Page 14 - Matrix Analysis & Applied Linear Algebra

6 Chapter 1 Linear Equations

Step 2. Select a new pivot.

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• For the time being, select a new pivot by moving down and to the right. If
this coefficient is not 0, then it is the next pivot. Otherwise, interchange
with an equation below this position so as to bring a nonzero number into
this pivotal position. In our example, −1 is the second pivot as identified
below:

2x + y + z = 1,
y − 2z = − 4,
-1
3y +2z = 8.

Step 3. Eliminate all terms below the second pivot.

• Add three times the second equation to the third equation so as to produce
the equivalent system:

2x + y + z = 1,
y − 2z = − 4, (1.2.5)
-1
− 4z = − 4 (E 3 +3E 2 ).

• In general, at each step you move down and to the right to select the next
pivot, then eliminate all terms below the pivot until you can no longer pro-
ceed. In this example, the third pivot is −4, but since there is nothing below
the third pivot to eliminate, the process is complete.
At this point, we say that the system has been triangularized. A triangular
system is easily solved by a simple method known as back substitution in which
the last equation is solved for the value of the last unknown and then substituted
back into the penultimate equation, which is in turn solved for the penultimate
unknown, etc., until each unknown has been determined. For our example, solve
the last equation in (1.2.5) to obtain

z =1.

Substitute z = 1 back into the second equation in (1.2.5) and determine

y =4 − 2z =4 − 2(1) = 2.
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The strategy of selecting pivots in numerical computation is usually a bit more complicated
than simply using the next coefficient that is down and to the right. Use the down-and-right
strategy for now,and later more practical strategies will be discussed.