The system of equations can be written as a matrix and then used to solve for the variables x, y, and z.
The augmented matrix of the system is:
-3 -2 1 | -8
0 4 -2 | -8
0 0 7 | 26
We can use row operations to find the row reduced echelon form of the matrix, which will give us the solutions to the system.
Starting with the first row:
1. Multiply row 1 by -3 to get a leading 1 in the first column:
3 6 -3 | 24
0 4 -2 | -8
0 0 7 | 26
2. Add row 1 to row 2:
3 6 -3 | 24
0 4 0 | 16
0 0 7 | 26
3. Add 3 times row 1 to row 3:
3 6 -3 | 24
0 4 0 | 16
0 0 7 | 26
The row reduced echelon form of the matrix is:
1 2 -1 | 8
0 1 0 | 4
0 0 1 | 9
From this, we can see that the solutions to the system are:
x = 8 - 2y
y = 4
z = 9
So the system has a unique solution: x = 8 - 2y = 8 - 2 * 4 = 0, y = 4, z = 9.