The Gauss-Jordan Method

• There are many variants to the Gaussian elimination scheme.
• 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.

$$(1) \left[ \begin{array}{rrrrr} 0 & 2 & 0 & 1 &0 \\ 2 & 2... ...& -3 & 0 & 1 &-7 \\ 6 & 1 &-6 &-5 &6 \\ \end{array} \right]$$

$$(2) \left[ \begin{array}{rrrrr} 1 & 0.1667 & -1&-0.8333 & 1 ... ...4 & 4.3334 & -11 \\ 0 & 2 & 0 & 1 & 0 \\ \end{array} \right]$$

$$(3) \left[ \begin{array}{rrrrr} 1 & 0 & -0.8182 &-0.6364 &0.... ... & -9 \\ 0 & 0 & 2.1818 & 3.3636 & -6 \\ \end{array} \right]$$

$$(4) \left[ \begin{array}{rrrrr} 1 & 0 & 0 & 0.04 &-0.58\\ 0... ...0 & -1.32\\ 0 & 0 & 0 & 1.5599 & -3.12\\ \end{array} \right]$$

$$(5) \left[ \begin{array}{rrrrr} 1 & 0 & 0 & 0 &-0.5\\ 0 & 1... ... & 1 & 0 & 0.3333\\ 0 & 0 & 0 & 1 & -2\\ \end{array} \right]$$

1. Interchanging rows 1 and 4, dividing the new first row by 6, and reducing the first column gives
2. 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
3. No interchanges now are required. We divide the third row by 6.8182 and create zeros below and above.
4. We complete by dividing the fourth row by 1.5599 and create zeros above.
5. 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.