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Sec. 2.1 x 17. P = y z x Q = y z 0 1 0 x 0 0 1 y 1 0 0 z 0 0 1 y 1 0 0 z 0 1 0 y = , z x x = y x z You might notice that Q = P . Does this work in general for a permutation matrix? Why? T x 19. E = y z x E −1 = y z 1 0 0 x 0 1 0 y 1 0 1 z 1 0 0 x 0 1 0 y −1 0 1 z 3 E −1 E 3 = E −1 4 5 4 x = , y z+x x = y , z−x 3 = 4 8 5 Note, of course, that E −1 E = EE −1 = I Sec. 2.2 2 x + 3y + z = 8 12. 4x + 7y + 5z = 20 → 2 x + 3y + z = 8 1 y + 3z = 4 Eqn2=Eqn2−2Eqn1 −2y + 2z = 0 → Eqn3=Eqn3+2Eqn2 2 x + 3y + z = 8 1 y + 3z = 4 −2y + 2z = 0 8z = 8 Back substitution: 8z = −8 , z = 1; y + 31 = 4 , y = 1; 2x + 31 + 1 = 8 , x = 2 x, y, z = 2, 1, 1 2 x − 3y =3 13. 4x − 5y + z = 7 → 2x − y − 3z = 5 2 x − 3y =3 1y + z = 1 → 2y − 3z = 2 2 x − 3y =3 1y + z = 1 -5 z = 0 z = 0 , y = 1 , x = 3 by back substitution Operations: Subtract 2 times row 1 from row 2; Subtract 1 times row 1 from row 3; Subtract 2 times row 2 from row 3; 1 14. A row interchange is required here if d = 10. 2x + 5y + z = 0 2x 4x + dy + z = 2 → + 5y + z = 0 d − 10y − z = 2 y−z = 3 2x + 5y + z = 0 → y−z = 3 d=10 y−z = 3 −z = 2 If d = 11 we obtain a singular system. 2x + 5y + z = 0 2x 4x + dy + z = 2 → + 5y + z = 0 d − 10y − z = 2 y−z = 3 2x + 5y + z = 0 → y−z = 2 d=11 y−z = 3 0 = −1 15. Row interchange: x + by =0 x + by =0 x − 2y − z = 0 → −2 + by − z = 0 y+z = 0 x − 2y → =0 y+z = 0 b=−2 y+z = 0 −z = 0 Singular: x + by =0 x + by =0 x − 2y − z = 0 → −2 + by − z = 0 y+z = 0 x−y → b=−1 =0 −y − z = 0 y+z = 0 0=0 z can be anything, e.g. if z = 1 we obtain y = −1 , x = −1 16. a) Two row interchanges: y−z = 0 z=3 → x+y+z = 1 x+y+z = 1 z=3 → y−z = 0 x+y+z = 1 y−z = 0 z=3 z = 3, y = 3, x = −5 b) A row interchange, but ultimate breakdown y−z = 0 y−z = 3 x+y+z = 1 2 19. x + 4y − 2z = 1 x + 4y − 2z = 1 x + 7y − 6z = 6 → 3y − 4z = 5 3y + qz = t 3y + qz = t → x + 4y − 2z = 1 3y − 4z = 5 q=−4 0 = t−5 So the system is singular if q = −4, but if t = 5 there are infinitely many solutions. With z = 1, we obtain y = 3 , x = −9. 21. 2x + y =0 x + 2y + z =0 y + 2z + t = 0 → 2x + y =0 3/2y + z =0 4/3z + t = 0 z + 2t = 5 5/4t = 5 (You can do this all in one shot) Note that the pivots are 2/1 , 3/2 , 4/3 , 5/4 and the solution is t = 4, z = −3, y = 2, x = −1 2x − y −x + 2y − z =0 =0 − y + 2z − t = 0 2x − y → =0 3/2y − z =0 4/3z − t = 0 − z + 2t = 5 5/4t = 5 The pivots are 2/1 , 3/2 , 4/3 , 5/4 and the solution is t = 4, z = 3, y = 2, x = 1 25. a 2 3 a a 4 a a a a → 2 0 a−2 3 1 0 a−2 a−3 a → 2 0 a−2 0 0 3 1 a−4 Three pivots will not exist if a = 0, 2, 4 3