Question 3.3.2: Find the reduced echelon form of the matrix A = [1 2 1 4 3 8...
Find the reduced echelon form of the matrix
A = \left [ \begin{matrix} 1 & 2 & 1 & 4 \\ 3 & 8 & 7 & 20 \\ 2 & 7 & 9 & 23\end{matrix} \right ].
In Example 3 of Section 3.2 we found the echelon form
\left [ \begin{matrix} 1 & 2 & 1 & 4 \\ 0 & 1 & 2 & 4 \\ 0 & 0 & 1 & 3\end{matrix} \right ] ,
which already satisfies Property 3. To clear out columns 2 and 3 (in order to satisfy Property 4), we continue the reduction as follows.
\left [ \begin{matrix} 1 & 2 & 1 & 4 \\ 0 & 1 & 2 & 4 \\ 0 & 0 & 1 & 3\end{matrix} \right ] \underrightarrow{(- 2) R_{2} + R_{1}} \left [ \begin{matrix} 1 & 0 & – 3 & -4 \\ 0 & 1 & 2 & 4 \\ 0 & 0 & 1 & 3\end{matrix} \right ]
\underrightarrow{(- 2) R_{3} + R_{2}} \left [ \begin{matrix} 1 & 0 & – 3 & -4 \\ 0 & 1 & 0 & – 2 \\ 0 & 0 & 1 & 3\end{matrix} \right ]
\underrightarrow{(3) R_{3} + R_{1}} \left [ \begin{matrix} 1 & 0 & 0 & 5 \\ 0 & 1 & 0 & – 2 \\ 0 & 0 & 1 & 3\end{matrix} \right ]
For instance, we see immediately from this reduced echelon form that the linear system
x + 2y + z = 4
3x + 8y + 7z = 20
2x + 7y + 9z = 23
with augmented coefficient matrix A has the unique solution x = 5, y = – 2, z = 3.