Chapter 1 Solutions to Review Problems

1 Chapter 1 Solutions to Review ProblemsChapter 1 Exercise 42 Which of the following equations are not linear and why:(a)x21+ 3x2 2x3= 5.(b)x1+x1x2+ 2x3= 1.(c)x1+2x2+x3= (a) The given equation is linear by (??).(b) The equation is not linear because of the termx1x2.(c) The equation is nonlinear becausex2has a negative powerExercise 43 Show that(2s+ 12t+ 13, s, s 3t 3, t)is a solution to the system{2x1+ 5x2+ 9x3+ 3x4= 1x1+ 2x2+ 4x3= these values forx1, x2, x3,andx4in each + 5x2+ 9x3+ 3x4= 2(2s+ 12t+ 13) + 5s+ 9( s 3t 3) + 3t= 1x1+ 2x2+ 4x3=(2s+ 12t+ 13) + 2s+ 4( s 3t 3)= both equations are satisfied, then it is a solution for allsandt12 Chapter 1. Solutions TO Review PROBLEMSE xercise 44 Solve each of the following systems using the method of elimination:(a){4x1 3x2= 02x1+ 3x2= 18(b){4x1 6x2= 106x1 9x2= 15(c){2x1+x2= 32x1+x2= 1 Which of the above systems is consistent and which is inconsistent?Solution.(a) Adding the two equations to obtain 6x1= 18 orx1= value forx1in one of the given equations and then solving forx2we findx2= system is consistent.}}}}

2 (b) The augmented matrix of the given system is(4 6 106 9 15)Divide the first row by 4 to obtain(1 32526 9 15)Now, add to the second row 6 times the first row to obtain(1 32520 0 0)Hence,x2=sis a free variable. Solving forx1we findx1=5+ systemis consistent.(c) Note that according to the given equation 1 = 3 which is the given system is inconsistent3 Exercise 45 Find the general solution of the linear system{x1 2x2+ 3x3+x4= 32x1 x2+ 3x3 x4= augmented matrix of the given system is(1 2 3 1 32 1 3 1 0)A corresponding row-echelon matrix is obtained by adding negative two timesthe first row to the second row.(1 2 3 1 30 3 3 3 6)Thusx3=sandx4=tare free variables. Solving for the leading variablesone findsx1= 1 s+tandx2= 2 +s+tExercise 46 Finda, b,andcso that the system x1+ax2+cx3= 0bx1+cx2 3x3= 1ax1+ 2x2+bx3= 5has the solutionx1= 3, x2= 1, x3= substitute these values into the given system to obtain a+ 2c= 33b c= 73a+ 2b= 7 The augmented matrix of the system is 1 0 2 30 3 1 73 2 0 7 4 Chapter 1.}

3 Solutions TO Review PROBLEMSA row-echelon form of this matrix is obtained as 1:r1 r1 1 0 2 30 3 1 73 2 0 7 Step 2:r3 r3 3r1 1 0 2 30 3 1 70 2 6 2 Step 3:r2 r2 r3 1 0 2 30 1 7 90 2 6 2 Step 4:r3 r3 2r2 1 0 2 30 1 7 90 0 20 20 Step 5:r3 120r3 1 0 2 30 1 7 90 0 1 1 The corresponding system is a 2c= 3b 7c= 9c= 1 Using back substitution we find the solutiona= 1, b= 2, c= 1 Exercise 47 Find a relationship between a,b,c so that the following system is consistent x1+x2+ 2x3=ax1+x3=b2x1+x2+ 3x3= augmented matrix of the system is 1 1 2a1 0 1b2 1 3c We reduce this matrix into row-echelon form as 1:r2 r2 r1andr3 r3 2r1 1 1 2a0 1 1b a0 1 1c 2a Step 2:r2 r2 1 1 2a0 1 1a b0 1 1c 2a Step 3:r3 r3+r2 1 1 2a0 1 1a b0 0 0c a b The system is consistent provided thata+b c= 0 Exercise 48 For which values of a will the following system have (a) no Solutions ? (b)exactly one solution? (c) infinitely many Solutions ? x1+ 2x2 3x3= 43x1 x2+5x3= 24x1+x2+ (a2 14)x3=a+ augmented matrix is 1 2 343 1524 1a2 14a+ 2 6 Chapter 1.

