04 Mar 2026
Gauss Elimination Method
Solving linear simultaneous equations by forward elimination and back substitution.
Many physical problems lead to simultaneous linear equations. Examples include balance of forces, electrical networks, normal-mode equations after approximation, and fitting constants from data. In such problems several unknowns must be found together.
Gauss elimination is a direct method for solving a system of linear equations:
\[A\mathbf{x}=\mathbf{b}.\]The method converts the coefficient matrix into upper triangular form and then solves the unknowns by back substitution.
The idea is the same as ordinary elimination in algebra: remove one unknown from lower equations, then remove the next unknown, until the last equation contains only one unknown.
Augmented matrix
For three equations, write
\[\left[ \begin{array}{ccc|c} a_{11} & a_{12} & a_{13} & b_1\\ a_{21} & a_{22} & a_{23} & b_2\\ a_{31} & a_{32} & a_{33} & b_3 \end{array} \right].\]Row operations are then used to remove the coefficients below the diagonal.
The allowed row operations are interchange of two rows, multiplication of a row by a non-zero number, and addition of a multiple of one row to another row. These operations do not change the solution of the system.
Forward elimination
The goal is to obtain
\[\left[ \begin{array}{ccc|c} u_{11} & u_{12} & u_{13} & c_1\\ 0 & u_{22} & u_{23} & c_2\\ 0 & 0 & u_{33} & c_3 \end{array} \right].\]This corresponds to the triangular system
\[\begin{aligned} u_{11}x+u_{12}y+u_{13}z&=c_1,\\ u_{22}y+u_{23}z&=c_2,\\ u_{33}z&=c_3. \end{aligned}\]Back substitution
First find
\[z=\frac{c_3}{u_{33}}.\]Then substitute $z$ into the second equation to find $y$, and finally substitute $y$ and $z$ into the first equation to find $x$.
Pivoting
If a pivot element is zero or very small, rows should be interchanged. This improves numerical stability.
Three-equation example
Solve
\[\begin{aligned} x+y+z&=6,\\ 2x+3y+z&=11,\\ x-y+2z&=5. \end{aligned}\]The augmented matrix is
\[\left[ \begin{array}{ccc|c} 1&1&1&6\\ 2&3&1&11\\ 1&-1&2&5 \end{array} \right].\]Use $R_2\to R_2-2R_1$ and $R_3\to R_3-R_1$:
\[\left[ \begin{array}{ccc|c} 1&1&1&6\\ 0&1&-1&-1\\ 0&-2&1&-1 \end{array} \right].\]Then use $R_3\to R_3+2R_2$:
\[\left[ \begin{array}{ccc|c} 1&1&1&6\\ 0&1&-1&-1\\ 0&0&-1&-3 \end{array} \right].\]Back substitution gives
\[z=3,\qquad y-z=-1\Rightarrow y=2, \qquad x+y+z=6\Rightarrow x=1.\]Hence
\[\boxed{x=1,\quad y=2,\quad z=3.}\]Key points
- Gauss elimination is efficient for small and medium linear systems.
- It is sensitive to round-off error when pivots are small.
- Partial pivoting is usually safer than plain elimination.
Practice questions
- What is an augmented matrix?
- Distinguish forward elimination and back substitution.
- Solve two simultaneous linear equations by Gauss elimination.
- Why is pivoting needed when a pivot element is very small?
- Verify a solution of linear equations by substitution.
Discussion