Final answer:
To find the solution to the system of linear equations using Gaussian elimination or Gauss-Jordan elimination, start by writing the augmented matrix of the given system. Then, perform row operations to transform the matrix into row-echelon form or reduced row-echelon form. Finally, use back substitution to find the values of the variables and determine the unique solution.
Step-by-step explanation:
To find the solution to the system of linear equations using Gaussian elimination or Gauss-Jordan elimination, we start by writing the augmented matrix of the given system:
[1 1 6 | 3]
[1 1 3 | 3]
[1 2 4 | 7]
Now, we perform row operations to transform the matrix into row-echelon form or reduced row-echelon form:
Step 1: Subtract the first row from the second row:
[1 1 6 | 3]
[0 1 -3 | 0]
[1 2 4 | 7]
Step 2: Subtract the first row from the third row:
[1 1 6 | 3]
[0 1 -3 | 0]
[0 1 -2 | 4]
Step 3: Subtract twice the second row from the third row:
[1 1 6 | 3]
[0 1 -3 | 0]
[0 0 4 | 4]
Now, we can use back substitution to find the values of x, y, and z:
Step 1: From the third row, we get 4z = 4, which gives z = 1.
Step 2: Substituting z = 1 into the second row, we get y - 3(1) = 0, which gives y = 3.
Step 3: Substituting z = 1 and y = 3 into the first row, we get x + 3 + 6(1) = 3, which gives x = -6.
Therefore, the unique solution to the system of linear equations is x = -6, y = 3, and z = 1.