Lecture Notes Course à MA 190 Numerical Mathematics, First ...

The system then takes the form⎛⎜⎝1 −1 1 −10 0.5 −0.75 .8750 1.5 −0.75 1.1250 2 0 2∣0.36790.23861.28082.3504⎞⎟⎠The coefficient in front of a 2 has the largest absolute value in equation 4 andhence this equation is interchanged with the second one, i.e. we perform apivoting operation before carrying out the elimination:⎛⎞1 −1 1 −10.3679⎜ 0 2 0 22.3504⎟⎝ 0 1.5 −0.75 1.1251.2808 ⎠0 0.5 −0.75 .875 ∣ 0.2386After the interchange we eliminate the coefficient in front of a 2 in the thirdand fourth equations. Thus we subtract the second equation multiplied by1.5/2 = 0.75 from the third equation and by 0.5/2 = 0.25 from the fourthequation. The coefficients and the right hand side in the third and fourthequations change according to:Third equation: −0.75 − 0.75 · 0 = −0.75, 1.125 − 0.75 · 2 = −0.375, 1.2808 −0.75 · 2.3504 = −0.4820Fourth equation: −0.75 − 0.25 · 0 = −0.75, 0.875 − 0.25 · 2 = 0.375, 0.2386 −0.25 · 2.3504 = −0.3490 Then we obtain:⎛⎞1 −1 1 −10.3679⎜ 0 2 0 22.3504⎟⎝ 0 0 −0.75 −0.375−0.4820 ⎠0 0 −0.75 0.375 ∣ −0.3490Before the last elimination step no pivoting is required and we obtain a triangularsystem after subtracting the third equation from the fourth.⎛⎞1 −1 1 −10.3679⎜ 0 2 0 22.3504⎟⎝ 0 0 −0.75 −0.375−0.4820 ⎠0 0 0 0.75 ∣ 0.1330Now back-substitution gives the solution:a 4 = 0.1330/0.75 = 0.17733a 3 = (−0.4820 + 0.375 · a 4 )/(−0.75) = 0.55400a 2 = (2.3504 − 2a 4 )/2 = 0.99787a 1 = 0.3679 + a 4 − a 3 + a 2 = 0.98910Hence we have a 1 = 0.98910, a 2 = 0.99787, a 3 = 0.55400, a 4 = 0.1773332