Answer to Question #282692 in Algebra for Atith

Question

Solve the following Non-Homogeneous systems by Gauss elimination and Gauss

Jordan Method.

a) 2x + y + 4z = 12

8x - 3y + 2z = 20 4x + 11y - z = 33

Answer 1

Solution:

By Gauss elimination (or Gauss jordan):

$\left\{\begin{array}{l} 2 x+y+4 z=12 \\ 8 x-3 y+2 z=20 \\ 4 x+11 y-z=33 \end{array}\right.$

Rewrite the system in matrix form and solve it by Gaussian Elimination (Gauss-Jordan elimination)

$\left(\begin{array}{ccc|c} 2 & 1 & 4 & 12 \\ 8 & -3 & 2 & 20 \\ 4 & 11 & -1 & 33 \end{array}\right)$

$\mathrm{R}_{1} / 2 \rightarrow \mathrm{R}_{1}$ (divide the 1 row by 2 )

$\left(\begin{array}{ccc|c} 1 & 0.5 & 2 & 6 \\ 8 & -3 & 2 & 20 \\ 4 & 11 & -1 & 33 \end{array}\right)$

$\mathrm{R}_{2}-8 \mathrm{R}_{1} \rightarrow \mathrm{R}_{2}$ (multiply 1 row by 8 and subtract it from 2 row);

$\mathrm{R}_{3}-4 \mathrm{R}_{1} \rightarrow \mathrm{R}_{3}$ (multiply 1 row by 4 and subtract it from 3 row)

$\left(\begin{array}{ccc|c} 1 & 0.5 & 2 & 6 \\ 0 & -7 & -14 & -28 \\ 0 & 9 & -9 & 9 \end{array}\right)$

$\mathrm{R}_{2} /-7 \rightarrow \mathrm{R}_{2}$ (divide the 2 row by -7 )

$\left(\begin{array}{ccc|c} 1 & 0.5 & 2 & 6 \\ 0 & 1 & 2 & 4 \\ 0 & 9 & -9 & 9 \end{array}\right)$

$\mathrm{R}_{1}-0.5 \mathrm{R}_{2} \rightarrow \mathrm{R}_{1}$ (multiply 2 row by 0.5 and subtract it from 1 row);

$\mathrm{R}_{3}-9 \mathrm{R}_{2} \rightarrow \mathrm{R}_{3}$ (multiply 2 row by 9 and subtract it from 3 row)

$\left(\begin{array}{ccc|c} 1 & 0 & 1 & 4 \\ 0 & 1 & 2 & 4 \\ 0 & 0 & -27 & -27 \end{array}\right)$

$\mathrm{R}_{3} /-27 \rightarrow \mathrm{R}_{3}$ (divide the 3 row by -27 )

$\left(\begin{array}{lll|l} 1 & 0 & 1 & 4 \\ 0 & 1 & 2 & 4 \\ 0 & 0 & 1 & 1 \end{array}\right)$

$\mathrm{R}_{1}-1 \mathrm{R}_{3} \rightarrow \mathrm{R}_{1}$ (multiply 3 row by 1 and subtract it from 1 row);

$\mathrm{R}_{2}-2 \mathrm{R}_{3} \rightarrow \mathrm{R}_{2}$ (multiply 3 row by 2 and subtract it from 2 row)

$\left(\begin{array}{lll|l} 1 & 0 & 0 & 3 \\ 0 & 1 & 0 & 2 \\ 0 & 0 & 1 & 1 \end{array}\right)$

Thus, $x=3,y=2,z=1$