Question

4. Solve the following system of linear equations using Gauss-Jordan elimination: X1 + 32 - 2x3 + 24 + 3x5 = 1 2x 1 - X2 + 2x3 + 2x4 + 6x5 = 2 3x1 + 2x2 - 4x3 - 3.24 - 9.25 = 3

Answer 1

Solved using Gaussian Jordan Elimination method

x, + 02 - 2x + x4 + 3 X5 = 1 2x, x₂+2x₂+2x4+ 605=2 3x1+2x₂ - 4x₂ - 3X4 - 9X5=3. 1 1 -2 13 2-1 2 2 6 3 2 -4-3-9 X2 X3 X4 1 2 3 X5 .:: The augmented mateix is. - 1 1 -2 13 2-1 2 6 - 3 2 -4-3-9 R₂-23, Rz - BR 2 2 3 1 R₂ (-1/ 0 -3 -2 3 6 0 O 0 -1 2-6-18 1 1-2 1 3 O - 2 O o LO -1 2 -6 -18 1 0 0 R, R2 R₃th 1 1 1 0 0 3 0 1 -2 0 0 бо 0-6-18 R 31-46) 0 1 R-R3 1 3 -2 0 0 ОО 0 13 0

O 01-20 0 LO O O 13 1 0 0 ОО - 2 0 0 0 0 1 3 1 O 2 X2 X3 X4 X5 0 0-223 [D4 +3X5 7x, = 1,x₂= 2x3, x=-3X5 X3, X are arbitrary .. The system has infinitely many solutions