When using Gaussian elimination to solve a system of linear equations, explain how you can recognize that the system has no solution. Give an example that illustrates your answer.
1 x +1 y = 2
1 x +1 y = 3
1 1 2
1 1 3 subtract top one from this
1 1 2
0 0 1 oh my! 0x+0y=1 ?
To recognize that a system of linear equations has no solution, you need to identify a contradiction or inconsistency during the process of Gaussian elimination. Here's a step-by-step explanation:
1. Write down the augmented matrix of the system, which combines the coefficients and constant terms of the variables.
2. Apply row operations to the augmented matrix to reduce it to its row echelon form or reduced row echelon form.
3. Look for any row with all zeros (except for the right-most column). This row represents an equation like 0 = constant. This equation is a contradiction, as it implies that there is no possible solution.
4. If the last row in the reduced row echelon form is a row of zeros, it indicates an underdetermined system with infinitely many solutions.
5. If there are no contradictions or all-zero rows but there are additional variables (free variables) remaining after the elimination process, it implies that the system has infinitely many solutions.
6. If none of the above conditions are met, and there are no contradictions, the system has a unique solution.
Example:
Consider the following system of linear equations:
2x + 3y - z = 4 (Equation 1)
4x - y + 2z = 6 (Equation 2)
6x - y + 2z = 8 (Equation 3)
By performing Gaussian elimination, we can set up the augmented matrix and apply row operations to reduce it:
[ 2 3 -1 | 4 ]
[ 4 -1 2 | 6 ]
[ 6 -1 2 | 8 ]
Performing row operations, we can transform it into row echelon form:
[ 2 3 -1 | 4 ]
[ 0 -7 4 | -2 ]
[ 0 0 0 | 0 ]
From the third row, we notice that all variables become zero, but the constant term is non-zero (0 ≠ 0). Hence, we have a contradiction. Therefore, this system of linear equations has no solution.
To recognize that a system of linear equations has no solution when using Gaussian elimination, you need to identify an inconsistency in the system. Gaussian elimination involves row operations to transform the augmented matrix of the system into row-echelon form. If, during this procedure, you encounter a row of the form [0 0 0 ... 0 | c], where c is nonzero, it implies that the system is inconsistent and has no solution.
Here's an example to illustrate this:
Consider the following system of equations:
1) x + 2y + 3z = 6
2) 2x + y + z = 4
3) 3x + 2y + 2z = 7
First, we write the augmented matrix for the system:
[ 1 2 3 | 6 ]
[ 2 1 1 | 4 ]
[ 3 2 2 | 7 ]
Next, we perform row operations to convert the matrix into row-echelon form. I'll show the intermediate steps:
[ 1 2 3 | 6 ] (R2 = R2 - 2R1)
[ 0 -3 -5 | -8 ]
[ 3 2 2 | 7 ] (R3 = R3 - 3R1)
[ 1 2 3 | 6 ] (R3 = R3 - 3R1)
[ 0 -3 -5 | -8 ]
[ 0 -4 -7 | -11 ] (R3 = R3 - R2)
At this point, we can observe that the last row of the augmented matrix is of the form [0 0 0 | c], where c = -11. Since c is nonzero (-11), it indicates that the system of equations is inconsistent and has no solution.
Therefore, in this example, the system of equations has no solution because the matrix obtained during Gaussian elimination contains a row where all the coefficients are zero, except for the constant term.