2 Partial pivoting

Partial pivoting is a refinement of the Gaussian elimination procedure which helps to prevent the growth of rounding error.

2.1 An example to motivate the idea

Consider the example

$\left[\begin{array}{cc}\hfill 10^{-4}\hfill & \hfill 1\hfill \\ \hfill -1\hfill & \hfill 2\hfill \end{array}\right]\left[\begin{array}{c}\hfill x_1\hfill \\ \hfill x_2\hfill \end{array}\right]=\left[\begin{array}{c}\hfill 1\hfill \\ \hfill 1\hfill \end{array}\right].$

First of all let us work out the exact answer to this problem

Now we compare this exact result with the output from Gaussian elimination. Let us suppose, for sake of argument, that all numbers are rounded to 3 significant figures. Eliminating the one non-zero element below the diagonal, and remembering that we are only dealing with 3 significant figures, we obtain

$\left[\begin{array}{cc}\hfill 10^{-4}\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 10^4\hfill \end{array}\right]\left[\begin{array}{c}\hfill x_1\hfill \\ \hfill x_2\hfill \end{array}\right]=\left[\begin{array}{c}\hfill 1\hfill \\ \hfill 10^4\hfill \end{array}\right].$

The bottom equation gives $x_2=1$ , and the top equation therefore gives $x_1=0$ . Something has gone seriously wrong, for this value for $x_1$ is nowhere near the true value 0.9998…found without rounding.The problem has been caused by using a small number ( $10^{-4}$ ) to eliminate a number much larger in magnitude ( $-1$ ) below it.

The general idea with partial pivoting is to try to avoid using a small number to eliminate much larger numbers.

Suppose we swap the rows

$\left[\begin{array}{cc}\hfill -1\hfill & \hfill 2\hfill \\ \hfill 10^{-4}\hfill & \hfill 1\hfill \end{array}\right]\left[\begin{array}{c}\hfill x_1\hfill \\ \hfill x_2\hfill \end{array}\right]=\left[\begin{array}{c}\hfill 1\hfill \\ \hfill 1\hfill \end{array}\right]$

and proceed as normal, still using just 3 significant figures. This time eliminating the non-zero below the diagonal gives

$\left[\begin{array}{cc}\hfill -1\hfill & \hfill 2\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]\left[\begin{array}{c}\hfill x_1\hfill \\ \hfill x_2\hfill \end{array}\right]=\left[\begin{array}{c}\hfill 1\hfill \\ \hfill 1\hfill \end{array}\right]$

which leads to $x_2=1$ and $x_1=1$ , which is an excellent approximation to the exact values, given that we are only using 3 significant figures.

2.2 Partial pivoting in general

At each step the aim in Gaussian elimination is to use an element on the diagonal to eliminate all the non-zeros below. In partial pivoting we look at all of these elements (the diagonal and the ones below) and swap the rows (if necessary) so that the element on the diagonal is not very much smaller than the other elements.

In practice on a computer we swap rows to ensure that the diagonal entry is always the largest possible (in magnitude). For calculations we can carry out by hand it is usually only necessary to worry about partial pivoting if a zero crops up in a place which stops Gaussian elimination working. Consider this example

$\left[\begin{array}{cccc}\hfill 1\hfill & \hfill -3\hfill & \hfill 2\hfill & \hfill 1\hfill \\ \hfill 2\hfill & \hfill -6\hfill & \hfill 1\hfill & \hfill 4\hfill \\ \hfill -1\hfill & \hfill 2\hfill & \hfill 3\hfill & \hfill 4\hfill \\ \hfill 0\hfill & \hfill -1\hfill & \hfill 1\hfill & \hfill 1\hfill \end{array}\right]\left[\begin{array}{c}\hfill x_1\hfill \\ \hfill x_2\hfill \\ \hfill x_3\hfill \\ \hfill x_4\hfill \end{array}\right]=\left[\begin{array}{c}\hfill -4\hfill \\ \hfill 1\hfill \\ \hfill 12\hfill \\ \hfill 0\hfill \end{array}\right].$

