Question 3.6.5: Let A = [1 2 1 2 4 4]. Find a sequence of elementary matrice...

We row reduce A keeping track of our elementary row operations:

\left [ \begin{matrix} 1 & 2 & 1 \\ 2 & 4 & 4 \end{matrix} \right ] \begin{matrix} \\ R_{2} – 2R_{1}\end{matrix} \sim \left [ \begin{matrix} 1 & 2 & 1 \\ 0 & 0 & 2 \end{matrix} \right ] \begin{matrix} \\ \frac{1}{2}R_{2} \end{matrix} \sim \left [ \begin{matrix} 1 & 2 & 1 \\ 0 & 0 & 1 \end{matrix} \right ] \begin{matrix} R_{1} – R_{2} \\ \\ \end{matrix} \sim \left [ \begin{matrix} 1 & 2 & 0 \\ 0 & 0 & 1 \end{matrix} \right ]

The first elementary row operation is R_{2} – 2R_{1}, so E_{1} = \left [ \begin{matrix} 1 & 0 \\ -2 & 1 \end{matrix} \right ].

The second elementary row operation is \frac{1}{2}R_{2}, so E_{2} = \left [ \begin{matrix} 1 & 0 \\ 0 & 1/2 \end{matrix} \right ] .

The third elementary row operation is R_{1} – R_{2}, so E_{3} = \left [ \begin{matrix} 1 & -1 \\ 0 & 1 \end{matrix} \right ] .

Thus, E_{3} E_{2} E_{1} A= \left [ \begin{matrix} 1 & -1 \\ 0 & 1 \end{matrix} \right ] \left [ \begin{matrix} 1 & 0 \\ 0 & 1/2 \end{matrix} \right ]\left [ \begin{matrix} 1 & 0 \\ -2 & 1 \end{matrix} \right ]\left [ \begin{matrix} 1 & 2 & 1 \\ 2 & 4 & 4 \end{matrix} \right ] = \left [ \begin{matrix} 1 & 2 & 0 \\ 0 & 0 & 1 \end{matrix} \right ] .