Asgn

10.6 Use LU decomposition to determine the matrix inverse for the same system as in Prob. 10.2. Do not use a pivoting strategy, and check your results by verifying that [A][A] ' = 1. 7x1+2x2-3x3=-12, 2x1 +5x2-3x3 =-20 , x1 -x2-6x3=-26. Writing the given system in matrix form yields: [A] = 7 2 − 3 2 5 − 3 1 − 1 − 6 Transforming the given matrix into an upper triangular one using Gauss eliminations. First, multiplying the first row by f 21 = 2/7 and subtract it from the second one. Also, multiplying the first row by f 31 =3/7 subtract it from the third one. 7 2 − 3 0 4.43 − 2.143 0 − 1.28 − 6.43 Now multiplying the second row by f 32 = -1.28/4.43= -0.148148 and subtracting it from the third one to obtain 7 2 − 3 0 4.43 − 2.143 0 0 − 6.193 similarly, [L]= 1 0 0 − 0.286 1 0 0.143 − 0.29 1 Hence, [A] = [L] [U] The solutions of systems [A] {X i } = {e i } are the columns of the matrix inverse of [A]. They can be determined by forward and back substitution using the LU decomposition above. The forward substitution will be used to find {Y i } from [L] {Y i } = {e i }, while back substitution will be employed to solve [U] { Xi} = {Yi} i = 1 y1=1 - 0.286y1 + y2 = 0 → y2 = 0.286 0.143y1 - 0.29y2 + У3 = 0 → y3 = -0.06 -6.193x3 = - 0.06 → x3 = 0.00795 4.43x2 -2.143X3 = 0.286 → x2 = 0.0713 7x1 + 2x2 - 3X3 = 1 → x1 = -0.8335

i = 2 y1=0 - 0.286y1 + y2 = 1 → y2 = 1 0.143y1 - 0.29y2 + У3 = 0 → y3 = 0.29 -6.193x3 =0.29 → x3 = -0.0468 4.43x2 -2.143X3 = 1 → x2 = 0.2032 7x1 + 2x2 - 3X3 = 0 → x1 = 0.11987 i= 3 y1=0 - 0.286y1 + y2 = 0 → y2 = 0 0.143y1 - 0.29y2 + У3 = 1 → y3 = 1 -6.193x3 =1 → x3 = -0.1615 4.43x2 -2.143X3 = 0 → x2 = -0.078 7x1 + 2x2 - 3X3 = 0 → x1 = -0.0915 [A] −1 = − 0.8335 0.11987 − 0.0915 0.0713 0.2032 − 0.078 0.00795 − 0.0468 − 0.1615 Performing multiplication [A] [A] −1 yields 0.6998691425 -0.085795545 0.071644485 -0.085795545 0.0578492969 -0.012919665 0.071644485 -0.012919665 0.03664474 Now that the system can now be solved as: {x} = [A] −1 {B} x 1 x 2 x 2 = − 0.8335 0.11987 − 0.0915 0.0713 0.2032 − 0.078 0.00795 − 0.0468 − 0.1615 x − 12 20 − 26 x 1 x 2 x 2 = 14.78 5.24 3.167 [A] −1 ≈ − 0.8335 0.11987 − 0.0915 0.0713 0.2032 − 0.078 0.00795 − 0.0468 − 0.1615

Following the same procedure which was used in part (a) using the formulas above to obtain an approximation for Хі, i = 1, 2, 3 in each step, but then altering the approximations using for the given value of A. First iteration x1= (-20 - x2 + 2х3) / -8 = -20/-8 = 2.5 x1 = 1.2 • 2.5 + (1 - 1.2) • 0 = 1.2 • 2.5 - 0.2 • 0 = 3 x2= (-38 -2x1 + x3) / -6 =(-38 - 2.3) /-6 = 7.3333 x2=2=1.2 ・ 7.33333- 0.2 ・ 0 = 8.8 x3=(-34 + 3x1 + x2)/ 7 = (-34 + 3 .25 + 7.16667)/7 = -2.31429 x 3 =1.2 ・ (-2.31429) - 0.2 ・ 0= － 2.77715 → |E a,1 | = 1, i = 1,2, 3 Second iteration x1=(-20-8.8+2 ・ (-2.77715))/ -8 = 4.29428 x1= 1.2 ・ 4.29428-0.2 ・ 3 ＝ 4.553145 → |E a,1 |= |(4.553145 - 3)/ 4.553145| = 0.3407 x2= (-38-2 ・ 4.553145- 2.77715)/6 = 8.3139 x2 = 1.2 • 8.3139 - 0.2 • 8.8 = 8.21668 8.21668 - 8.8 → |E a,1 |= |(8.21668-8.8)/8.21668| = 0.071 x3 =(-34+3 ・ 4.553145+ 8.21668)/ 7 = -1.73198 x3 = 1.2 ・ (-1,73198) － 0.2 ・ (-2.77715) = -1.52295 → |E a,1 |= |(-1.52295+2.77715) /-1.52295 = 0.8235 iteration | x1 X2 X3 | |E a,1 | |E a,2 | |E a,3 | ----------------------------------------------------------------------------------------------- -------- 1 | 3 8.8 -2.77715 | 1 1 1 2 | 4.553145 8.21668 1.52295 | 0.3407 0.071 0.8235 3 | 3.77875 7.77275 -2.24815 | 0.204 9 0.0571 0.3226 4 | 4.0846 8.12892 -1.88476 | 0.0749 0.0430 0.1928 5 | 3.96785 7.93831 -2.05016 | 0.0294 0.024 0.0807

6 | 4.01222 8.02726 -1.97901 | 0.0111 0.0111 0.036 →(X1, X2, X3) = (4.01222,8.02726,-1.97901) The value of 1 seems to have been chosen improperly, as this method demonstrates a slower convergence that the version without relaxation in part (a). However, the solution is approximately the same after rounding off: Therefore, (X1, X2, X3) ≈ (4,8, -2)