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Second, we may operate on a linear system transforming it into a new system that has the same solution space. For instance, given the system

\begin{equation*} \begin{alignedat}{4} -x \amp {} + {} \amp 2y \amp {}-{} \amp z \amp {}={} \amp -3 \\ \amp \amp 3y \amp {}+{} \amp z \amp {}={} \amp -1. \\ \amp \amp \amp \amp 2z \amp {}={} \amp 4, \\ \end{alignedat} \end{equation*}we may multiply the third equation by \(1/2\) to obtain

\begin{equation*} \begin{alignedat}{4} -x \amp {} + {} \amp 2y \amp {}-{} \amp z \amp {}={} \amp -3 \\ \amp \amp 3y \amp {}+{} \amp z \amp {}={} \amp -1. \\ \amp \amp \amp \amp z \amp {}={} \amp 2. \\ \end{alignedat} \end{equation*}Any solution to this system of equations must then have \(z=2\text{.}\)

Once we know that, we may substitute \(z=2\) into the first and second equation and simplify to obtain a new system of equations having the same solutions:

\begin{equation*} \begin{alignedat}{4} -x \amp {} + {} \amp 2y \amp {}{} \amp \amp {}={} \amp -1 \\ \amp \amp 3y \amp {}{} \amp \amp {}={} \amp -3. \\ \amp \amp \amp \amp z \amp {}={} \amp 2. \\ \end{alignedat} \end{equation*}Continuing in this way, we eventually obtain a decoupled system showing that there is exactly one solution, which is \((x,y,z)=(-1,-1,2)\text{.}\)

Our original system,

\begin{equation*} \begin{alignedat}{4} -x \amp {} + {} \amp 2y \amp {}-{} \amp z \amp {}={} \amp -3 \\ \amp \amp 3y \amp {}+{} \amp z \amp {}={} \amp -1 \\ \amp \amp \amp \amp 2z \amp {}={} \amp 4, \\ \end{alignedat} \end{equation*} is called a *triangular* system due to the shape formed by the coefficients. As this example demonstrates, triangular systems are easily solved by a process called *back substitution*.