Section 1.3 Reduced row echelon form
The process of solving a linear system can be simplified by transforming its augmented matrix into a more manageable form known as the reduced row echelon form (RREF). This form allows us to easily identify the solutions to the system, whether they are unique, infinite, or nonexistent. The RREF is achieved through a series of elementary row operations that systematically eliminate variables and simplify the equations represented by the matrix. By converting the augmented matrix into RREF, we can directly read off the solutions or determine the relationships between variables, making it a powerful tool in linear algebra for solving systems of equations efficiently and effectively.
Exercise: Find the reduced row echelon form of the matrices \(A=\begin{pmatrix}
0 \amp 2 \amp -8 \amp 8 \\
1 \amp -2 \amp 1 \amp 0 \\
5 \amp 0 \amp -5 \amp 10
\end{pmatrix}\) and \(B=\begin{pmatrix}
0 \amp 3 \amp -6 \amp 6 \amp 4 \amp -5 \\
3 \amp -7 \amp 8 \amp -5 \amp 8 \amp 9 \\
3 \amp -9 \amp 12 \amp -9 \amp 6 \amp 15
\end{pmatrix}\text{,}\) respectively.
If the matrix \(C\) is an augmented matrix of a linear system, it can be interpreted as
\begin{align*}
x_1\amp=1\\
x_2\amp=0\\
x_3\amp=-1
\end{align*}
Input the matrix \(B\) below, and find the reduced echelon form of \(B\text{.}\) In next section, we will find the general solution of a linear system whose augmented matrix is \(B\)
