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

Question

Want to see more full solutions like this?

Answer 1

Question

Chapter B, Problem 10P

To determine

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

Expert Solution & Answer

Answer to Problem 10P

The inverse of the matrix is A−1=[0.48180.05450.28180.34550.0545−0.1636−0.3455−0.03640.2818−0.3455−0.11820.14550.3455−0.03640.14550.4364]_.

Explanation of Solution

Given information:

A=[420−323−400−42−1−30−15]

Calculation:

Write the given matrix in augmented matrix form:

A=AI=[420−323−400−42−1−30−15|1000010000100001]

Divide the first row by 4.

A=[1120−3423−400−42−1−30−15|14000010000100001]

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

A=[1120−3423−400−42−1−30−15|14000010000100001]R2→R2−2R1=[1120−342−2×13−2×12−40−2×(−34)0−42−1−30−15|140000−2×(14)10000100001]=[1120−3402−4320−42−1−30−15|14000−1210000100001]

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

A=[1120−3402−4320−42−1−30−15|14000−1210000100001]R4→R4−(−3)R1=[1120−3402−4320−42−1−3−(−3×1)0−(−3×12)−15−(−3×(−34))|14000−1210000100−(−3×(14))001]=[1120−3402−4320−42−1032−1114|14000−12100001034001]

Divide the second row by 2.

A=[1120−3401−2340−42−1032−1114|14000−141200001034001]

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

A=[1120−3401−2340−42−1032−1114|14000−141200001034001]R1→R1−(12R2)=[112−(12×1)0−(12×(−2))−34−(12×34)01−2340−42−1032−1114|14−(12×(−14))0−(12×(12))00−141200001034001]=[101−9801−2340−42−1032−1114|38−1400−141200001034001]

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

A=[101−9801−2340−42−1032−1114|38−1400−141200001034001]R3→R3−(−4)R2=[101−9801−2340−4−(−4)×12−(−4)×(−2)−1−(−4)×(34)032−1114|38−1400−1412000−(−4)×(38)0−(−4)×(−14)1034001]=[101−9801−23400−62032−1114|38−1400−141200−121034001]

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

A=[101−9801−23400−62032−1114|38−1400−141200−121034001]R4→R4−(32)R2=[101−9801−23400−62032−(32)×1−1−(32)×(−2)114−(32)×(−34)|38−1400−141200−121034−(32)×(−14)0−(32)×(−12)01]=[101−9801−23400−62002138|38−1400−141200−121098−3401]

Divide the third row by –6.

A=[101−9801−234001−13002138|38−1400−14120016−13−16098−3401]

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

A=[101−9801−234001−13002138|38−1400−14120016−13−16098−3401]R1→R1−R3=[101−1−98−(−13)01−234001−13002138|38−(16)−14−(−13)0−(−16)0−14120016−13−16098−3401]=[100−192401−234001−13002138|524112160−14120016−13−16098−3401]

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

A=[100−192401−234001−13002138|524112160−14120016−13−16098−3401]R2→R2−(−2R3)=[100−192401−2−(−2×1)34−(−2×(−13))001−13002138|524112160−14−(−2×(16))12−(−2×(12))0−(−2×(−16))016−13−16098−3401]=[100−1924010112001−13002138|524112160112−16−13016−13−16098−3401]

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

A=[100−1924010112001−13002138|524112160112−16−13016−13−16098−3401]R4→R4−(−1R3)=[100−1924010112001−13002−(−1×1)138−(−1×(−13))|524112160112−16−13016−13−16098−(−1×(16))−34−(−1×(−13))0−(−1×(−16))1]=[100−1924010112001−130005524|524112160112−16−13016−13−1601924−112131]

Divide the fourth row by 5524.

A=[100−1924010112001−130001|524112160112−16−13016−13−1601955−2558552455]

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

A=[100−1924010112001−130001|524112160112−16−13016−13−1601955−2558552455]R1→R1−(−1924R4)=[100−1924−(−1924×1)010112001−130001|524−(−1924×1955)112−(−1924×(−255))16−(−1924×855)0−(−1924×2455)112−16−13016−13−1601955−2558552455]=[1000010112001−130001|53110355311101955112−16−13016−13−1601955−2558552455]

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

A=[1000010112001−130001|53110355311101955112−16−13016−13−1601955−2558552455]R2→R2−(112R4)=[1000010112−(112×1)001−130001|53110355311101955112−(112×1955)−16−(112×(−255))−13−(112×855)0−(112×2455)16−13−1601955−2558552455]=[10000100001−130001|53110355311101955355−955−1955−25516−13−1601955−2558552455]

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

A=[10000100001−130001|53110355311101955355−955−1955−25516−13−1601955−2558552455]R3→R2−(−13R4)=[10000100001−13−(−13×1)0001|5311053110311101955355−955−1955−25516−(−13×1971)−13−(−13×(−271))−16−(−13×871)0−(−13×2471)1955−2558552455]=[1000010000100001|53110355311101955355−955−1955−25531110−1955−131108551955−2558552455]=[1000010000100001|0.48180.05450.28180.34550.0545−0.1636−0.3455−0.03640.2818−0.3455−0.11820.14550.3455−0.03640.14550.4364]=[I|A−1]

Thus, the inverse of the matrix is A−1=[0.48180.05450.28180.34550.0545−0.1636−0.3455−0.03640.2818−0.3455−0.11820.14550.3455−0.03640.14550.4364]_.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

The floor system of an apartment building consists of a 4-inch-thick reinforced concrete slab resting on three steel floor beams, which are in turn supported by two steel girders, as shown below. The areas of cross section of the floor beams and the girders are 18.3 in.2 and 32.7 in.², respectively. Determine the dead load acting on the beam CD. Steel girder (A-32.7 in.2) At I- 25 ft B I Steel column C D Steel floor beam (A =18.3 in.2) I a. 600.0 lb/ft b. 662.3 lb/ft c. 62.3 lb/ft d. 1347.3 lb/ft E 4 in. concrete slab 2 at 12 ft = 24 ft

What is the classification of the structure shown below? Hinge a. Internally unstable, statically indeterminate b. Internally stable, statically determinate c. Internally stable, statically indeterminate Od. Internally unstable, statically determinate

ship construction question. Sketch a bilge keel, garboard strake and sheer strake.

Answer 2