Gauss Jorden Method
Let a₁x + b₁y + c₁z = d₁, a₂x + b₂y + c₂z = d₂, a₃x + b₃y + c₃z = d₃ these equation are system linear equations. If the augmented matrix \(\left[ \begin{matrix}{{a}_{1}} & {{b}_{1}} & {{c}_{1}} & {{d}_{1}} \\{{a}_{2}} & {{b}_{2}} & {{c}_{2}} & {{d}_{2}} \\{{a}_{3}} & {{b}_{3}} & {{c}_{3}} & {{d}_{3}} \\{{a}_{4}} & {{b}_{4}} & {{c}_{4}} & {{d}_{4}} \\\end{matrix} \right]\) can be reduced to the to the form \(\left[ \begin{matrix}1 & 0 & 0 & \alpha \\0 & 1 & 0 & \beta \\0 & 0 & 1 & \gamma \\\end{matrix} \right]\) by using elementary row transformation then x = α , y = β and z = γ is the solution of linear equation.
Example: Solve the equation x + y + z = 9, 2x + 5y + 7z = 52, and 2x + y – z = 0 by GAUSS JORDENMETHOD
Solution: Given system linear equations are x + y + z = 9, 2x + 5y + 7z = 52, and 2x + y – z = 0.
Now we can written as matrix form and apply elementary row transformation.
A = \(\left[ \begin{matrix}1 & 1 & 1 & 9 \\2 & 5 & 7 & 52 \\2 & 1 & -1 & 0 \\\end{matrix} \right]\).
R₂ → R₂ – R₁
R₃ → R₃ – R₁
\(\sim \left[ \begin{matrix}1 & 1 & 1 & 9 \\0 & 3 & 5 & 34 \\0 & -1 & -3 & -18 \\\end{matrix} \right]\).
R₁ → 3R₁ – R₂
R₃ → 3R₃ + R₂
\(\sim \left[ \begin{matrix}3 & 0 & -2 & -7 \\0 & 3 & 5 & 34 \\0 & 0 & -4 & -20 \\\end{matrix} \right]\).
R₃ → -1/4 R₃
\(\sim \left[ \begin{matrix}3 & 0 & -2 & -7 \\0 & 3 & 5 & 34 \\0 & 0 & 1 & 5 \\\end{matrix} \right]\).
R₁ → R₁ +2 R₃
R₂ → R₂ – 5 R₃
\(\sim \left[ \begin{matrix}3 & 0 & 0 & 3 \\0 & 3 & 0 & 9 \\0 & 0 & 1 & 5 \\\end{matrix} \right]\).
R₁→R₁/3
R₂→R₂/3
\(\sim \left[ \begin{matrix}1 & 0 & 0 & 1 \\0 & 1 & 0 & 3 \\0 & 0 & 1 & 5 \\\end{matrix} \right]\).
Therefore x = 1, y = 3 and z = 5 is the solution.