The first step is to use the 1 in the top left corner to eliminate all the non-zeros below it in the augmented matrix

$\left[\begin{array}{ccccc}\hfill 1\hfill & \hfill -3\hfill & \hfill 2\hfill & \hfill 1\hfill & \hfill -4\hfill \\ \hfill 2\hfill & \hfill -6\hfill & \hfill 1\hfill & \hfill 4\hfill & \hfill 1\hfill \\ \hfill -1\hfill & \hfill 2\hfill & \hfill 3\hfill & \hfill 4\hfill & \hfill 12\hfill \\ \hfill 0\hfill & \hfill -1\hfill & \hfill 1\hfill & \hfill 1\hfill & \hfill 0\hfill \end{array}\right]\begin{array}{c}\hfill \\ R2-2\times R1\hfill \\ R3+R1\hfill \\ \hfill \\ \hfill \end{array}\Rightarrow \left[\begin{array}{ccccc}\hfill 1\hfill & \hfill -3\hfill & \hfill 2\hfill & \hfill 1\hfill & \hfill -4\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill -3\hfill & \hfill 2\hfill & \hfill 9\hfill \\ \hfill 0\hfill & \hfill -1\hfill & \hfill 5\hfill & \hfill 5\hfill & \hfill 8\hfill \\ \hfill 0\hfill & \hfill -1\hfill & \hfill 1\hfill & \hfill 1\hfill & \hfill 0\hfill \end{array}\right].$

What we would like to do now is to use the boxed element to eliminate all the non-zeros below it. But clearly this is impossible. We need to apply partial pivoting. We look down the column starting at the diagonal entry and see that the two possible candidates for the swap are both equal to $-1$ . Either will do so let us swap the second and fourth rows to give

$\left[\begin{array}{ccccc}\hfill 1\hfill & \hfill -3\hfill & \hfill 2\hfill & \hfill 1\hfill & \hfill -4\hfill \\ \hfill 0\hfill & \hfill -1\hfill & \hfill 1\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill -1\hfill & \hfill 5\hfill & \hfill 5\hfill & \hfill 8\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill -3\hfill & \hfill 2\hfill & \hfill 9\hfill \end{array}\right].$

That was the partial pivoting step. Now we proceed with Gaussian elimination

$\left[\begin{array}{ccccc}\hfill 1\hfill & \hfill -3\hfill & \hfill 2\hfill & \hfill 1\hfill & \hfill -4\hfill \\ \hfill 0\hfill & \hfill -1\hfill & \hfill 1\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill -1\hfill & \hfill 5\hfill & \hfill 5\hfill & \hfill 8\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill -3\hfill & \hfill 2\hfill & \hfill 9\hfill \end{array}\right]\begin{array}{c}\hfill \\ \hfill \\ R3-R2\hfill \\ \hfill \end{array}\Rightarrow \left[\begin{array}{ccccc}\hfill 1\hfill & \hfill -3\hfill & \hfill 2\hfill & \hfill 1\hfill & \hfill -4\hfill \\ \hfill 0\hfill & \hfill -1\hfill & \hfill 1\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 4\hfill & \hfill 4\hfill & \hfill 8\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill -3\hfill & \hfill 2\hfill & \hfill 9\hfill \end{array}\right].$

The arithmetic is simpler if we cancel a factor of 4 out of the third row to give

$\left[\begin{array}{ccccc}\hfill 1\hfill & \hfill -3\hfill & \hfill 2\hfill & \hfill 1\hfill & \hfill -4\hfill \\ \hfill 0\hfill & \hfill -1\hfill & \hfill 1\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill & \hfill 1\hfill & \hfill 2\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill -3\hfill & \hfill 2\hfill & \hfill 9\hfill \end{array}\right].$

And the elimination phase is completed by removing the $-3$ from the final row as follows