4 Solutions TO Review PROBLEMSThe reduction of this matrix to row-echelon form is outlined 1:r2 r2 3r1andr3 r3 4r1 1 2 340 7 14 100 7a2 2a 14 Step 2:r3 r3 r2 1 2 340 7 14 100 0a2 16a 4 The corresponding system is x1+ 2x2 3x3= 4 7x2+14x3= 10(a2 16)x3=a 4(a) Ifa= 4 then the last equation becomes 0 = 8 which is , the system is inconsistent.(b) Ifa6= 4 then the system has exactly one solution, namely,x1=8a+157(a+4), x2=10a+547(a+4), x3=1a+4.(c) Ifa= 4 then the system has infinitely many Solutions . In this case,x3=tis the free variable andx1=8 7t7andx2=10+14t7 Exercise 49 Find the values of A,B,C in the following partial fractionx2 x+ 3(x2+ 2)(2x 1)=Ax+Bx2+ 2+C2x multiplying both sides of the equation by (x2+ 2)(2x 1) and thenequating coefficients of like powers ofxwe obtain the following system 2A+C= 1 A+ 2B= 1 B+ 2C= 37 Replacer3by 2r3+r2to obtain 2A+C= 1 A+ 2B= 1 A+ 4C= 5 Next, replacer3by 2r3+r1to obtain 2A+C= 1 A+ 2B= 19C= 11 Solving backward, we findA= 19, B= 59,andC=119 Exercise 50 Find a quadratic equation of the formy=ax2+bx+cthat goes through thepoints( 2,20),(1,5),and(3,25).

5 Components of these points satisfy the given quadratic equation. Thisleads to the following system 4a 2b+c= 20a+b+c= 59a+ 3b+c= 25 Apply Gauss algorithm as r2 a+b+c= 54a 2b+c= 209a+ 3b+c= 25 Step r2 4r1andr3 r3 9r1 a+b+c= 5 6b 3c= 0 6b 8c= 208 Chapter 1. Solutions TO Review PROBLEMSStep r3 r2 a+b+c= 5 6b 3c= 0 5c= 20 Solving by the method of backward substitution we finda= 3, b= 2,andc= 4 Exercise 51 For which value(s) of the constantkdoes the following system have (a) nosolutions? (b) exactly one solution? (c) infinitely many Solutions ?{x1 x2= 32x1 2x2=kSolution.(a) The system has no Solutions ifk26= 3, 6.(b) The system has no unique solution for any value ofk.(c) The system has infinitely many solution ifk= general solution isgiven byx1= 3 +t, x2=tExercise 52 Find a linear equation in the unknownsx1andx2that has a general solutionx1= 5 + 2t, x2= 5 + 2x2that isx1 2x2= 5 Exercise 53 Consider the linear system 2x1+ 3x2 4x3+x4= 5 2x1+x3= 73x1+ 2x2 4x3= 3(a) Find the coefficient and augmented matrices of the linear system.}

