Solve the system of equations using matrices. Use Gaussian elimination with back- substitution.
x+y+z = -5
x-y+3z = -1
4x+y+z = -2
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To solve the system of equations using matrices and Gaussian elimination with back-substitution, we will first write the system of equations in matrix form.
The given system of equations can be written as:
[1 1 1][x] = [-5]
[1 -1 3][y] = [-1]
[4 1 1][z] = [-2]
We will perform Gaussian elimination to transform the coefficient matrix into row-echelon form. Here are the steps involved:
Step 1: Write the augmented matrix.
We represent the coefficients of the equations in matrix form, appending the constants on the right side of the vertical line:
[1 1 1 | -5]
[1 -1 3 | -1]
[4 1 1 | -2]
Step 2: Perform row operations to eliminate the variables below the main diagonal.
We want to create zeros below the main diagonal by performing row operations.
- Add (-1) times the first row to the second row.
- Subtract 4 times the first row from the third row.
The matrix becomes:
[1 1 1 | -5]
[0 -2 2 | 4]
[0 -3 -3 | 18]
Step 3: Perform row operations to create zeros above the main diagonal.
- Multiply the second row by (-1/2).
- Add 3 times the second row to the third row.
The matrix becomes:
[1 1 1 | -5]
[0 1 -1 | -2]
[0 0 0 | 12]
Step 4: Solve for the variables using back-substitution.
From the third row, we obtain the equation 0z = 12, which means 0 = 12. This indicates that there is no solution to the system of equations.
Therefore, the given system of equations is inconsistent (no solution).