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For the following augmented matrix, write the system of equations it represents.
The system is readily obtained as below.
Once a system is expressed as an augmented matrix, the Gauss-Jordan method reduces the system into a series of equivalent systems by employing the row operations. This row reduction continues until the system is expressed in what is called the reduced row echelon form . The reduced row echelon form of the coefficient matrix has 1's along the main diagonal and zeros elsewhere. The solution is readily obtained from this form.
The method is not much different form the algebraic operations we employed in the elimination method in the first chapter. The basic difference is that it is algorithmic in nature, and, therefore, can easily be programmed on a computer.
We will next solve a system of two equations with two unknowns, using the elimination method, and then show that the method is analogous to the Gauss-Jordan method.
Solve the following system by the elimination method.
We multiply the first equation by – 3, and add it to the second equation.
By doing this we have transformed our original system into an equivalent system as follows.
We divide the second equation by – 5, and we get the next equivalent system.
Now we multiply the second equation by – 3 and add to the first, we get
Solve the following system from [link] by the Gauss-Jordan method, and show the similarities in both methods by writing the equations next to the matrices.
The augmented matrix for the system is as follows.
We multiply the first row by – 3, and add to the second row.
We divide the second row by – 5, we get,
Finally, we multiply the second row by – 3 and add to the first row, and we get,
Now we list the three row operations the Gauss-Jordan method employs.
One can easily see that these three row operation may make the system look different, but they do not change the solution of the system.
The first row operation states that if any two rows of a system are interchanged, the new system obtained has the same solution as the old one. Let us look at an example in two equations with two unknowns. Consider the system
We interchange the rows, and we get,
Clearly, this system has the same solution as the one above.
The second operation states that if a row is multiplied by any non-zero constant, the new system obtained has the same solution as the old one. Consider the above system again,
We multiply the first row by –3, we get,
Again, it is obvious that this new system has the same solution as the original.
The third row operation states that any constant multiple of one row added to another preserves the solution. Consider our system,
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