Final answer:
To put the system of equations in matrix form, we rewrite it as an augmented matrix and perform row operations to transform it into row-echelon form. Then, we use back substitution to find the values of X₁, X₂, and X₃. The solution is X₁=1/7, X₂=-3/7, and X₃=8/7.
Step-by-step explanation:
To put the system of equations in matrix form, we can rewrite the equations as follows:



This can be represented as:
| 3 -1 1 | | X₁ | | 3 |
| 1 3 1 | x | X₂ | = | 0 |
| 1 1 3 | | X₃ | | 1 |
To solve this system, we can use row operations to transform the augmented matrix into row-echelon form, and then use back substitution to find the values of X₁, X₂, and X₃.
The step-by-step process would involve:
- Multiply the first row by 1/3 to make the leading coefficient of the first row equal to 1.
- Add -1 times the first row to the second row to eliminate the X₁ term.
- Add -1 times the first row to the third row to eliminate the X₁ term.
- Multiply the second row by 1/4 to make the leading coefficient of the second row equal to 1.
- Add -1/2 times the second row to the third row to eliminate the X₂ term.
- Multiply the third row by 2/7 to make the leading coefficient of the third row equal to 1.
- Back substitute to find the values of X₁, X₂, and X₃.
The solution to the system of equations is X₁ = 1/7, X₂=-3/7, and X₃=8/7.