag+ch) cf−bg ],then the sum of the terms on the diagonal is (bg−cf) + (cf−bg) = 0. 33 powers of the matrixAare given by A 2 =1 0 00 1 00 0 1, A 3 =1 0 00 −1 00 0 1, A 4 =1 0 00 1 00 0 1,andA 5 = 1 0 00 −1 00 0 1.We can see that ifnis even, thenAnis the identity matrix, so in particularA 20 = 1 0 00 1 00 0 1.Notice also that, ifnis odd, thenAn= 1 0 00 −1 00 0 1.34 (A+B)(A−B) =A 2 −AB+BA−B 2 ,then (A+B)(A−B) =A 2 −B 2 whenAB=BA. 35 can first rewrite the expressionA 2 BasA 2 B=AAB=BA,thenA 2 B=AAB=ABA= BAA=BA 2.
commute. b= [1 00 1],so thatAcommutes with every 2×2 matrix. Then select any two matrices that do not commute. For example, letB= [1 01 0]andC= [0 10 1].
10...0gives the first column vector of the matrixA Ax= 0 forces the first column vector ofAto be the zero vector. Then letx= 01...0and so on, to show that each column vector ofAis the zero vector. Hence,Ais the zero matrix. 38= [1 −n −n n 1 +n ]andAm= [1 −m −m m 1 +m ].Then AnAm= [(1−n)(1−m)−nm (1−n)(−m)−(1 +m)n n(1−m) +m(1 +n) −mn+ (1 +n)(1 +m) ]=[1 −(m+n) −(m+n) m+n 1 + (m+n) ]=Am+n. 39= [a b c d ],so thatAt= [a c b d ].Then AAt= [a b c d ][a c b d ]=[a 2 +b 2 ac+bd ac+bd c 2 +d 2 ]=[0 00 0]if and only ifa 2 +b 2 = 0, c 2 +d 2 = 0,andac+bd= 0 only solution to these equations isa=b=c=d= 0, so the only matrix that satisfiesAAt= 0 is the 2×2 zero matrix. 16 Chapter 1 Systems of Linear Equations and Matrices − 2 −2 1 1 0 01 − 1 − 2 0 1 02 1 − 2 0 0 1→1 0 0 −4 3 − 50 1 0 2 −2 30 0 1 −3 2 − 4︸ ︷︷ ︸A− 1 but 0 − 2 − 2 1 0 0− 1 −1 0 0 1 02 1 − 1 0 0 1→1 0 − 1 0 1 10 1 1 0 − 2 − 10 0 0 1 − 4 − 2.The inverse of the product of two invertible matricesAandBcan be found from the inverses of the individual matricesA− 1 andB− 1 .But as in the case of the transpose operation, the order of multiplication is reversed, that is, (AB)− 1 =B− 1 A− 1. Solutions to Exercises
A− 1 = 15 [−1 2−3 1].
[2 − 1− 1 − 3].3 (−2)(−4)−(2)(4) = 0,then the matrix is not invertible.
0 1 − 1 1 0 03 1 1 0 1 01 2 − 1 0 0 1R 1 ↔R 3−−−−−−→1 2 − 1 0 0 13 1 1 0 1 00 1 − 1 1 0 0(−3)R 1 +R 2 →R 2−−−−−−−−−−−−−−→1 2 − 1 0 0 10 −5 4 0 1 − 30 1 − 1 1 0 0R 2 ↔R 3−−−−−−→1 2 − 1 0 0 10 1 − 1 1 0 00 −5 4 0 1 − 3(−2)R 2 +R 1 →R 1−−−−−−−−−−−−−−→1 0 1 −2 0 10 1 − 1 1 0 00 −5 4 0 1 − 3(5)R 2 +R 3 →R 3−−−−−−−−−−−−−→1 0 1 −2 0 10 1 − 1 1 0 00 0 − 1 5 1 − 3(−1)R 3 →R 3−−−−−−−−−−→1 0 1 −2 0 10 1 − 1 1 0 00 0 1 − 5 −1 3(−1)R 3 +R 1 →R 1−−−−−−−−−−−−−−→1 0 0 3 1 − 20 1 − 1 1 0 00 0 1 − 5 −1 3(1)R 3 +R 2 →R 2−−−−−−−−−−−−−→1 0 0 3 1 − 20 1 0 − 4 −1 30 0 1 − 5 −1 3.Since the original matrix has been reduce to the identity matrix, the inverse exists andA− 1 = 3 1 − 2− 4 −1 3− 5 −1 3.6 0 2 1 1 0 0−1 0 0 0 1 02 1 1 0 0 1reduces to −−−−−−−→ 1 0 0 0 −1 00 1 0 1 − 2 − 10 0 1 −1 4 2the matrix is invertible andA− 1 = 0 −1 01 − 2 − 1−1 4 2.1 The Inverse of a Matrix 17
matrix 1 −1 00 0 10 0 0,the matrixAcan not be reduced to the identity and hence is not invertible.
9− 1 =1 / 3 − 1 −2 1/ 20 1 2 − 10 0 −1 1/ 20 0 0 − 1 / 2 10− 1 =1 −3 3 00 1 −1 1/ 20 0 1/2 1/ 20 0 0 1/ 211− 1 = 133 0 0 0−6 3 0 01 − 2 −1 01 1 1 1 12.A− 1 =1 0 0 02 1 0 0− 1 / 2 − 1 / 2 − 1 /2 01 1 0 1/ 213 matrixAis not invertible. 14 matrixAis not invertible. 15− 1 =0 0 −1 01 − 1 −2 11 − 2 −1 10 − 1 −1 1 16 matrixAis not invertible. 17 the operations, we have thatAB+A= [3 810 − 10]=A(B+I) andAB+B= [2 96 − 3]=(A+I)B.18 the distributive property holds for matrix multiplication and addition, we have that (A+I)(A+I) = A 2 +A+A+I=A 2 + 2A+I. 19= [1 2−2 1].a 2 = [−3 4− 4 − 3]and− 2 A= [− 2 − 44 − 2],thenA 2 − 2 A+ 5I= 0. b. Since (1)(1)−(2)(−2) = 5,the inverse exists andA− 1 = 15 [1 − 22 1]= 15 (2I−A).c 2 − 2 A+5I= 0,thenA 2 − 2 A=− 5 I,so thatA ( 15 (2I−A))= 25 A− 15 A 2 =− 15 (A 2 − 2 A) =− 15 (− 5 I) =I.HenceA− 1 = 15 (2I−A). 20. Applying the operations (−3)R 1 +R 2 →R 2 and (−1)R 1 +R 3 →R 3 gives 1 λ 0 3 2 0 1 2 1 reduces to −−−−−−−→ 1 λ 0 0 2− 3 λ 0 1 2 1 .So ifλ= 2 3 ,then the matrix can not be reduced to the identity and hence, will not be invertible.
1 λ 0 0 3−λ 1 0 1− 2 λ 1 .Ifλ=− 2 ,then the second and third rows are identical, so the matrix can not be row reduced to the identity and hence, is not invertible.
1 2 10 λ− 4 − 1 0 4− 2 λ 0 ,ifλ= 2,then the matrix can not be row reduced to the identity matrix and hence, is not invertible.
bλ 6 = 1 the inverse matrix isA− 1 = −λ− 11 λ−λ 1 −λλ− 1 1 λ− 1 − 1 λ− 1 1 λ− 1 0 0 1 . |
Linear algebra with applications 10th edition solutions
Related Posts
Copyright © 2024 muatrau Inc.
