Civil Engineering Reference
In-Depth Information
With the following variables:
a
k
−1
original elements
a
k
new elements
i
row (
n
)
j
column (
m
)
k
pivotal row number
The following is the back substitution procedure in algorithmic form:
a
a
x
=
nm
n
nn
∑
n
a
−
a x
im
ij
j
ji
=+
1
x
=
where
in in
=− −…
121
,
,
,
i
a
ii
Example 2.7
Gaussian elimination method
Find the solution set to the following nonhomogeneous linear algebraic
equations using Gaussian elimination.
xx xx
x x xx
xx xx
xx xx
+++=
+++=
−+−+=
+++
10
1
2
3
4
842
26
1
2
3
4
2
1
2
3
4
000
=
4
1
2
3
4
The reduction process is shown in Table 2.4. The numbers to the right of
each row outside the augmented matrix, [
A
|
C
], are the reduction multipli-
ers. The pivot row is multiplied by these numbers to reduce the rows below
the pivot row. As an example, the first row is multiplied by −8 and added
to the second row, producing a zero in the first column of the second row.
Then, the first row is multiplied by 1 and added to the third row, producing a
zero in the first column of the third row. Last, the first row is multiplied by 0
and added to the last row, producing a zero in the first column of the last
row. The result is the reduction of all the values below the pivot element in
the first column to zero. The second column is then reduced to zeros below
the pivot element, and lastly the third column is reduced to zeros below the
pivot element. The solution vector, [
x
], is also shown.
These values for
x
are solved after the final reduction. From row four,
the following equation can be written:
14 4
4
x
=∴ =
x
4
