Structural Analysis, 5th Edition
Structural Analysis, 5th Edition
5th Edition
ISBN: 9788131520444
Author: Aslam Kassimali
Publisher: Cengage Learning
Question
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Chapter B, Problem 10P
To determine

Find the inverse of the matrix using the Gauss Jordan method.

Expert Solution & Answer
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Answer to Problem 10P

The inverse of the matrix is A1=[0.48180.05450.28180.34550.05450.16360.34550.03640.28180.34550.11820.14550.34550.03640.14550.4364]_.

Explanation of Solution

Given information:

A=[4203234004213015]

Calculation:

Write the given matrix in augmented matrix form:

A=AI=[4203234004213015|1000010000100001]

Divide the first row by 4.

A=[112034234004213015|14000010000100001]

In the second row, subtract the first row multiplied with 2 from second row.

A=[112034234004213015|14000010000100001]R2R22R1=[11203422×132×12402×(34)04213015|1400002×(14)10000100001]=[1120340243204213015|140001210000100001]

In the fourth row, subtract the first row multiplied with ­3 from fourth row.

A=[1120340243204213015|140001210000100001]R4R4(3)R1=[1120340243204213(3×1)0(3×12)15(3×(34))|140001210000100(3×(14))001]=[1120340243204210321114|1400012100001034001]

Divide the second row by 2.

A=[1120340123404210321114|14000141200001034001]

In the first row, subtract the second row multiplied with 12 from first row.

A=[1120340123404210321114|14000141200001034001]R1R1(12R2)=[112(12×1)0(12×(2))34(12×34)0123404210321114|14(12×(14))0(12×(12))00141200001034001]=[101980123404210321114|381400141200001034001]

In the third row, subtract the second row multiplied with ­4 from third row.

A=[101980123404210321114|381400141200001034001]R3R3(4)R2=[101980123404(4)×12(4)×(2)1(4)×(34)0321114|3814001412000(4)×(38)0(4)×(14)1034001]=[101980123400620321114|381400141200121034001]

In the fourth row, subtract the second row multiplied with 32 from the fourth row.

A=[101980123400620321114|381400141200121034001]R4R4(32)R2=[10198012340062032(32)×11(32)×(2)114(32)×(34)|381400141200121034(32)×(14)0(32)×(12)01]=[10198012340062002138|3814001412001210983401]

Divide the third row by –6.

A=[101980123400113002138|3814001412001613160983401]

In the first row, subtract the third row from the first row.

A=[101980123400113002138|3814001412001613160983401]R1R1R3=[101198(13)0123400113002138|38(16)14(13)0(16)01412001613160983401]=[10019240123400113002138|5241121601412001613160983401]

In the second row, subtract the third row multiplied with ­2 from the second row.

A=[10019240123400113002138|5241121601412001613160983401]R2R2(2R3)=[1001924012(2×1)34(2×(13))00113002138|52411216014(2×(16))12(2×(12))0(2×(16))01613160983401]=[100192401011200113002138|524112160112161301613160983401]

In the fourth row, subtract the third row multiplied with ­1 from the fourth row.

A=[100192401011200113002138|524112160112161301613160983401]R4R4(1R3)=[100192401011200113002(1×1)138(1×(13))|52411216011216130161316098(1×(16))34(1×(13))0(1×(16))1]=[1001924010112001130005524|5241121601121613016131601924112131]

Divide the fourth row by 5524.

A=[1001924010112001130001|52411216011216130161316019552558552455]

In the first row, subtract the fourth row multiplied with 1924 from the first row.

A=[1001924010112001130001|52411216011216130161316019552558552455]R1R1(1924R4)=[1001924(1924×1)010112001130001|524(1924×1955)112(1924×(255))16(1924×855)0(1924×2455)11216130161316019552558552455]=[1000010112001130001|5311035531110195511216130161316019552558552455]

In the second row, subtract the fourth row multiplied with 112 from the second row.

A=[1000010112001130001|5311035531110195511216130161316019552558552455]R2R2(112R4)=[1000010112(112×1)001130001|53110355311101955112(112×1955)16(112×(255))13(112×855)0(112×2455)161316019552558552455]=[10000100001130001|531103553111019553559551955255161316019552558552455]

In the third row, subtract the fourth row multiplied with 13 from the third row.

A=[10000100001130001|531103553111019553559551955255161316019552558552455]R3R2(13R4)=[1000010000113(13×1)0001|5311053110311101955355955195525516(13×1971)13(13×(271))16(13×871)0(13×2471)19552558552455]=[1000010000100001|5311035531110195535595519552553111019551311085519552558552455]=[1000010000100001|0.48180.05450.28180.34550.05450.16360.34550.03640.28180.34550.11820.14550.34550.03640.14550.4364]=[I|A1]

Thus, the inverse of the matrix is A1=[0.48180.05450.28180.34550.05450.16360.34550.03640.28180.34550.11820.14550.34550.03640.14550.4364]_.

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