$\left[\begin{array}{ccccc}\hfill 1\hfill & \hfill -3\hfill & \hfill 2\hfill & \hfill 1\hfill & \hfill -4\hfill \\ \hfill 0\hfill & \hfill -1\hfill & \hfill 1\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill & \hfill 1\hfill & \hfill 2\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill -3\hfill & \hfill 2\hfill & \hfill 9\hfill \end{array}\right]\begin{array}{c}\hfill \\ \hfill \\ \hfill \\ R4+3\times R3\hfill \end{array}\Rightarrow \left[\begin{array}{ccccc}\hfill 1\hfill & \hfill -3\hfill & \hfill 2\hfill & \hfill 1\hfill & \hfill -4\hfill \\ \hfill 0\hfill & \hfill -1\hfill & \hfill 1\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill & \hfill 1\hfill & \hfill 2\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 5\hfill & \hfill 15\hfill \end{array}\right].$

This system is upper triangular so back substitution can be used now to work out that $x_4=3$ , $x_3=-1$ , $x_2=2$ and $x_1=1$ .

The Task below is a case in which partial pivoting is required.

[For a large system which can be solved by Gauss elimination see Engineering Example 1 on page 62].

Task!

Transform the matrix

$\left[\begin{array}{ccc}\hfill 1\hfill & \hfill -2\hfill & \hfill 4\hfill \\ \hfill -3\hfill & \hfill 6\hfill & \hfill -11\hfill \\ \hfill 4\hfill & \hfill 3\hfill & \hfill 5\hfill \end{array}\right]$

into upper triangular form using Gaussian elimination (with partial pivoting when necessary).

Answer

Exercises

1. Solve the following system by back substitution $$\begin{array}{rcll}{x}_1+2x_2-x_3& =& 3& \\ 5x_2+6x_3& =& -2& \\ 7x_3& =& -14& \end{array}$$
2. 1. Show that the exact solution of the system of equations

      $\left[\begin{array}{cc}\hfill 10^{-5}\hfill & \hfill 1\hfill \\ \hfill -2\hfill & \hfill 4\hfill \end{array}\right]\left[\begin{array}{c}\hfill x_1\hfill \\ \hfill x_2\hfill \end{array}\right]=\left[\begin{array}{c}\hfill 2\hfill \\ \hfill 10\hfill \end{array}\right]$  is  $\left[\begin{array}{c}\hfill x_1\hfill \\ \hfill x_2\hfill \end{array}\right]=\left[\begin{array}{c}\hfill -0.99998\hfill \\ \hfill 2.00001\hfill \end{array}\right]$ .

   2. Working to 3 significant figures, and using Gaussian elimination without pivoting, find an approximation to $\left[\begin{array}{c}\hfill x_1\hfill \\ \hfill x_2\hfill \end{array}\right]$ . Show that the rounding error causes the approximation to $x_1$ to be a very poor one.
   3. Working to 3 significant figures, and using Gaussian elimination with pivoting, find an approximation to $\left[\begin{array}{c}\hfill x_1\hfill \\ \hfill x_2\hfill \end{array}\right]$ . Show that the approximation this time is a good one.
3. Carry out row operations (with partial pivoting if necessary) to reduce these matrices to upper triangular form.
   1. $\left[\begin{array}{ccc}\hfill 1\hfill & \hfill -2\hfill & \hfill 4\hfill \\ \hfill -4\hfill & \hfill -3\hfill & \hfill -3\hfill \\ \hfill -1\hfill & \hfill 13\hfill & \hfill 1\hfill \end{array}\right]$ ,
   2. $\left[\begin{array}{ccc}\hfill 0\hfill & \hfill -1\hfill & \hfill 2\hfill \\ \hfill 1\hfill & \hfill -4\hfill & \hfill 2\hfill \\ \hfill -2\hfill & \hfill 5\hfill & \hfill -4\hfill \end{array}\right]$ ,
   3. $\left[\begin{array}{ccc}\hfill -3\hfill & \hfill 10\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill -3\hfill & \hfill 2\hfill \\ \hfill -2\hfill & \hfill 10\hfill & \hfill -4\hfill \end{array}\right]$ .

   (Hint: before tackling (c) you might like to consider point 2(b) in Key Point 4.)

Answer