91Ó°ÊÓ

A matrix in reduced row echelon form is used to solve systems of linear equations. There are four prerequisites for the reduced row echelon form:

• The number 1 is the first non-zero integer in the first row (the leading entry).
• The second row begins with the number 1, which is more to the right than the first row's leading item. The number 1 must be further to the right in each consecutive row.
• Each row's first item must be the sole non-zero number in its column.
• Any rows that are not zero are pushed to the bottom of the matrix.

A linear system of 3 equations is,

$\left[\begin{array}{c}{a}_{1}x+b_{1}y+c_{1}z=0\\ a_{2}x+b_{2}y+c_{2}z=0\\ a_{3}x+b_{3}y+c_{3}z=0\\ a_{4}x+b_{4}y+c_{4}z=0\end{array}\right]$

The matrix form of the linear system of equations is,

$\left[\begin{array}{ccc}{a}_1& b_1& c_1\\ a_2& b_2& c_2\\ a_3& b_3& c_3\\ a_4& b_4& c_4\end{array}\right]$

The reduced row-echelon form of the matrix is,

$\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\\ 0& 0& 0\end{array}\right]$

The last row must be zero. There are no free variables present in the matrix, thus, the system will have unique solution.

$\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\\ 0& 0& 0\end{array}\right]$ is the reduced row-echelon form of the linear system of 4 equations with 3 unknowns.