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Question 1144115: find inverse of matrix by elementary transformation
2 2 1
1 0 3
1-4-4
Answer by ikleyn(52794)   (Show Source):
You can put this solution on YOUR website! .
Not every specialist, even if he (or she) knows the theory, is able to complete it by hands.
Fortunately, in our days (years) there is computer software for it, and in the Internet you can find
free of charge online solvers that solve the problem accompanying the solution with very detailed comments.
I used the online code/solver
https://matrix.reshish.com/inverCalculation.php
The solution by the solver is below.
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Your matrix
A1 A2 A3
1 2 2 1
2 1 0 3
3 1 -4 -4
Determinant is not zero, therefore inverse matrix exists
Very detailed solution
Write the augmented matrix
A1 A2 A3 B1 B2 B3
1 2 2 1 1 0 0
2 1 0 3 0 1 0
3 1 -4 -4 0 0 1
Find the pivot in the 1st column and swap the 2nd and the 1st rows
A1 A2 A3 B1 B2 B3
1 1 0 3 0 1 0
2 2 2 1 1 0 0
3 1 -4 -4 0 0 1
Multiply the 1st row by 2
A1 A2 A3 B1 B2 B3
1 2 0 6 0 2 0
2 2 2 1 1 0 0
3 1 -4 -4 0 0 1
Subtract the 1st row from the 2nd row and restore it
A1 A2 A3 B1 B2 B3
1 1 0 3 0 1 0
2 0 2 -5 1 -2 0
3 1 -4 -4 0 0 1
Subtract the 1st row from the 3rd row3-ะตะน
A1 A2 A3 B1 B2 B3
1 1 0 3 0 1 0
2 0 2 -5 1 -2 0
3 0 -4 -7 0 -1 1
Make the pivot in the 2nd column by dividing the 2nd row by 2
A1 A2 A3 B1 B2 B3
1 1 0 3 0 1 0
2 0 1 -5/2 1/2 -1 0
3 0 -4 -7 0 -1 1
Multiply the 2nd row by -4
A1 A2 A3 B1 B2 B3
1 1 0 3 0 1 0
2 0 -4 10 -2 4 0
3 0 -4 -7 0 -1 1
Subtract the 2nd row from the 3rd row and restore it
A1 A2 A3 B1 B2 B3
1 1 0 3 0 1 0
2 0 1 -5/2 1/2 -1 0
3 0 0 -17 2 -5 1
Make the pivot in the 3rd column by dividing the 3rd row by -17
A1 A2 A3 B1 B2 B3
1 1 0 3 0 1 0
2 0 1 -5/2 1/2 -1 0
3 0 0 1 -2/17 5/17 -1/17
Multiply the 3rd row by 3
A1 A2 A3 B1 B2 B3
1 1 0 3 0 1 0
2 0 1 -5/2 1/2 -1 0
3 0 0 3 -6/17 15/17 -3/17
Subtract the 3rd row from the 1st row and restore it
A1 A2 A3 B1 B2 B3
1 1 0 0 6/17 2/17 3/17
2 0 1 -5/2 1/2 -1 0
3 0 0 1 -2/17 5/17 -1/17
Multiply the 3rd row by -5/2
A1 A2 A3 B1 B2 B3
1 1 0 0 6/17 2/17 3/17
2 0 1 -5/2 1/2 -1 0
3 0 0 -5/2 5/17 -25/34 5/34
Subtract the 3rd row from the 2nd row and restore it
A1 A2 A3 B1 B2 B3
1 1 0 0 6/17 2/17 3/17
2 0 1 0 7/34 -9/34 -5/34
3 0 0 1 -2/17 5/17 -1/17
There is the inverse matrix on the right
A1 A2 A3 B1 B2 B3
1 1 0 0 6/17 2/17 3/17
2 0 1 0 7/34 -9/34 -5/34
3 0 0 1 -2/17 5/17 -1/17
Result:
B1 B2 B3
1 6/17 2/17 3/17
2 7/34 -9/34 -5/34
3 -2/17 5/17 -1/17
Solved.
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