MATH 218 Problems Solution Manual Chapter 1.6, Fall 2011

1.6 Determinants 23 A is invertible < det(A) 4 0 Ax = b has a unique solution «> Ax = 0 has only the trivial solution <> A is row equivalent to J. One useful observation that follows is that if the determinant of the coefficient matrix is 0, then the linear system is inconsistent or has infinitely many solutions. W Solutions to Exercises 1. Since the matrix is triangular, the determinant is the product of the diagonal entries. Hence the determinant is 24, ‘3. Since the matrix is triangular, the determinant is the product of the diagonal entries, Hence the determinant is —10, 5. Since the determinant is 2, the matrix is in- vertible. : 7. Since the matrix is triangular the determinant is -6 and hence, the matrix is invertible. 9. a. Expanding along row one 2. Since the matrix has two identical rows, the determinant is 0, 4, Since the second row is twice the first, the determinant 0. 6. Since the determinant is -17, the matrix is invertible. 8. Since there are two identical rows the determi- nant is 0, and hence the matrix is not invertible. -l1 4 3 4), 3-1 aer(a) =2| 1 -2 |- | 4-2 }+ | 404 | —5. b. Expanding along row two ‘ 0 1 2 1 2 0 det(A) = —3 1 2f Een] 3 3 [+o i [=-5 c. Expanding along column two _ 3 4 2 1 21|_, eaia)=-0| 8 £]ecn] 2 Afew|2 tens -4 1 ~2 d. -det 3 -1 4 =5 e, Let B denote the matrix in part (d) and B’ denote the new matrix. 2 0 1 Then det(B’) = —2det(B) = —10, Hence, det(A) = $det(B’), f Let B” denote the new matrix. The row operation does not change the determinant, so det(B”) = det(B’) = —10. g. Since det(A) 4 0, the matrix A does have an inverse. 10. a. Expanding along row four 1 1 2 -1 1 -1 1 2 -l 1 21 det(A)=-3} -2 0 -1/+3]/ 8 0 —3) 3 ~2 -1]+3}) 3 ~2 0 | =—165. 1 0 1 0 0 0 1 1 0 1 0 b, Expanding along row three the signs alternate in the pattern +,—,-+,—, and the determinant is again —15. ¢, Expanding along column two the signs alternate in the pattern —,+,—~,+ and the determinant is again —15, d Since row two contains two zeros this is the preferred expansion. nonzero, the matrix is invertible. 11. Determinant: 13; Invertible 13. Determinant: —16; Invertible 12, Determinant: ~19; Invertible 14, Determinant: 5; Invertible e. Since the determinant is 24 , : Chapter 1 Systems of Linear Equations and Matrices 15. Determinant: 0; Not invertible 16, Determinant: 0; Not invertible 17. Determinant: 30; Invertible 18. Determinant: -76; Invertible 19, Determinant: —90; Invertible 20. Determinant: 8; {nvertible 21. Determinant: 0; Not invertible 22. Determinant: 0; Not invertible 23. Determinant: —32; Invertible 24, Determinant: 59; Invertible 25. Determinant: 0; Not invertible , 26. Determinant: 5; Invertible i 27. Since multiplying a matrix by a scalar mul- 28, det(2A~') = 2 det(A!) = agftay = ¢ tiplies each row by the scalar, we have that det(3.A) = 39 det(A) = 270. 29. 30. Since the matrix is the transpose of the orig 1 1 1 inal matrix but with two columns interchanged, det, “lh. = = he determinant is —i.0. det((2A)~*) det(DA) ~ Waoi(A) * BO the determinant is -10 31. Expanding along row 3 a 2 2 2 2 2 1 1 |= (-5))", | | =—-(@)@? — 22) = 5a? + 102, 0 0 -5 Then the determinant of the matrix is 0 when —5a? + 10a = —5a(a — 2) = 0, that is@=0ora=2 32. Since 1 111i 111i ii 0o1i1d1i Ooi11%241 10 1 1 1 |]reducestg}]0 0 1 1 1 110141 000 81 11101 000 01 using only the operation of adding a multiple of one row to another, the determinants of the two matrices are equal. Since the reduced matrix is triangular and the product of the diagonal entries is 1, then the determinant of the original matrix is also 1. 33, Since the determinant { is a1b9 — b1a2 — bg + vag + yb — yar, then the determinant will be zero precisely when y = pan0a w+ bigamaiba This equation describes a straight line in the plane. 41 11 14 34, a. A= [: 2 | 2 = [: 2 | c= E 2 | b, det(A) # 0,det(B) = 0,det(C) = ~4 . . “4 3 1 1] 3 oe . Only the matrix C has an inverse. d. Since | hp reduces tg 0 0 | 5 , the linear system is 1 1 2 2/1 1 1738 0 inconsistent. e. Since E 3 p ts reduces to |. there are infinitely many solutions given by 1 2 0 w= 3— ys ye Rf Since— [a 1 tt | seus: ti ce , 2 S10 4] - ior solution @ = aN ll WY 1-1 -2 -1 2 8 b, det(A) = 2 c, Since the determinant of the coefficient matrix is not zero 2 -—2 —2 35.a, A= 8 it is invertible and hence, the linear system has a unique solution. d. The unique solution is x = 8 . : —4 | tne tnear system has the unique ~ 1.6 Determinants . 