In: Advanced Math
solve for matrix B
Let I be Identity matrix
(I-2B)-1=
1 | -3 | 3 |
-2 | 2 | -5 |
3 | -8 | 9 |
Matrix I is
A1 A2 A3
1 1 0 0
2 0 1 0
3 0 0 1
Matrix 2B is
B1 B2 B3
1 1 -6 6
2 -4 4 -10
3 6 -16 18
now on subtracting 2B from I ,we get following matrix
C1 C2 C3
1 0 6 -6
2 4 -3 10
3 -6 16 -17
now we have find inverse of above matrix
Your matrix is
A1 | A2 | A3 | |
---|---|---|---|
1 | 0 | 6 | -6 |
2 | 4 | -3 | 10 |
3 | -6 | 16 | -17 |
Determinant is not zero, therefore inverse matrix exists
Write the augmented matrix
A1 | A2 | A3 | B1 | B2 | B3 | |
---|---|---|---|---|---|---|
1 | 0 | 6 | -6 | 1 | 0 | 0 |
2 | 4 | -3 | 10 | 0 | 1 | 0 |
3 | -6 | 16 | -17 | 0 | 0 | 1 |
Make the pivot in the 1st column by dividing the 2nd row by 4 and swap the 2nd and the 2nd rows
A1 | A2 | A3 | B1 | B2 | B3 | |
---|---|---|---|---|---|---|
1 | 1 | -0.75 | 2.5 | 0 | 0.25 | 0 |
2 | 0 | 6 | -6 | 1 | 0 | 0 |
3 | -6 | 16 | -17 | 0 | 0 | 1 |
Multiply the 1st row by -6
A1 | A2 | A3 | B1 | B2 | B3 | |
---|---|---|---|---|---|---|
1 | -6 | 4.5 | -15 | 0 | -1.5 | 0 |
2 | 0 | 6 | -6 | 1 | 0 | 0 |
3 | -6 | 16 | -17 | 0 | 0 | 1 |
Subtract the 1st row from the 3rd row and restore it
A1 | A2 | A3 | B1 | B2 | B3 | |
---|---|---|---|---|---|---|
1 | 1 | -0.75 | 2.5 | 0 | 0.25 | 0 |
2 | 0 | 6 | -6 | 1 | 0 | 0 |
3 | 0 | 11.5 | -2 | 0 | 1.5 | 1 |
Make the pivot in the 2nd column by dividing the 2nd row by 6
A1 | A2 | A3 | B1 | B2 | B3 | |
---|---|---|---|---|---|---|
1 | 1 | -0.75 | 2.5 | 0 | 0.25 | 0 |
2 | 0 | 1 | -1 | 0.16666666666666666666 | 0 | 0 |
3 | 0 | 11.5 | -2 | 0 | 1.5 | 1 |
Multiply the 2nd row by -0.75
A1 | A2 | A3 | B1 | B2 | B3 | |
---|---|---|---|---|---|---|
1 | 1 | -0.75 | 2.5 | 0 | 0.25 | 0 |
2 | 0 | -0.75 | 0.75 | -0.125 | 0 | 0 |
3 | 0 | 11.5 | -2 | 0 | 1.5 | 1 |
Subtract the 2nd row from the 1st row and restore it
A1 | A2 | A3 | B1 | B2 | B3 | |
---|---|---|---|---|---|---|
1 | 1 | 0 | 1.75 | 0.125 | 0.25 | 0 |
2 | 0 | 1 | -1 | 0.16666666666666666666 | 0 | 0 |
3 | 0 | 11.5 | -2 | 0 | 1.5 | 1 |
Multiply the 2nd row by 11.5
A1 | A2 | A3 | B1 | B2 | B3 | |
---|---|---|---|---|---|---|
1 | 1 | 0 | 1.75 | 0.125 | 0.25 | 0 |
2 | 0 | 11.5 | -11.5 | 1.9166666666666666665 | 0 | 0 |
3 | 0 | 11.5 | -2 | 0 | 1.5 | 1 |
Subtract the 2nd row from the 3rd row and restore it
A1 | A2 | A3 | B1 | B2 | B3 | |
---|---|---|---|---|---|---|
1 | 1 | 0 | 1.75 | 0.125 | 0.25 | 0 |
2 | 0 | 1 | -1 | 0.16666666666666666666 | 0 | 0 |
