Solutions of Simultaneous Linear Algebraic Equations Using Matrix Algebra - Numerical Structural Analysis - page 85

Table 2.14. ( Continued )
2.110

2.110 −3.040 1.110 1.000 0.000 0.000 0.000

A|I = 0.000 1.250 2.191 1.031 0.009 1.000 0.000 0.000

0.000 1.412 0.000 1.000 0.000

−0.200 −1.154 −0.066

0.000 1.902 3.206 0.000 0.000 1.000

−1.120 −0.626

2.1100 0.0000 −6.7387 −0.6295 0.9840 −1.6880 0.0000 0.0000

A|I = 0.0000 1.2500 2.1912 1.0305 0.0095 1.0000 0.0000 0.0000

0.0000 0.0000 1.7623 −0.9888 −0.0648 0.1600 1.0000 0.0000

0.0000 0.0000 3.8651 4.1289 −0.6171 0.8960 0.0000 1.0000

Note that partial pivoting should be performed and row three is now
swapped with row four. Columns 3 and 4 are then eliminated as shown in
Table 2.15.

Table 2.15. Example 2.13 Matrix inversion method

2.1100 0.0000 −6.7387 −0.6295 0.9840 −1.6880 0.0000 0.0000

A|I = 0.0000 1.2500 2.1912 1.0305 0.0095 1.0000 0.0000 0.0000

0.0000 0.0000 3.8651 4.1289 − 0.6171 0.8960 0.0000 1.0000

0.0000 0.0000 1.7623 − 0.9888 − 0.0648 0.1600 1.0000 0.0000

2.1100 0.0000 0.0000 6.5692 −0.0919 −0.1258 0.0000 1.7435

A|I = 0.0000 1.2500 0.0000 −1.3102 0.3593 0.4920 0.0000 −0.5669

0.0000 0.0000 3.8651 4.1289 −0.6171 0.8960 0.0000 1.0000

0.0000 0.0000 0.0000 −2.8714 0.2165 −0.2485 1.0000 −0.4560

2.1100 0.0000 0.0000 0.0000 0.4035 −0.6944 2.2878 0.7003

A|I = 0.0000 1.2500 0.0000 0.0000 0.2605 0.6055 −0.4563 −0.3589

0.0000 0.0000 3.8651 0.0000 −0.3057 0.5386 1.4380 0.3444

0.0000 0.0000 0.0000 −2.8714 0.2165 −0.2485 1.0000 −0.4560

Now the third and fourth columns are swapped back for the inverse
matrix shown in Table 2.16.

Table 2.16. Example 2.13 Matrix inversion method

1.0000 0.0000 0.0000 0.0000 0.1912 −0.3291 1.0843 0.3319

I|A -1 = 0.0000 1.0000 0.0000 0.0000 0.2084 0.4844 − 0.3651 − 0.2871

0.0000 0.0000 1.0000 0.0000 −0.0791 0.1394 0.3720 0.0891

0.0000 0.0000 0.0000 1.0000 −0.0754 0.0866 − 0.3483 0.1588

( Continued )