Start with the linear system
where is a coefficient. Organize the linear system by variable for columns and equations for rows:
Now drop the variables and operators and draw matrix brackets around the values:
Example: Start with the linear system
There are row operations that can be done on a matrix that, if done correctly, will help transform the matrix into a solution to the linear system represented by the matrix. These operations are:
Scalar multiplication of a row is like scalar multiplication of the entire matrix, but it is done to only one row. In the following example, row 1 is multiplied by 2.
When adding rows, corresponding elements of each row are added together. The results are placed into either of the rows being added. In the following example, row 1 is added to row 2 and the result placed into row 2.
Any two rows in the matrix can be swapped without changing the linear system represented by the matrix. While row swapping is not necessary for solving this sample system, it is demonstrated here by swapping rows 2 and 3.
Any of the operations can be combined into a single transformation of the matrix. Be careful though, the more complicated combinations of transformations can lead to errors. Any error will give false results. In the next example, row 1 is multiplied by negative 1 and added to row 2.
Each of the steps in this algorithm are completed using the row operations of scalar multiplication of rows, addition of rows, and swapping rows.
The first step in solving a linear equation using a matrix is to get all zeros below the diagonal. The order in which these are conventionally done is illustrated here:
The second step is to transform the matrix so that the coefficients in the diagonal are all ones. This is illustrated by the following matrix.
The third step is to transform the matrix so entries above the diagonal are all zeros. The following matrix illustrates this step.
Divide both row 1 and row 2 by 2. This makes them easier to work with.
Multiply row 2 by 3 and add it to row 3.
Multiply row 3 by -1.
Multiply row 3 by -2 and add it to row 2 with the result in row 2.
Multiply row 3 by 2 and add it to row 1 with the result in row 1.
Read the solution from the matrix.
Exactly one solution: A linear system with exactly one solution will have a diagonal of ones, and the rest zeros except for the rightmost column:
Infinite solutions: A linear system with infinite solutions will have a row of all zeros, and can not be reduced to reduced eschelon form:
No solution: A linear system with no solution will have a row with all zeros except for the last entry. This is the mathematical statement 0=7, which is impossible. So there is no solution.
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