Using a row operation, compute the inverse of the matrix A= ( 1 1 1

Question

0 2 3
5 5 1)

Answer 1

To compute the inverse of matrix A using row operations, we start by augmenting A with the identity matrix of the same size.

Augmented Matrix:
[ 1 1 1 | 1 0 0 ]
[ 0 2 3 | 0 1 0 ]
[ 5 5 1 | 0 0 1 ]

Next, we perform row operations to transform the left-hand side of the augmented matrix into the identity matrix. The right-hand side will then become the inverse of A.

1. R2 = R2 - 2R1:
[ 1 1 1 | 1 0 0 ]
[ 0 0 1 | -2 1 0 ]
[ 5 5 1 | 0 0 1 ]

2. R3 = R3 - 5R1:
[ 1 1 1 | 1 0 0 ]
[ 0 0 1 | -2 1 0 ]
[ 0 0 -4 | -5 0 1 ]

3. R3 = -R3/4:
[ 1 1 1 | 1 0 0 ]
[ 0 0 1 | -2 1 0 ]
[ 0 0 1 | 5/4 0 -1/4 ]

4. R1 = R1 - R3 and R2 = R2 - R3:
[ 1 1 0 | -1/4 0 1/4 ]
[ 0 0 0 | -3/4 1/4 1/4 ]
[ 0 0 1 | 5/4 0 -1/4 ]

5. R2 = -4/3R2:
[ 1 1 0 | -1/4 0 1/4 ]
[ 0 0 0 | 1 -1/3 -1/3 ]
[ 0 0 1 | 5/4 0 -1/4 ]

6. R1 = R1 - R2:
[ 1 1 0 | -5/12 1/3 5/12 ]
[ 0 0 0 | 1 -1/3 -1/3 ]
[ 0 0 1 | 5/4 0 -1/4 ]

7. R1 = -R1 and R2 = -R2:
[ -1 -1 0 | 5/12 -1/3 -5/12 ]
[ 0 0 0 | -1 1/3 1/3 ]
[ 0 0 1 | 5/4 0 -1/4 ]

8. R1 = -1/2R1:
[ 1 1 0 | -5/12 1/3 5/12 ]
[ 0 0 0 | -1 1/3 1/3 ]
[ 0 0 1 | 5/4 0 -1/4 ]

9. R1 = R1 + R3 and R2 = R2 - R3:
[ 1 1 0 | 0 1/3 3/6 ]
[ 0 0 0 | -1 1/3 1/3 ]
[ 0 0 1 | 5/4 0 -1/4 ]

10. R1 = R1 - R2:
[ 1 1 0 | 1 0 1/6 ]
[ 0 0 0 | -1 1/3 1/3 ]
[ 0 0 1 | 5/4 0 -1/4 ]

Therefore, the inverse of matrix A is:
[ 1 0 1/6 ]
[ -1 1/3 1/3 ]
[ 5/4 0 -1/4 ]