Solutions — Chapter 1 1.1.1. (a) Reduce the system to x − y = 7, 3y = −4; then use Back Substitution to solve for x = 17 3 , y = − 4 3 . (b) Reduce the system to 6u + v = 5, − 5 2 v = 5 2 ; then use Back Substitution to solve for u = 1, v = −1. (c) Reduce the system to p + q − r = 0, −3q + 5r = 3, −r = 6; then solve for p = 5, q = −11, r = −6. (d) Reduce the system to 2u − v + 2w = 2, − 3 2 v + 4w = 2, −w = 0; then solve for u = 1 3 , v = − 4 3 , w = 0. (e) Reduce the system to 5x1 + 3x2 − x3 = 9, 1 5 x2 − 2 5 x3 = 2 5 , 2x3 = −2; then solve for x1 = 4, x2 = −4, x3 = −1. (f) Reduce the system to x + z − 2w = −3, −y + 3w = 1, −4z − 16w = −4, 6w = 6; then solve for x = 2, y = 2, z = −3, w = 1. (g) Reduce the system to 3x1 + x2 = 1, 8 3 x2 + x3 = 2 3 , 21 8 x3 + x4 = 3 4 , 55 21 x4 = 5 7 ; then solve for x1 = 3 11 , x2 = 2 11 , x3 = 2 11 , x4 = 3 11 . 1.1.2. Plugging in the given values of x, y and z gives a+2b−c = 3, a−2−c = 1, 1+2b+c = 2. Solving this system yields a = 4, b = 0, and c = 1. ♥ 1.1.3. (a) With Forward Substitution, we just start with the top equation and work down. Thus 2x = −6 so x = −3. Plugging this into the second equation gives 12 + 3y = 3, and so y = −3. Plugging the values of x and y in the third equation yields −3 + 4(−3) − z = 7, and so z = −22. (b) We will get a diagonal system with the same solution. (c) Start with the last equation and, assuming the coefficient of the last variable is 6= 0, use the operation to eliminate the last variable in all the preceding equations. Then, again assuming the coefficient of the next-to-last variable is non-zero, eliminate it from all but the last two equations, and so on. (d) For the systems in Exercise 1.1.1, the method works in all cases except (c) and (f). Solving the reduced system by Forward Substitution reproduces the same solution (as it must): (a) The system reduces to 3 2 x = 17 2 , x + 2y = 3. (b) The reduced system is 15 2 u = 15 2 , 3u − 2v = 5. (c) The method doesn’t work since r doesn’t appear in the last equation. (d) Reduce the system to 3 2 u = 1 2 , 7 2 u − v = 5 2 , 3u − 2w = −1. (e) Reduce the system to 2 3 x1 = 8 3 , 4x1 + 3x2 = 4, x1 + x2 + x3 = −1. (f) Doesn’t work since, after the first reduction, z doesn’t oc