The Gauss-Jordan Method

• There are many variants to the Gaussian elimination scheme. The back-substitution step can be performed by eliminating the elements above the diagonal after the triangularization has been finished.
• The diagonal elements may all be made ones as a first step before creating zeros in their column; this performs the divisions of the back-substitution phase at an earlier time.
• One variant that is sometimes used is the Gauss-Jordan scheme. In it, the elements above the diagonal are made zero at the same time that zeros are created below the diagonal.
• Usually, the diagonal elements are made ones at the same time that the reduction is performed; this transforms the coefficient matrix into the identity matrix.
• When this has been accomplished, the column of right-hand sides has been transformed into the solution vector. Pivoting is normally employed to preserve arithmetic accuracy.
  $$\left[ \begin{array}{rrrrr} 0 & 2 & 0 & 1 &0 \\ 2 & 2 & 3... ...& -3 & 0 & 1 &-7 \\ 6 & 1 &-6 &-5 &6 \\ \end{array} \right]$$
  Interchanging rows 1 and 4, dividing the new first row by 6, and reducing the first column gives
  $$\left[ \begin{array}{rrrrr} 1 & 0.1667 & -1&-0.8333 & 1 \\ ... ...4 & 4.3334 & -11 \\ 0 & 2 & 0 & 1 & 0 \\ \end{array} \right]$$
  Interchanging rows 2 and 3, dividing the new second row by -3.6667, and reducing the second column (operating above the diagonal as well as below) gives
  $$\left[ \begin{array}{rrrrr} 1 & 0 & -0.8182 &-0.6364 &0.5 \\... ... & -9 \\ 0 & 0 & 2.1818 & 3.3636 & -6 \\ \end{array} \right]$$
  No interchanges now are required. We divide the third row by 6.8182 and create zeros below and above.
  $$\left[ \begin{array}{rrrrr} 1 & 0 & 0 & 0.04 &-0.58\\ 0 & 1... ...0 & -1.32\\ 0 & 0 & 0 & 1.5599 & -3.12\\ \end{array} \right]$$
  We complete by dividing the fourth row by 1.5599 and create zeros above:
  $$\left[ \begin{array}{rrrrr} 1 & 0 & 0 & 0 &-0.5\\ 0 & 1 & 0... ... & 1 & 0 & 0.3333\\ 0 & 0 & 0 & 1 & -2\\ \end{array} \right]$$

• the fourth column is now the solution.
• While the Gauss-Jordan method might seem to require the same effort as Gaussian elimination, it really requires almost 50% more operations.