To solve the given system of equations through triangularization, we start with the following equations:
- 1) 3x + y + 6z = 28
- 2) 2x + 4y + 4z = 8
- 3) 2x + 3y + 4z = 4
We begin by writing the augmented matrix for this system:
Augmented Matrix:
| 3 1 6 | 28 |
| 2 4 4 | 8 |
| 2 3 4 | 4 |
We will perform row operations to convert this matrix into an upper triangular form.
Step 1: We can start by making the first column below the first row zero. To do this, we can subtract a suitable multiple of the first row from the second and third rows.
Let’s make the first element of the second row zero:
- Row2 = Row2 – (2/3) * Row1
This leads to:
- Row 2 becomes:
- 2 – (2/3)(3) = 0,
- 4 – (2/3)(1) = 10/3,
- 4 – (2/3)(6) = -8/3,
- 8 – (2/3)(28) = -40/3.
Now we perform the same operation for Row3:
- Row3 = Row3 – (2/3) * Row1
After computation, the new augmented matrix becomes:
| 3 1 6 | 28 |
| 0 10/3 -8/3 | -40/3 |
| 0 3/3 -8/3 | -20/3 |
Step 2: Now let’s simplify Row2 to make it easier to handle:
- Multiply Row2 by 3/10:
After this simplification, we get:
| 3 1 6 | 28 |
| 0 1 -8/10 | -12 |
| 0 1 -8/3 | -20/3 |
Now we can eliminate the y variable from Row 3 to create an upward triangular form:
- Row 3 = Row 3 – Row 2
By calculating this, we obtain:
| 3 1 6 | 28 |
| 0 1 -0.8 | -12 |
| 0 0 -1.6 | -8 |
Step 3: At this point, we perform back substitution. From Row 3, we have:
-1.6z = -8
Thus, z = 5.
Substituting z back into Row 2:
y – 0.8(5) = -12
y – 4 = -12
y = -8.
Finally, substitute y and z back into the first equation:
3x + (-8) + 6(5) = 28
3x – 8 + 30 = 28
3x + 22 = 28
x = 2.
Thus, the solution to the system of equations is:
x = 2, y = -8, z = 5