25 3-2 1 36, a A= jj} 2 5 1 b, det(A) = 0 c. The system will have infinitely many solutions or is 26 —4 1 38 -2|-—1 1 38 -2)-1 inconsistent. d. Since | 2 5 1 | 2 | reducestg| 0 ~1 5 | 4 |, the linear system has infinitely 2 6 —4|~-2 0 0 0] @ many solutions given by # = 11 — 18z,y= —-4+5z2,zER. -l1 0 -1 87,.a,A=| 2 0 2 b, Expanding along column three, then 1 -3 ~3 1-1 = 0. c, Since the determinant of the coefficient matrix is 0, A is not invertible. det(A) = | D8 Therefore, the linear system has either no solutions or infinitely many solutions. | d. Since the augmented matrix reduces to -1 0 -1/-1 1 0 2 0 2/1 }—-]o1 1 -3 -3| 1 0 0 Owns te no! the linear system is inconsistent. 38. a, Since b. woe y il 00 3 1/9 LoL Lp | 282? = 27m = 12y + 36, 164 -2 1 the equation of the parabola is 8a? — 27a — 12y + 36. 39. a. Since b. yow yl . 4 -2 -2 1] _ 2 OV . ane 4 3 2 1/= —29y* 4 20x — 25y + 106, 9 4 -—3 1 the equation of the parabola is ~29y? + 20a — 2y + 106 = 0. 40. a. Since b. ety «2 y i ay 18 -3 -3 1] 2 94,2 5 5 12 11% —240°—24y*—6a—60y+234, | 9 3.0 «42 the equation of the circle is 24m? — 24y? — 6x ~ 6Oy + 234 = 0. 26 ‘ Chapter 1 Systems of Linear Equations and Matrices 41. a, Since b. we ya oy dl 0 140 —4 1 0 16 0 4 1 | = 1360? —16y? ~ 328a + 256, 14 1-2 1 4 9 2 3 1 the equation of the hyperbola is 186a? — 16y? ~ 3282 + 256 = 0. 42, a. Since b. we yr oew yl ay 9 4 -3 2 1 s+ 1 9 -1 8 1 | =—84n?~294y?+840-+630y+924, 1 ot i 1 1 4 > 164 4 2 1 | -5 5 oY the equation of the ellipse is [ 84x? — 294y? + 84a + 630y + 924 = 0, 43, a. Since b, vw ey yy we y l ay 1 0 0 .-1 0 1 s+ 0 0 1 0 1 2f_ 2 an, 2 ] 100 1 01; —124+122°—36xy+42y°—30y, 4 4 4 2 2 1 HH z > 9 38 1 8 11 ° > * the equation of the ellipse is _s] ~12 + 12x — 36xy + 42y? — 30y = 0. 44, 45. 43 7-5 w-l42_, _|6 -3}] 9 “To 3] ° e=15 oa 8? 2 2 Q 24 507 »a 2 4l_, [26] 16 YS la 3|~ vS)5 5] 8 2 2 2 -3 46. ar. 45 3-4 aoe _|-10 5] 25 “To 5| 8° e= 7) -9 4] > 73° 41 75 2 4 -9 8 Wa 4 3] 5 i -7 -10} 1 yl? 6] 9 ye) -9 4] 41 ~7 5 1.7 Elementary Matrices and LU Factorization 27: 48, , 49. -12 -7 4-3 5 ll 97 3 4 25 c= io Us poo =o, -10 ~7 26 —-1 —8 28 : 12 11 —~8 4 -10 ~12 14 _ 12 5 _ 47 ye ~8 83 _ 29 ye) -i0 -7] > 38 ye TIT 23 28 12 il -8 4 -§ 1 —-4 -2 -8 -4 —-2 1 —8 3-4 «1 0 3 1 0. -4 8 -8 0 -1 1 4-8 -l 3 4 0 -8| 45 50, B= 5 = ae = 3,2 = = = 0 -4 1 0 -4 1 0 -4 1 4 0 =-l 4 0 -1 4 0 -1 -2 3 2 2 -2 2 2 3 -2 —2 -3 -8 -1 -2 ~8 -1 -3 ~2] 2 2 -7 -3 2 -7 -3 2 2 | a a) = Top = 2 3 2] 02 = 37 a] = its j; -1 -3 -8 -1 -3 -8 -1 -3 -8 -3 2 -7 -3 2 —7 -3 2 —-7 52. Suppose At = —A, Then det(A) = det(A*) = det(—A) = (—1)"det(A). If n is odd, then det(A) = ~—det(A) and hence det(A) = 0. Therefore, A is not invertible. : 53. Expansion of the determinant of A across row one equals the expansion down column one of A’, so det(A) = det(A‘). . 54, If A = (a,j) is upper triangular, then det(A) = a11429:++@nn. But A® is lower triangular with the same diagonal entries, so det(A) = a11422+++dnn = det(A). Exercise Set 1.7 A factorization of a matrix, like factoring a quadratic polynomial, refers to writing a matrix as the product of other matrices. Just like the resulting linear factors of a quadratic are useful and provide information about the original quadratic polynomial, the lower triangular and upper triangular factors in an LU factorization are easier to work with and can be used to provide information about the matrix, An elementary matrix is obtained by applying one row operation to the identity matrix. For example, 10 07. 1 00 0 1 0 | (-1)Ri+Rs3 > Rs] 0 1 °0 0014)0>0—7 J] +1 081 Nn ee Elementary Matrix Ifa matrix A is multiplied by an elementary matrix E, the result is the same as applying to the matrix A the corresponding row operation that defined E. For example, using the elerfientary matrix above 1 0 0 13 1 1 3 41 EFA=/] 0 1 0 21 0;/=]2 1 0), -l1 01 121 0 -1 0 Also since each elementary row operation can be reversed, elementary matrices are invertible. To find an LU factorization of A: