SOLUTION OF EQUATIONS AND EIGENVALUE PROBLEMS

UNIT-J SOLUTION OF EQUATIONS AND EIGENVALUE PROBLEMS method of False position (or) Regula Falsi method (or) linear interpolation method. If Lo, then X2 lies between 2, and a X2 af(x1) - x, fla) = f(x) - t(a) 1 Find the positive root of X-32-520 by the Regula Falsi method Soln: Let f(x) = x3-22-5=0 There is only one poistive reot by Descarte's rule of signs Ho)====== f(1)=1-2-5=-6=-ve f(2) = 8-4-5 = -1 =-ve f(3) = 27-6-5= 16= +ve Therefore the positive root lies between 2 and 3. It is closer to 2 also. at(5)-bbla) x, = f(b)-t(a) = 2 ((3)-6(2) = 2(16)-3(-1) 16-(-1) 32+3_35 = 14 T7 2.05.88 [ correct to 4 decimal places] = f(x) = f(2.0588) = (2.0588) - 2(0.0588)-5 = 8.7265 - 4.1176-5 = - 0.3911 : The root lies between 2.0588 and 3

2.0588 +(3) - 3 f (2.0588) x2 = f(3) - *(2.0588) = 2.0588 (16)-31-0.3911) 16-(-0.3991) 32.9408 + 1.1733 34-1141 = = 16.3911 16.3911 = 2.0813 f(x2)= +(2.0813) = (2.0813)3 - 2(2.0813)-5 = 9-0158 - 4.1626-5 = -0.1468 T he root lies between 2.0813 and 3 2.0813 f(3) - x3 = f(3) -(12.0813) 2.0813(16)-3(-0-1468) = 16 -(--0.1468) 33.3008 + 0.4404 33.7412 = 16.1468 16.1468 2.08965 = = 2.0897 ( four derimal places) f(x3) = f (2.0897) = (2.0897)- 2(2.0897)-5 = -0.054 = -ve The root lies between 2.0897 and 3 x4 = (2.0897) f(3) - 36(2.0897) f(3) - f(2.0897) = (2.0897)(16)-31-0-054) 16-(-0.054)

3 = 33.4352+0.162 16.054 = 33.15972 = 2.0928 16.054 3 6(x4) = 6(2.0928) = (2.0928) - 2(2.0928)-5 9.1661 - 4.1856-5 = - 0.0195 =- ve The root lies between 2.0928 and 3 as = 2.0928f(3)-36(2-0928) - f(3) - +(2.0928) (2-0928) (16)-3(-0.0195) = 16 -(-0.0155) 33.4848 + 0.0585 = 16.0195 33.5433 = 16.995 = 2.0939 f(x5) = +(2.0939) = (2.0939) - 2(2.0939) -5 + ++ 1805 - 4 1 878 - 5 - =-0.0073 --ve is The root lies between 2-0939 and 3 A6 = (2.0939) 0f(3)-3t(2.0939) f(3) - fl 2.0939) = (2.0939) (16)-3/-0.0073) 16-(-0.0073) = 33.5024 + 0.0219 = 33.5243 16.0073 16.0073 = 2.0943

f(x6) =f (2.0943) = (2.0943)3 12(2.0943)-5 = 9.1868 - 4.1886-5 = 0.0028 =-ve : The root lies between 2.0943 and 3 Xy = (2.0943) f(3)-3t(2.0943) f(3) - +(2.0943) = (2.0943) (16) - 3(-0.0028) 16 - (-0.0028) 33.5088 + 0.0084 = 16.0028 33.15172 = 2.0945 = 16.0028 f(x)) = f(2.0945) = (2.0945)-2(2.0945)-5 = 9-1884-4-189-5 = - 0.0006 :- ve :: The root lies between 2.0945 and 3 X8 = 2.0945 f(3) - 3 ( 2.0945) f(3) - t(2.0945) = (2.0945) (16)-3(-0.000) 16 - (-0.0006) 33.512 + 0.0018 = 16.0006 = 33.5138 16.0006 = 2.0945

5 we observe that Xy=X8 = 2.0945 correct to 4 places of decimals. Hence the required root correct to four places of decimals is 2.0945 The results of the complete working are tabulated below Iteration(s) a b ar sign of A/tr) I 2 3 2.0588 -0.3911 2 2.0588 3 2.0813 -0.1468 3 2.0813 3 2.0897 -0.054 4 2.0897 3 2.0928 -0.0195 5 2.0928 3 2.0939 -0.0073 6 2.0939 3 2.0943 -0.0028 7 2.0943 3 2.0945 -0.0006 8 2.0945 3 2.0945 Formula xr = af(b)-bfla) (f(5)-fla) we serve that xy = x8=2.0945 Hence the required root is 2-0945 2 using method of false position find a root of the equation 3 x -32-5-0 soln:- -Given f(x) = - flo) = 0-0-5=-5= - ve f(1) = 1-3-5-1-8--7 ve f(2) = =8-6-5=8-11-3.ve f(3) = 27-9-5-27-142132 +ve

one root lies between 2 and 3 set a = 2, b= 3 a ((5) fla) x = f(b) - fla) = 2(13) - 3(-3) 13-(-3) = 26+9 35 31.2.1875 f(x,) = 6(2.1875) = (2.1875) - 3(2.1875)-5 = 10.4675 - 6.5625-5 = -1.095 : one root lies between 2.1875 and 3 X2 = 6(3) - ( ( 2 - 1875) - (2.1875) (1)-3(-1.095) = . 13 - (-1.095) 28.4375 + 3.285 = 14.095 = 31.7225 = 2.2506 14.095 f1x2) = 6(2.2506) = (2.2506)3 - 3(2.2506)- = 11.3997-6.7518-5 - = - 0.3521 = -ve The root lies between 2.2506 and 3 X3 = 2.2506 (((s) . - 38(2.2506) f(3) - H2-2506) (2.2506) (13) 3(-0.3521) = 13-(-0.3521) 29.2578 +1.0563 = 30.314) 13.3521 = 2.2704 = 13.3521

f(x3) = f(2.2704) = (2.2704) 3 - 3(2-2704)-5 7 = 11.7033 - 6-8112-5 = -0.1079 --ve The root lies between 2.2704 and 3 04 = 2-27046(3) - 3f (2.2704) ((s) - f(2-2704) = (2.2704) (13) - 3(-0.1079) 13-(-0.1079) = 29.5152 - 0.3237 29.8389 = 2.2764 13.1079 13.1079 3 f(24) = f(2.2764) = (2.2764) - 3(2-2764) -5 = 11.7963 - 6.8292-5 -0.0329 =-ve The root lies between 2.2764 and 3 as = (2-2764) (+13)-38(2-2764) f(3) - ( ( 2 - 2764) = (2.2764) (13) - - 3 - 0.03929) 13-(-0.0329) = 25.5932 + 0.0987 = 29.6919 2.2782 13.0329 13.0329 f(75) = f(2.2782) = (2.2782) 3(2.2782) -5 = 11.8243 - 6.8346 -5 = 10.0103= The root lies between 2.2782 and 3 9/6 = (2-2782) f(3) - -3f (2.2782) f(3) - f(2-2782)

= (2.2782)(12)-3(-0.0103) 13 - (-0.0103) 29.6166 + 0.0309 29.6475 = = 13.0103 = 2-2788 13.0103 of (x6) = ((2.2788) = (2.2788) - 3(2.2788) -5 = 11.8336 - 6.8304 -0.0028 =-ve The root lies between 2.2788 and 3 x7 = 2.2788f(3)-3t(2-2788) f(3) - +(2-2788) = (2-2188)(13)-3(-0.0028) 13-(-0.0028) = 29-6244 0.0084 13.0028 = 29.6328 13.0028 = 2.2790 B f(x7) = $(2.2790) = (2.2790) - 3(2.2790)-5 = 11.8367-6.837-1 = -0.0003 --ve : The root lies between 2-279 and 3 x8= f(3) - 6(2.279) = (2-279)(13)-3(-0.0003) 13-(-0.0003) = 129.627 +0.000 13.0003

9 = 29.6279 13.0003 = 2.2790 we observe that x7=x8=2.2790 correct to tour places of decimals. Hence the required root is 2.2790 TABLE f(a) =23-3x-5 Formula xr = af(b)-hfla) f(b)-fla) Iteration(s) a b tr f(xr) 1 2 3 2.1873 2.1875 3 2 2.2506 -1.095 2.2506 3 3 2.2704 -0.3521 2.2704 3 4 2.2764 -0.1079 5 2.2764 3 2.2782 -0.0329 6 2.2782 3 2.2788 0.0103 7 2.2788 3 2.2790 0.0028 2-2790 -0.0003 8 2-2790 3 Hence the required root is 2.2790 3 Find an approximate root of a loy, x - 1.2=0 by Regula Falsi method. Sohn:- Let fox) = -1.2 f(1) = 0-1.2=-1.2 - - ve f(2) = 2(0.30103) -1-2 = -0.5979 = - ve f(3) = 3. (0.47712)+ = 0.2314 = +ve Hence a root lies between 2 and 3 x, 26(3)-36(2) = f(s) - f12) = 2(0.2314) - 3(-05979) I 0.4628 + 1.7937 0.2314 - (-0.5979) 0.8293

= 2.2565 = 2.7210 0.8293 f(x) = 6(2.7210) = (2.7210) log,10 2.7210-1.2 = -0.0171 = -ve Therefore the root lies between 2.7210 and 3 N2 = (2.7210) (H1)-35(2-210) - (33) - 6(2.7210) = 0.68094 = 2.7402 0.2485 2-7402-1.2 f(x2) = f(2-7402) = (2.7402) 10910 = -0.0004 =- ve : The root lies between 2.7402 and 3 X3:= 2.7402 6(8)-38(7-7402) f(3) - +(2.7402) - = (0.2314) - (-0.0004) = (2-7407) (0.2314) - (-0.0004) (0.2314) 0 - ( - 0.0004) - = 2-7407 f(x3) = f(2.7407) = 2.7407 10910 2.7467-1.2 =0.0001 =+ve The root lies between 2.7402 and 2.7407 14 = 2.7402 6( 2.7407) - (2.7407) f (2-7402) ((2.7407) - 6(2.7402) = (2.7402)(10.0001) - -(2.7407)(-0.0004) (0.0001) - (-0.0004)

= (2.7402)(0.0001) + (2-7409) (0.0004) 11 0.000/ + 0.0004 = 2.7406 f(x4) = (2.7406) = 2.7406leg 10 2.7406-1.2 = -0.0004 = ve 2.7406 ((2.7407) - 2.7407 6(2.7406) x5= (12-7407) - 6(2-7406) (2.7406) (0.0001) - (2.7407)(-0.00004) = 0.0001-(-0.00044) = (2.7406)(0.0001) + (2.7407) (0.00004) =2.7406 0.0001 + 0.00004 we observe that 74=25=2-7406 = correct to Four places of decimals Hence on required root is 2.7406 f(x)= legiox -1.2 Formula xr = a ((b) - 6 tla) f(5)-fla) Flaration(r) a to xr f(xs) 1 2 3 2.7210 2 -0.0171 2 2.7210 3 2.7402 -0.0004 3 2.2402 3 2.7407 0.0001 4 2-7402 2.7407 2.7406 -0.00004 5 2.7406 2.7406 - 2.7407 we find that H2.7406) is approaching zero Hence the required root is 2.7406.

NEWTON'S METHOD [ Newton. - RaPhson method] Formula:- Xn+1 = in - f(x) / n= 0,1,2 f (xn) D Find the positive root of x4-x = 10 correct to three decimal places using Newton - Raphson method. Sohn'- let f(x)= x4 x -10 6'cx) = 4x3-1 f(1) = 1-1-10= -10 = -ve f(2) = 24-2-10=16-2-10 =4 +ve i: a root lies between land2. Take Xo=2 anti an flan) = - 61(xn) 24 = to - f(x)o) = 2- f(2) 6'(to) f((2) = 2-[29-2-16 24-2-10 = 2-14- = 1.8709 = 1.871 [correct to three decimal places] X2 = x, - 61x1) f (x4) = 1.871 - 6(1.871) 6'(1881) 1.871 - (1.871)4 - 1.871-10 = 4(1.871)5-1 1.871 - 0.3835 =1.856 25.199

x3 = x2- - H(22) 13 ('XX2) = 1.856 - ((1.856) &'(1.856) = 1.856 - 4(1886)3-1 = 1.856 - 0.010 24.574 =1.856 The better approximate root is 1.856 ounl 2 using Newton's verative method find the root between of x(3 = 6x-4 correct to two decimal places Sohn Let f(x) = 2-6x+4 f(cx) = 3x(-6 Ho7=4=tve f(1) D=1-6+4=-15- = -ve a root lies between o and I 16101 > 16(1) : This rat is nearer to I = Xn- f(x) rn+1 6'(x) net Xo =1 the = No- - f(xo) f(colo) = 1-611) 1- 3(1)2-6 ((1) = 1-(-1) = 1-1/3 =0.666 10.67 [correct to two derimal places]

X2 = x, - f(2,) f'(x1) =0.67 - 6(0.67) 8'(0-67) =0.67 +4 3(0.67)2-1 =0.67 0.28 -4-65 = 0.67 + 0.28 4.65 = 0.73 x3 = X2 - 6(x2) ('CX2) = 0-73- 6(0.73) 6'(10.73) = 0.73 - (6.73)3-6(0.73) +4 3(0.73) = 0.73- 0.009 -4.4013 = 0.73 + 0.009 4.4013 0.7320 20.73 [ correct to two decimal places] Here X2 =xs = 0.73 : The root is 0.73 correct to two decimalplaces 3 Find the real positive root of 3x - - cas x - 1 =0 by Newton's method correct to 6 decimal places. john let f(a) = 3x-loss-1 +10) - 0-142-25 -ve

((1) =3-c01-12-cs1 = = 1.459698 = +ve 15 :- a root lies between 0 and \ 161011>16001 Hance the root is nearer to 1. let 20=0.6 x1= 20- 6(*0) 6'(ro) = 0.6 - (6.0.6) 6'(1.0)6) - cos(0.6) -I =0-6-530. 3(0-6) 3+ Sin (0-b) = 0.6-(-0.007101) = 0.607108 X2=21-6(91) 3(0.607108) - Cas (0.607108). = 0.607108 - 3+ Sin (0.607/08) = 0.607108 - (0.000006) = 0.607102 = f(c)2) = 0.607102 - 6(0.607102) 6'(0.607102) =0.607102 - 3(0.607102)- less (0.607102) 3 + sin (0.607102) 0.607102 - 0.0000004 = 0.607102 Here The root is 0.607102 correct to SIX decimals

4 Find by NR method, the root of 10 x = 12.34 Start with No=10. Sohn: let f(x) = x 109,08-12-34 6'cx) = x 1/x logice + logion = logo + log 10 x Given Xo = 10 an+1 = Xn - f(x)n) f(cxn) 4=x - f(ro) f(106) =10- f(10) f((10) = 10- 10/09,00 - 12-34 leg10° + log 10 lo -2.34 = 10- 1.4343 = 10+ 2.34 1.4343 = 11.6315 X2 = 21- - f(x) f'(xx) 11.6315 - ((11.6315) = f((11.6315) = 11-6315 - 11.6315 logio 11.6315 -12.34 logice + logio 11.6315 - 11.6315 - 0.0649 - 1.5 = 11.5949

x3 = X2 - 6(22) 7 f((x) = 11.5949 - f(11.5949) t'(11.5949) = 11.5949 - 11.5949 log 10 11.5949 12.34 logice + 10910 11. 5949 = 11.5949 - 0.00006 1.4986 = 11.5949 from x2 and x3 we find out the root is 11.5949. FIXED POINT ITERATION x = 9(x) method T solve the equation x2 - 2x-3 =0 for the positive root by lttration method 80hr! let for) is easy to factor to show roots at x=-1 and x=3 Rearrange equation (1) x 9(x) = 22+3 let do = 4 x = = 9(x) 2%+3 = 8+3 - IT =3.31662 =3.10375 x2 = g(a,) = 224+3 = 9.63325 A3 = 9(d2) = 222+3 = 19,20750 = 9.20750 = 3.03439 24 = g(xs) = 22s+3 = 9.06877 = 3.01144 x5 = 9(94) = 214+3 = 9.02288 = 3.00381 No = 9126) = 27663 = 3.00127 917 = 9(27) = 2973 = 3.00042

(8=9127) = V2X7+3 = 3.00014 019=9(x8) = 2xg+3=3.00005 X10=9(x9) = 299+3 = 3.00002 x11=9(x10) = 2x(10+3 : 3.00001 x12 = g(x(1) = 2x11+3 = 3.00000 X13 = 9(242)= 2212+3 = 3.00000 Here x12 = x13 = 3 [correct 5 decimal places] Hence the root is 3. 2 Find a real root of the equation x +x2 -100=0 som Let f(x) = x + x 2 - 100 - 0 f(a) = -100=-ve f(1) f(2) = 8+4-100 --ve f(3) = 27+9-100=-64, --ve f14) = 64716-100==20==== --ve f(5) = 125+25-100= 50 ++ve So there is a real root between 4 and 5 The given equation can be written as x2(xx1) = 100 x 10 - 9cx) Va+1 g'(x) = 10 [-1/2] -5 (24+1)3 = (2+1) 4/2 19'cxsl = 5 (2+1)3/2 18'14) = 53/2

19 19'(5)] = 5.00

Gaussian Elimination method. & Gauss -Jordan method 9 solve the system of equations by (i) Gauss elimination method (ii) Ganss -Jordan method. 10 X-24 +32 =23 2x +104-52--33 3 x -44 tloz 41 soln! (i) Garss elimination method The given system is equivalent to a 10 -2 3 23 = -5 y 2 10 -33 Z 41 3 -4 10 3 23 10 -2 Here [A, B] = 2 10 -5 -33 3 -4 10 41 NOW, we will make the matrix A as a upper triangular Fix the first row, change 2 and 3 row with row -2 3 23 R2 (-) 5R2 R1 10 [A,B]~ -188 0 52 -28 R3 10 R3 -3R, -32 91 341 0 FIX 1 and 2 row, change 3 row with 2nd row -2 3 23 10 N -188 R3 t 52 R3 + 34 R2 0 52 -28 0 0 3780 11340 This is an upper triangular matria from (1) we get [ by back Substitution 3780 2 - 11340 z=3

21 52Y-282 = - 188 524-28(3) = - 188 Y=-2 10x-2y +32=23 10 x-21-2) + 3(3) = 23 10x+4+9=23 10x+13=23 102=23-13 10x = 10 x=1 Hence the Solution is x=1, y= -2, 2=3 (ii) Gauss 1 Jordan method. Take the equation (1) 10 -2 3 23 R, I 1260 R, - R3 (A, B) 2 o 52 -28 -188 R2 0 135 R2 + R3 0 3780 11340 diagonal matrix Now we will make the matrix A a Fix the third row and change 2nd row and first row 12600 -2520 017640 N 0 70200-14040 0 0 3780 11340 Rix the 2 and 3 row change 1 row with 2 nd row N 88452000 0 O 88452000 7020 -14040 R1 7020R1+2520R2 o O 3780 11340 (ie) 1 N , 1010-2 10013 : x =1, y=-2, Z=3

2 solve the system of equations by Gauss - elimination method 5x, = 4 x1 + 7x2 x3 +x4 = 12 a + X2 +693 +X4 = -5 x, +x2 + X3 444 = -6 - som The given system is equivalent to 5 a, 4 x2 1 = 12 1 61 x3 -5 I 14 214 -6 [A, B] = 5 I 4 1 7 1 12 1 6 ) -5 1 I 4 -6 5 1 1 4 R2 5R2 -R1 a o 34 4 4 56 R3 5R3 - R1 o 4 29 4 -29 0 4 19 -34 R2 5 R4 -R, 4 5 I 1 1 4 R3 34R3 -4R2 ~ 4 56 0 34 4 -1210 970 120 R4 34R4 -4R2 0 120 630 -1380 0 o R2 R2 5 1 1 4 2 I a 2 2 28 17 R3 R3 0 10 -121 0 0 97 12 0 -138 R4 R4 0 12 63 10 4 5 1 1 1 N 28 0 17 2. R4 97R4 - 12R3 -121 o 97 12 0 o 0 5967 -11934 5967 94 = - 11934 = -2

23 97% + 1224 = -121 97x3 12 (-2) = -121 9773 -24 =421 97x3==121424 97%3==97 19 17 17.12 +2(+1)+2(-2)=28 17x2 12-2-4=28 1722 -6=28 17x2 = 28+6 17x2=34 x2=2 5x4 +x2++3+X4=4 524 5x1-104 5x4=5 p4=1 Hence the solution is x1=1,212 =2, x3=-1, 242-2 3 using the Gams-Jordan method Solve the following equations lox 2+4+2=12 2x +104+2= 13 x + Y+5Z= 7 Soln:- Interchanging the first and the last equation than [A,B] = I 2 10 1/13 10,112 Fix the phot element row and make the other elements zero in the pivot element column. 51 R2 R2-2R1 R3 ( ) R3-10R, 0 -9 -491-581

5 7 ~ -1.125 -49 -0.125 R2 in) R2 8 -58 7.125 R, has Rr-R2 1 0 6-125 ) -1.125 -0.125 R3 R3+9R2 2 -59.125 -59.125 7.125 6.125 R3 R34- -1.125 -0.125 - 59.125 1 R1 (mm R1-6.125R3 R1 R2 +1.125 R3 :. 21=1,4=1,221 ITERATIVE METHODS (a) Galess Jacobi method (b) Galuss. - seidal method. 1 solve the following system of equations by Gauss -Jacobi method and Gauss - seidal method 27x +64-Z-85 x+y+542= 110 6x+15y+az=72 Sdm: As the coefficient matrix is not diagonally domiant we rewrite the equations. 27x +by-z=85 S. x+ y +542 = 110 diagnolly dominant 6x + 15y +22 = 72 byo! 10241 +1623 less since the diagonal elements are dominant in the coefficient matrix we write x,y, Z as follows: r= 1/4 [85-6y+z] y= 15 [ 72-6x-2z) z 2 4/4 [110-4-4]

(1) Gauss Jacobi method 25 Let the Initial values be x20, 420, 220 First itelation:- = [85]=3.148 y(1) = 15 [72]=4.8 = second x2' 1=1/14 [85 - = 1/2 [85 - 6.(4-8) +(2.037) = 2.157 y(2) = 1/5 [72-6x"22"]] - = 1/5 [72 - 6(3.148). -2 (2.037)] 0 3.269 z(2) = 1/4 = - 3.148-4.8] =1.890 Third iteration:- x (3) = 17 = +1.890] 2 2.492 y(3) = 1/15 = 1s[12 - = (3) = 54 = 1/4 [no -2.157-3.269] = 1.937 Fourth iteration:- x4 = = = 2.401 y(4) = = =3.545 z 2(4) - = [110 - 2.492-3.685] = 1-923 54 Fifth iteration:- a (5) 1/T [85-6964" (4)] = 1/4 [85-6(3-545) +1.923] = 2.432 (5) = =ts [72-62-22", = ts [92-6(260)-2(1920)=3 = 3.583 (5) = 1/4 1/4 [110 - 2.401-3.645] = 1.927 54 sixth Iteration x6 1/7 [85_6y6tz"], [ 85 - 6 ( 3.583) + -1.927] = 2.423 y(b) = 1/6 72-625-2255 = 1/5 - 2 = 3.570

2(6) = 1/4 = seventh iteration: 7 = 1/7 x y(7) = 1/5 [12 - = [72 - 6(2.423)-2(1-926) = 3.574 2 (7) = 1/[10- xb -y"]= 1 54 [110 - = Eighth Iteration:- a a(8) = 1/2 [85-6y(7). + 217)] = 1/2 [85-6(3.574)+1.926)=2.425 y(8) = 15 E = 1/5 - = 3.573 2 = 1/4 54 [ 110-x"-y(7) - = 1/4 [.110 - 12.426-3.574)-1.926 - Nineth Iteration:- (9) = 1/7 [85 ((8) t >(8)] = 27 1/2 [85 - 6(3.573) + 1.926] 4 2.426 y(9) = 1/5 172-628-220: = [72-6(2.425)-2(1.926)]=3.57 - >19 = y(8)] = [110-2.425-3.573)=1.926 - 54 Tenth Iteration:- - = e.426 (ao) = ts 1/5 [72 6(2.426) - 2 2(1.926) = 3.573 2(10) = 1 [110-2-426-3-5737-11926 54 Hence 12.426, y = 3.573, Z = 1.926 [ correct to three decimal places] 2 Gauss - seidel method Let the initial values be y 20 Z=0

27 First ltoration a(1) = 1/7 5/17 [ 85-610)+03=3.148 yli) = 1/5 [72-6x-gzo]] = 15 [ 1-6(3.148)-07=3.54 = 2(1) = 54 - = 1/4 54 [110 - 3.148-3.5417=1-913 second iteration:- x2 = 1/7 [85-6y(") +z(") = 1/4 [85-6(2-541)41.913]. = 2.432 y(2) = 15 15 [72 - 6(2-432)-2(1+913), - = 3.572 (2) = 1/4 1/4 [110 - 2.432-3.572]= 1.926 T hird iteration:- x(s) = 14[85-692 - (22) = [85 6(3.572) +1.92b] A 2.426 y(3) = [72 - bx3 -2z(2) = 15 [12 - b(2.426) - 2(1.926)] = 3.573 z 2(3) = - = 154 [110 - 2.426-3.573] = 1.926 Fourth iteration:- x4 = = - 6(3.573) 71.926]-2.426 y(4) = - = 1/5 [72 - 6(2.426). - 2(1.926)] = 3.573 2(4) = 54 - = 1 54 [110 - 2.426 - 3.5737-1.92 Hence x 2.426 y= 3.573, z=1.926 , This shows that the convergence is rapid in Gauss - seided method when compared to Gauss - Jacobi method. 2 solve the following equations by Gauss Seidel method 4x ++2Y+Z=14 x+15y-Z=10 x+y+82=20

Soln:- as the coefficient matrix is diagonally dominant solving for x,y, 2 we get x = 14 [ 44-24-2] y = 1/5 [10-x+z] z = % let the initial values be y=0, Z=0 First iteration:- x(1) = (14-210)-0) e 4=3.5 y(1) = = 1/5 [10-3.5+0]=1.3 2(1) = 1/8 = 1/8 [ 20-3.5-1.3]=1.9 second iteration x x(2) = 1/4 - 1/4 y(2) = 1sl 10 x(2) = 1/5 [10-2-375 +1.9]=1.905 q(2) = 1/502 -y(2) = 1/8 [20-2.375 1-905]-1-965 Third ltoration: 23 = - - = y(3) = 1/5 [10-x (3) + 2(2)] = 1/5 [10-2.056 +1.965]=1.981 2 = 1/3 [20-2.056 +1.9818] =1.995 Fourth lttration:- x = 1/4[14 - 2y(_) = 2-510 y = 1/5 [10 = Y/5 C 10- 2.510+1.995] = 1.897 2 4) = 1/8 [20 2.510-1.897 = 1.949

Fifth literation:- 29 x(5) = y(5) = +24]= 15 10-2.064+1.949)=11977 - z(5) = 1/8 / 20 - 2.064 -1-977]. 1-995 sixth iteration:- x(6) = - = y(6) = 10-2(+2(5) 1/[[10 - 2.013+1.995] = 1.996 2 (6) = [[20 P-2-013-1.996], 1.999 Seventh ltteration:- q(n) = 1/4 [14-261-996)-1-999] 2.002 y(7) =1/5 [10 = 15C10-2.002+1.999]-1.999 Z = 1/8 1/8 [20-2.002-1-999)=2.000 Eighth literation x = 454 "]. 14[14-2(1999)-2)=2.001 = y(8) = 15-[ = z(8) = 20-2.001-2] = 2.000 Ninth iteration:- a(9) = A[14-2(2)-2]=2 2(i) = 1/ [ 20 y(9) iteration:- 1/5 [ 10-x(9) tz(8) (5[10-242] is [20-2-2] =2 =2 = = Tenth x(10) = 4514-29-9 -zl9] )4[[14-2(2)-2]02 y(10) = 1/5 [ 10 - - 1/5 [10-242]=2 2(10) = Hence r==, Y=2, Z=2

INVERSE OF A MATRIX BY GAUSS JORDON METHOD Gauss - Jordan elimination method. ( using Gruss 3-Jordan - method, find the Inverse of the matrix. [ aid 2 2 135 John:- [A,I] = 2 1 35 001 1 / ~ 3/2 1/2 1 of R1 R1 2 2 1 3 5 1 I 3/2 1/2 R2 to R2 - 2R, N -2 -1 2 7/2 R3 is R3-R1 0 -1/2 1 ~ 3/2 1/2 0 R2 I 1 RT R2 2 2 1/2 -1/2 0 1 0 1/2 -1/2 1 R, I R1-R2 N I 2 1 o Rs + R3 -2R, -1/2 5/2 2 I 1 ~ 1/2 -1/2 1 2 1 . o) R3 I R3(-2) I 5 -4 -2 1 2 ~ -1 R1 I R, + 1 2 R3 -9 7 4 R2 to R2 - 2R3 -4 -2 Hence A -1 = 2 -1 -1 -9 7 4 5 -4 -2

verification:- 31 AA' 4 = 2 135 5 -4 -2 2 using Gauss - Jordon method, find the inverse of A= 14 I 13 13-3 2-2-4 somi- Let [A,I]= = 1 3 -3 -2 -4-4 N 1 :00 R3 tr R3+2R1 0 -2 2 20 3 , N [ : 0 -3 -1/2 of R2 th 1/2 Rz/ R2 2 -2 2 2 6 3/2 -1/2 R1HR1-R21 N -3 -1/2 1/2 R3 to Rs+2R2 : -4 I I 6 H 3/2 -1/2 N -1/2 10.00 R3 in R3 1/2 o -4 -1/4 -1/4 =1/4 3 N : o 3/2 R,H,R1-6R3 -5/4 -Y/4 -3/4 R2 " R2+3R3 -1/4 -1/4 -1/4 A = 3 3/2 -5/4 -1/4 -14 -14 -1/4 6 -1/4 =3

verification:- AA = I 3 LAE -5/4 -1/4 -3/4 -1/4 -1/4 -1/4 = EIGEN VALUE OF A MATRIX BY POWER METHOD The power method 1 Find the numerically largest eigenvalue of A= 1-32 4 4 -1 by 635 power method. Sohn, set X1 = be an arbitrary initial cigaretor 1 -3.27 0.167 Ax, = = 6x2 4 6 35 6.167 0.166 0.021 1 -3 2 = 2.336 = 8.003 AX2 = 0.292 = 8.003 X3 4 -1 0.667 8.003 .1 635 I 0.167 - 3(0.067)+2(1)-00-166 4(0.167) + (0.667)-1 = 2-336 6 (0.167) + (0.667)+5 = 8.003 1 6 3 5 -3 2 0.292 0.021 = 1.145 0.252 6.002 = 6.002 0.042 56.002X Axs = 4 1 110.021) - 3(0.292) +2(1) = 1.145 4(0.021) +4(0.292) -((1) = 0.252 6(0.021) + 310.292) +5(1)=6.002

33 1635 THE 0.042 -0.068 2.065 0.191 0.329 AX4 = 6.272 = -0.011 = 6.272x5 6.272 1 ( 0.191) - 3(0.042) +2(1) = 2.065 ( 0.191) + 4 (0.042) -1(1)==0.068 6(0.191) +3(0.042) +5(1)=6,272 2.362 0.34 Ax5 = \ -3 = 6.941 0.039 6.941 x6 4 4 0.272 \ 6 3 5 6.941 2 0.34 2.223 0.311 = = 7.157 AX6 = 0.516 0.072 =7.157X7 4 5 L7.157 2 10.311 2.095 10.296 1 -3 = 7.082 Axy = = 0.532 0.075 7.082X8 4 0.072 7.082 6 3 5 2.071 0.296 1 -3 2 Axq = 0.296 = 0.484 = 7.001 0.069 =7.001x9 4 4 -1 0.075 1 7.00) 63 5 This shows the largest eigenvalue =7. 2 Find the dominant oigen value and the corresponding eigen vector of A= (:- find also the least latent root and hence the third value also. solvic Let X1= be an approximate eigen value AX, = AX2= = 03 =7.x3 0 Ax3 = ) 3. = 1.8572 3.5714 0 3.5714 0.52] 5.3714 X4

0.52 1 = 2.04 4-12 6 4.12 [0.4951 =4.12 X5 Ax5 = 0.4951 1 0 = 3.9706 1.9902 = 3.9706 0.5012 0 3.9706.X6 D 1 0 0 3 HI 0.5012 0 = 4.0072 A x 6 2.0024 = 4.0072 = 4.0072X7 AX7 = 161 003 0.4997 0 = 1.9994 3:9982 = 3.9982 [005000] 3.9982 X8 AX8= 1003 0 4x9 Dominant eigen value = 4; corresponding eigen vector is (1,0.5,0) To find the least eigen value, let B=A - 4I, since X1=4 -3 : B= 0 0 2 = we will bind the dominant eigen value of B. set Y = [:] be the initial vector. BY, = -0.3333 3 Y/2 0 BY2 = -3 0 : -2 2 0.3333 o = 1.6666 -5 0 -5 1/3 1 BY/3 = 6 -5 = -5 0.3333 0 100-2 Dominant elgen value of B is -5 - Adding 4, smallest oigen value of A = -5+4=-1 sum of eigen values = trace of A = 1 +2+3 = 6 4 +(+1)+ 13=6, :- A3=3 All the three eigen values are, 4, 3,-1.