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
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