Civil Engineering Reference
In-Depth Information
And, from row one the following equation:
111110
1
xx xx
+++=
2
3
4
Substituting
x
2
=
2,
x
3
=
3, and
x
4
=
4 into this equation yields:
+
()
+
()
+
()
=∴=
1121314 0
x
x
1
1
1
It should be noted that the product the diagonal values of any triangular
matrix is the determinant. In this case, the determinant is |
A
| = (1)(−4)(−3)
(1) = 12.
Example 2.8
Gaussian elimination method
Determine the solution to the following set of equations using Gaussian
elimination and back substitution using partial pivoting. Include a deter-
minant check for uniqueness.
+++=
+++=
−+−+=
+++
x
4
xx xx
x x xx
xx xx
x
10
1
2
3
4
842
26
1
2
3
4
2
3210
1
2
3
4
x
x
=
10
1
2
3
The reduction process is shown in Table 2.5. Again, the numbers to the
right of each row outside the augmented matrix, [
A|C
], are the reduction
multipliers. Also noted is the partial pivoting. Note that the second row
has the largest number in the first column. Therefore, that row is swapped
with the first row, placing the largest element in the pivot element posi-
tion. Reduction is then performed on the first column. After reduction of
the first column, the largest number in the second column is in the third
row. That row is swapped with the second row, placing the largest number
in the pivot position. Reduction is then performed on the second column.
The third column does not require partial pivoting, since the largest num-
ber in the third column is already in the pivot position.
Since two partial pivots were performed, the product of the diagonal
must be multiplied by (−1)
2
to achieve the correct sign on the determinant,
|
A
| = (8)(1.5)(1)(−1) (−1)
2
= −12. Back substitution is performed to deter-
mine the solution vector, [
x
], which is shown in Table 2.5.
