Linear Algebra And Optimization

Q2.A] ANSWER:-

Q2 A

To find the inverse of matrix A using the Gauss-Jordan method, we first augment matrix A with the identity matrix I.

| 1 2 3 | | 1 0 0 |
| 2 5 5 | | 0 1 0 |
| 3 7 8 | | 0 0 1 |

We then perform the following row operations:

1. Subtract 2 times the first row from the second row.
2. Subtract 3 times the first row from the third row.

| 1 2 3 | | 1 0 0 |
| 0 1 1 | | -2 1 0 |
| 0 -1 -5 | | -3 0 1 |

3. Multiply the second row by -1.
4. Add the second row to the third row.

| 1 2 3 | | 1 0 0 |
| 0 -1 -1 | | 2 -1 0 |
| 0 0 4 | | -5 0 1 |

5. Divide the third row by 4.

| 1 2 3 | | 1 0 0 |
| 0 -1 -1 | | 2 -1 0 |
| 0 0 1 | | -5/4 0 1/4 |

6. Add the third row to the second row.

| 1 2 3 | | 1 0 0 |
| 0 1 -2 | | 2 -1 1/4 |
| 0 0 1 | | -5/4 0 1/4 |

7. Subtract 2 times the third row from the second row.
8. Subtract 3 times the third row from the first row.

| 1 0 7/4 | | 1 0 0 |
| 0 1 0 | | 13/4 -1 1 |
| 0 0 1 | | -5/4 0 1/4 |

The inverse of matrix A is now the right-hand side of the augmented matrix.

| 1 0 7/4 |
| 13/4 -1 1 |
| -5/4 0 1/4 |

Therefore, the answer is:

A^{-1} = \begin{pmatrix}
1 & 0 & 7/4 \\
13/4 & -1 & 1 \\
-5/4 & 0 & 1/4
\end{pmatrix}

C]ANSWER:-