Question 2.2.4: Solve the following system of equations by row reducing the ...

Our first pivot is already a leading one, so we place zeros beneath it.

\left [ \begin{matrix} 1 & 1 & 0 \\ 2 & 4 & 1 \\ 1 & 2 & 1 \end{matrix} \left | \begin{matrix} -7 \\ -16 \\ 9 \end{matrix} \right.\right ] \begin{matrix} \\ R_{2} – 2R_{1} \\ R_{3} – R_{1} \end{matrix} \thicksim \left [ \begin{matrix} 1 & 1 & 0 \\ 0 & 2 & 1 \\ 0 & 1 & 1 \end{matrix} \left | \begin{matrix} -7 \\ -2 \\ 16 \end{matrix} \right.\right ]

To make our next pivot a leading one, rather than introducing fractions, we use R_{2} – R_{3}.

\left [ \begin{matrix} 1 & 1 & 0 \\ 0 & 2 & 1 \\ 0 & 1 & 1 \end{matrix} \left | \begin{matrix} -7 \\ -2 \\ 16 \end{matrix} \right.\right ] \begin{matrix} \\ R_{2} – R_{3} \\ \\ \end{matrix} \thicksim \left [ \begin{matrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 1 & 1 \end{matrix} \left | \begin{matrix} -7 \\ -18 \\ 16 \end{matrix} \right.\right ]

We now need to get zeros above and below this new leading one.

\left [ \begin{matrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 1 & 1 \end{matrix} \left | \begin{matrix} -7 \\ -18 \\ 16 \end{matrix} \right.\right ] \begin{matrix} R_{1} – R_{2} \\ \\ R_{3} – R_{2} \end{matrix} \thicksim \left [ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \left | \begin{matrix} 11 \\ -18 \\ 34 \end{matrix} \right.\right ]

The matrix is now in reduced row echelon form. The reduced row echelon form corresponds to the system x_{1} = 11, x_{2} = -18, x_{3} = 34. Hence, the solution is \vec{x} = \left [ \begin{matrix} 11 \\ -18 \\ 34 \end{matrix} \right ].