When using Gaussian elimination to solve a system of linear equations, explain how you can recognize that the system has no solution.
man, this has been an agonizingly slow homework dump. Absolutely no work shown on your part, and general questions that indicate a total lack of preparation or study.
When using Gaussian elimination to solve a system of linear equations, you can recognize that the system has no solution by observing the reduced row echelon form (RREF) of the augmented matrix.
Gaussian elimination involves manipulating the augmented matrix representing the system of equations by performing row operations. The objective is to transform the matrix into a specific form known as RREF. Once in RREF, you can easily determine if the system has no solution.
To perform Gaussian elimination and identify if the system has no solution, follow these steps:
1. Write the system of linear equations in the form of an augmented matrix. This matrix consists of the coefficients of the variables on the left and the constants on the right.
For example, if you have the following system of equations:
2x + 3y = 5
4x - 6y = 10
The augmented matrix would be:
[2 3 | 5]
[4 -6 | 10]
2. Apply row operations to transform the augmented matrix into row echelon form (REF). These operations include swapping rows, multiplying rows by a constant, and adding/subtracting rows. The goal is to get zeros below the diagonal entries.
Using the example above, after performing the row operations, we get:
[2 3 | 5]
[0 -12 | 0]
3. Continue applying row operations to convert the matrix into reduced row echelon form (RREF). The goal is to obtain leading 1's in each row, with zeros above and below them.
In our example, to convert the matrix to RREF, we divide the second row by -12, resulting in:
[2 3 | 5]
[0 1 | 0]
4. Examine the RREF form of the augmented matrix. If there is a row of the form [0 0 ... 0 | c], where c is a non-zero constant, then the system has no solution. This occurs when you have contradictory equations.
In our example, there are no contradictory equations, so there is a solution for x and y. However, if the RREF were:
[2 3 | 5]
[0 1 | 0]
[0 0 | 7]
We would conclude that the system has no solution since the third row has a non-zero constant, which implies 0 = 7.
In summary, you can recognize that a system of linear equations has no solution by checking the RREF form of the augmented matrix. If there is a row with all zeros in the coefficients and a non-zero constant on the right, it indicates contradictory equations, and the system has no solution.