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

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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]_.

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Answer 2