6 (b) Find the matrix (a) IfAis the coefficient matrix andBis the augmented matrix thenA= 2 3 4 1 2 0 1 03 2 0 4 , B= 2 3 4 1 5 2 0 1 0 73 2 0 4 3 (b) The given system can be written in matrix form as follows 2 3 4 1 2 0 1 03 2 0 4 x1x2x3 = 08 9 Exercise 54 Solve the following system using elementary row operations on the augmentedmatrix: 5x1 5x2 15x3= 404x1 2x2 6x3= 193x1 6x2 17x3= augmented matrix of the system is 5 5 15 404 2 6 193 6 17 41 The reduction of this matrix to row-echelon form isStep 1:r1 15r1 1 1 3 84 2 6 193 6 17 41 Step 2:r2 r2 4r1andr3 r3 3r1 1 1 3 80 2 6 130 3 8 17 10 Chapter 1. Solutions TO Review PROBLEMSStep 3:r2 r2+r3 1 1 3 80 1 2 40 3 8 17 Step 4:r3 r3 3r2 1 1 3 80 1 2 40 0 2 5 Step 5:r2 r2andr3 12r3 1 1 3 80 1 2 40 0 1 52 It follows thatx3= 52, x2= 6,andx1= 112 Exercise 55 Solve the following system. 2x1+x2+x3= 1x1+ 2x2+x3= 03x1 2x3= augmented matrix is given by 2 1 1 11 2 1 03 0 2 5 The reduction of the augmented matrix to row-echelon form is as 1:r1 r2 1 2 1 02 1 1 13 0 2 5 Step 2:r2 r2 2r1andr3 r3 3r1 1 2 1 00 3 1 10 6 5 5 11 Step 3:r3 r3 2r2 1 2 1 00 3 1 10 0 3 7 Step 4:r2 13r2andr3 13r3 1 2 1 00 113130 0 1 73 The solution is given by:x1=19, x2=109, x3= 73 Exercise 56 Which of the following matrices are not in reduced row-ehelon from and why?

7 (a) 1 2 0 00 0 0 00 0 1 00 0 0 1 (b) 1 0 0 30 2 0 20 0 3 0 (c) 1 0 40 1 20 0 0 Solution.(a) No, because the matrix fails condition 1 of the definition. Rows of zerosmust be at the bottom of the matrix.(b) No, because the matrix fails condition 2 of the definition. Leading entryin row 2 must be 1 and not 2.(c) Yes. The given matrix satisfies conditions 1 - 4 Exercise 57 Use Gaussian elimination to convert the following matrix into a row-echelon12 Chapter 1. Solutions TO Review Problems matrix. 1 3 1 1 0 1 1 3 0 3 1 32 6 3 0 1 2 1 3 1 5 1 6 follow the steps in Gauss-Jordan 1:r2 r2+r1, r3 r3 2r1,andr4 r4+r1 1 3 1 1 0 10 0 1 2 1 20 0 1 2 1 40 0 2 4 1 5 Step 2:r3 r3 r2andr4 r4= 2r2 1 3 1 1 0 10 0 1 2 1 20 0 0 0 2 20 0 0 0 1 1 Step 3:r3 12r3 1 3 1 1 0 10 0 1 2 1 20 0 0 0 1 10 0 0 0 1 1 Step 4:r4 r4+r3 1 3 1 1 0 10 0 1 2 1 20 0 0 0 1 10 0 0 0 0 0 Exercise 58 Use Gauss-Jordan elimination to convert the following matrix into reducedrow-echelon form.

8 2 1 1 156 1 2 361 1 1 11 5 5 5 14 the Gauss-Jordan to bring the given matrix into reduced row-echelonform as 1:r1 r3 1 1 1 116 1 2 36 2 1 1 15 5 5 5 14 Step 2:r2 r2 6r1, r3 r3+ 2r1,andr4 r4+ 5r1 1 1 1 110 54300 1 1 70 10 10 69 Step 3:r2 r3 1 1 1 110 1170 54 300 10 10 69 Step 4:r3 r3 5r2andr4 r4+ 10r2 1 1 1 110 1 160 0 1 50 0 0 9 Step 5:r3 r3andr4 19r4 1 1 1 110 1 160 0 150 0 01 Step 6:r1 r1+r2 1 0 0 50 1 1 60 0 1 50 0 0 1 14 Chapter 1. Solutions TO Review PROBLEMSStep 7:r2 r2 r3 1 0 0 50 1 0 10 0 1 00 0 0 1 Step 8:r1 r1+ 5r4andr2 r2 6r4 1 0 0 00 1 0 10 0 1 00 0 0 1 Exercise 59 Solve the following system using Gauss-Jordan elimination. 3x1+x2+ 7x3+ 2x4= 132x1 4x2+ 14x3 x4= 105x1+ 11x2 7x3+ 8x4= 592x1+ 5x2 4x3 3x4= augmented matrix of the system is 3 1 7 2 132 4 14 1 105 11 7 8 592 5 4 3 39 The reduction of the augmented matrix into reduced row-echelon form isStep 1:r1 r1 r2 1 5 7 3 232 4 14 1 105 11 7 8 592 5 4 3 39 Step 2:r2 r2 2r1, r3 r3 5r1,andr4 r4 2r1 1 5 7 3230 14 28 7 560 14 28 7 560 5 10 9 7 15 Step 3:r3 r3 r2 1 5 7 3 230 14 28 7 560 0 0 000 5 10 9 7 Step 4:r4 r3andr2 114r2 1 5 7 3 230 1 21240 5 10 9 70 0 0 0 0 Step 5:r1 r1 5r2andr3 r3+ 5r2 1 0 30 1 40 0 0 130 0 00 0 Step 6:r3 213r3 1 0 30 1 40 0 0 1 20 0 0 0 0 Step 7:r1 r1.

