[3 2 6
1 1 2
2 2 5] find the inverse of matrix by row transformation
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Step-by-step explanation:
Given: [3 2 6
1 1 2
2 2 5] find the inverse of matrix
- So A = 3 2 6
- 1 1 2
- 2 2 5
- mod A = 3 (5 – 4) – 2(5 – 4) + 6 (2 – 2)
- = 3 – 2 + 0
- = 1 does not belong to 0
- So A^-1 exist
- [ A ij] = (- 1) ^I + j M ij
- Now A11 = (- 1)^1 + 1 M11
- 1 2
- 2 5
- 5 – 4 = 1
- A12 = (- 1)^1 + 2 M12 = (- 1) 1 2
- 2 5
- = - 1 (5 – 4)
- = - 1
- A 13 = (- 1)^1 + 3 M 13 = (1 ) 1 1
- 2 2
- = 2 – 2
- = 0
- A 21 = (- 1) ^2 + 1 M 21 = (- 1)^3 2 6
- 2 5
- = - (10 – 12)
- = 2
- A 22 = (- 1)^2 + 2 M22 = (- 1)^4 3 6
- 2 5
- = 15 – 12
- = 3
- A 23 = (- 1)^2 + 3 M 23 = - 3 2
- 2 2
- = - (6 – 4)
- = - 2
- A 31 = (- 1)^3 + 1 M 31 = 2 6
- 1 2
- = (4 – 6)
- = - 2
- A 32 = (- 1)^3 + 2 M 32 = (- 1) 3 6
- 1 2
- = - (6 – 6)
- = 0
- A 33 = (- 1)^3 + 3 M 33 = 3 2
- 1 1
- = (3 – 2)
- = 1
- So factor Matrix = 1 - 1 0
- 2 3 - 2
- -2 0 1
- Therefore (factor matrix)^T = adjoint A
- Adjoint A = 1 2 - 2
- -1 3 0
- 0 -2 1
- Or A^-1 = 1/mod A x Adjoint A
- = 1/1 x Adj A
- Or A^-1 = 1 2 - 2
- -1 3 0
- 0 -2 1
Reference link will be
https://brainly.in/question/40390636
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