3 | 0 | 0 | 9.5 | -1.9166666666666666665 | 1.5 | 1 |
Make the pivot in the 3rd column by dividing the 3rd row by 9.5
A1 | A2 | A3 | B1 | B2 | B3 | |
---|---|---|---|---|---|---|
1 | 1 | 0 | 1.75 | 0.125 | 0.25 | 0 |
2 | 0 | 1 | -1 | 0.16666666666666666666 | 0 | 0 |
3 | 0 | 0 | 1 | -0.20175438596491228068 | 0.15789473684210526315 | 0.1052631578947368421 |
Multiply the 3rd row by 1.75
A1 | A2 | A3 | B1 | B2 | B3 | |
---|---|---|---|---|---|---|
1 | 1 | 0 | 1.75 | 0.125 | 0.25 | 0 |
2 | 0 | 1 | -1 | 0.16666666666666666666 | 0 | 0 |
3 | 0 | 0 | 1.75 | -0.35307017543859649119 | 0.27631578947368421051 | 0.18421052631578947367 |
Subtract the 3rd row from the 1st row and restore it
A1 | A2 | A3 | B1 | B2 | B3 | |
---|---|---|---|---|---|---|
1 | 1 | 0 | 0 | 0.47807017543859649119 | -0.02631578947368421051 | -0.18421052631578947367 |
2 | 0 | 1 | -1 | 0.16666666666666666666 | 0 | 0 |
3 | 0 | 0 | 1 | -0.20175438596491228068 | 0.15789473684210526315 | 0.1052631578947368421 |
Multiply the 3rd row by -1
A1 | A2 | A3 | B1 | B2 | B3 | |
---|---|---|---|---|---|---|
1 | 1 | 0 | 0 | 0.47807017543859649119 | -0.02631578947368421051 | -0.18421052631578947367 |
2 | 0 | 1 | -1 | 0.16666666666666666666 | 0 | 0 |
3 | 0 | 0 | -1 | 0.20175438596491228068 | -0.15789473684210526315 | -0.1052631578947368421 |
Subtract the 3rd row from the 2nd row and restore it
A1 | A2 | A3 | B1 | B2 | B3 | |
---|---|---|---|---|---|---|
1 | 1 | 0 | 0 | 0.47807017543859649119 | -0.02631578947368421051 | -0.18421052631578947367 |
2 | 0 | 1 | 0 | -0.03508771929824561402 | 0.15789473684210526315 | 0.1052631578947368421 |
3 | 0 | 0 | 1 | -0.20175438596491228068 | 0.15789473684210526315 | 0.1052631578947368421 |
There is the inverse matrix on the right
A1 | A2 | A3 | B1 | B2 | B3 | |
---|---|---|---|---|---|---|
1 | 1 | 0 | 0 | 0.47807017543859649119 | -0.02631578947368421051 | -0.18421052631578947367 |
2 | 0 | 1 | 0 | -0.03508771929824561402 | 0.15789473684210526315 | 0.1052631578947368421 |
3 | 0 | 0 | 1 | -0.20175438596491228068 | 0.15789473684210526315 |
0.1052631578947368421 |
Result is
B1 | B2 | B3 | |
---|---|---|---|
1 | 0.47807017543859649119 | -0.02631578947368421051 | -0.18421052631578947367 |
2 | -0.03508771929824561402 | 0.15789473684210526315 | 0.1052631578947368421 |
3 | -0.20175438596491228068 | 0.15789473684210526315 |
0.1052631578947368421 |
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