9 5r3andr2 r2 .5rr 1 0 3 0 40 1 2 0 50 0 0 1 20 0 0 0 0 The corresponding system is given by x1+ 3x3= 4x2 2x3= 5x4= 2 The general solution is:x1= 4 3t, x2= 5 + 2t, x3=t, x4= 216 Chapter 1. Solutions TO Review PROBLEMSE xercise 60 Find the rank of each of the following matrices.(a) 1 1 0 00 0 2 34 0 2 13 1 0 4 (b) 1 1 32 0 4 1 3 1 Solution.(a) We reduce the given matrix to row-echelon 1:r3 r3+ 4r1andr4 r4+ 3r1 1 1 0 00 0 2 30 4 2 10 4 0 4 Step 2:r4 r4 r3 1 1 0 00 0 2 30 4 2 10 0 2 3 Step 3:r1 r1andr2 r3 1 1 0 00 4 2 10 0 2 30 0 2 3 Step 4:r4 r3 r4 1 1 0 00 4 2 10 0 2 30 0 0 0 17 Step 5:r2 14r2andr3 12r3 1 1 0 00 .250 0 1 0 0 0 Thus, the rank of the given matrix is 3.(b) Apply the Gauss algorithm as 1:r2 r2 2r1andr3 r3+r1 1 1 30 2 20 4 4 Step 2:r3 r3+ 2r2 1 1 30 2 20 0 0 Step 3:r2 12r2 1 1 30 1 10 0 0 Hence, the rank is 2 Exercise 61 Choosehandksuch that the following system has (a) no Solutions , (b)exactly one solution, and (c) infinitely many Solutions .

10 {x1 3x2= 12x1 hx2= augmented matrix of the system is(1 3 12 h k)18 Chapter 1. Solutions TO Review PROBLEMSBy performing the operationr2 r2 2r1we find(1 310 6 h k 2)(a) The system is inconsistent ifh= 6 andk6= 2.(b) The system has exactly one solution ifh6= 6 and for solutionis given by the formula:x1=3k h6 h, x2=k 26 h.(c) The system has infinitely many Solutions ifh= 6 andk= para-metric form of the solution is:x1= 1 + 3s, x2=sExercise 62 Solve the linear system whose augmented matrix is reduced to the followingreduced row-echelon form 1 0 0 7 80 1 0 3 20 0 1 1 5 free variable isx4= by back-substitution one findsx1=8 + 7s, x2= 2 3s,andx3= 5 sExercise 63 Solve the linear system whose augmented matrix is reduced to the followingrow-echelon form 1 3 7 10 1 4 00 0 0 1 of the last row the system is inconsistentExercise 64 Solve the linear system whose augmented matrix is given by 1 1 2 8 1 2 3 13 7 4 10 reduction of the augmented matrix to row-echelon form isStep 1:r2 r2+r1andr3 r3 3r1 1 1 280 1 590 10 2 14 Step 2:r2 r2 1 1 280 1 5 90 10 2 14 Step 3:r3 r3+ 10r2 1 1 280 1 5 90 0 52 104 Step 4:r3 152r3 1 1 2 80 1 5 90 0 1 2 Using backward substitution we find the solution.}