Solution

Example 4

In the previous two examples, we used Naïve GaussEliminationto solve

using five and six significant digits with chopping in the calculations. Using five significant digits with chopping, the solution found was

This is different from the exact solution

Find the solution using Gaussianelimination with partial pivotingusing five significant digits with chopping in your calculations.

=

Forward Elimination of Unknowns

Now for the first step of forward elimination, the absolute value of first column elements are

, ,

or

10, 3, 5

So the largest absolute value is in the Row 1. So as per GaussianElimination with partial pivoting, the switch is between Row 1 and Row 1 to give

=

Dividing Row 1 by 10 and multiplying by –3, that is, multiplying the Row 1 by -0.3, and subtract it from Row 2 would eliminate a₂₁,

=

Again dividing Row 1 by 10 and multiplying by 5, that is, multiplying the Row 1 by 0.5, and subtract it from Row 3 would eliminate a_31,

₌

This is the end of the first step of forward elimination.

Now for the second step of forward elimination, the absolute value of the second column elements below the Row 2 is

,

or

0.001, 2.5

So the largest absolute value is in Row 3. So the Row 2 is switched with the Row 3 to give

=

Dividing row 2 by 2.5 and multiplying by –0.001, that is multiplying by 0.001/2.5=-0.0004, and then subtracting from Row 3 gives

=

Back substitution

=1

Substituting the value of in Row 2

=

=

=

Substituting the value of and in Row 1

So the solution is

=

This, in fact, is the exact solution. By coincidence only, in this case, the round off error is fully removed.

Can we use Naïve GaussEliminationmethodsto find the determinantof a square matrix?

One of the more efficient ways to find the determinantof a square matrixis by taking advantage of the following two theorems on a determinant of matricescoupled with Naïve GaussElimination.

Theorem 1:

Let [ A ] be a n x n matrix. Then, if [ B ] is a matrix that results from adding or subtracting a multiple of one row to another row, then det(B) = det(A). (The same is true for column operations also).

Theorem 2:

Let [ A ] be a n x n matrixthat is upper triangular, lower triangular or diagonal, then det(A) = a₁₁* a₂₂*…….. * a_nn

This implies that if we apply the forward elimination steps of Naive GaussEliminationmethod, the determinantof the matrixstays the same according the Theorem 1. Then since at the end of the forward elimination steps, the resulting matrix is upper triangular, the determinant will be given by Theorem 2.

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