9. Solve the following system of equation, using the Gauss - Jordan & Inverse method if it exists.

X1+2x2+3x3 = 6
2x1-3x2+2x3 =14
-3x1-x2+x3 = 2

1 answer

To solve this system of equations using the Gauss-Jordan method, we first write the augmented matrix:

| 1 2 3 | 6 |
| 2 -3 2 | 14 |
|-3 -1 1 | 2 |

Next, we perform row operations to transform the matrix into row echelon form:

R2 - 2R1 -> R2
R3 + 3R1 -> R3

| 1 2 3 | 6 |
| 0 -7 -4 | 2 |
| 0 5 10 | 20 |

Now, we perform row operations to further transform the matrix into reduced row echelon form:

R2 * (-1/7) -> R2
R3 * (1/5) -> R3

| 1 2 3 | 6 |
| 0 1 4/7 | -2/7 |
| 0 1 2 | 4 |

R3 - R2 -> R3

| 1 2 3 | 6 |
| 0 1 4/7 | -2/7 |
| 0 0 8/7 | 30/7 |

R3 * (7/8) -> R3

| 1 2 3 | 6 |
| 0 1 4/7 | -2/7 |
| 0 0 1 | 15/2 |

R1 - 3R3 -> R1
R2 - (4/7)R3 -> R2

| 1 2 0 | -6 |
| 0 1 0 | -19/2 |
| 0 0 1 | 15/2 |

R1 - 2R2 -> R1

| 1 0 0 | 7/2 |
| 0 1 0 | -19/2 |
| 0 0 1 | 15/2 |

The reduced row echelon form of the augmented matrix shows that:

x1 = 7/2
x2 = -19/2
x3 = 15/2

Therefore, the solution to the system of equations is:

x1 = 7/2
x2 = -19/2
x3 = 15/2
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