- #1
karush
Gold Member
MHB
- 3,269
- 5
| $\left[ \begin{array}{rrrr|r} 1& -5& 4& 0&0\\ 0& 1& 0& 1&0\\ 0& 0& 3& 0&0\\ 0& 0& 0& 2&0 \end{array}\right] $ OK my first move on this is $r_3/3$ and $r_4/2$.
|
Last edited:
| $\left[ \begin{array}{rrrr|r} 1& -5& 4& 0&0\\ 0& 1& 0& 1&0\\ 0& 0& 3& 0&0\\ 0& 0& 0& 2&0 \end{array}\right] $ OK my first move on this is $r_3/3$ and $r_4/2$.
|
Row operations are used to simplify and solve systems of linear equations represented in a matrix form. By performing row operations, we can manipulate the matrix in order to find the solution to the system of equations.
The three types of row operations are: multiplying a row by a non-zero constant, swapping two rows, and adding a multiple of one row to another row. These operations do not change the solution of the system of equations.
A complete augmented matrix is in reduced row-echelon form when it satisfies the following conditions: all rows with all zero entries are at the bottom, the first non-zero entry in each row (called a leading entry) is a 1, each leading entry is the only non-zero entry in its column, and the leading entry in each row is to the right of the leading entry in the row above it.
No, row operations do not change the solution of a system of equations. They simply rearrange the matrix in order to make it easier to find the solution.
By performing row operations, we can transform the matrix into reduced row-echelon form. Once the matrix is in this form, we can easily read off the solution to the system of equations. The variables will correspond to the columns without leading entries, and the constants will be the values in the last column.