Example 1.2.2

Let's illustrate the use of these operations to find the solution space to the system of equations:

\begin{equation*} \begin{alignedat}{4} x \amp {}+{} \amp 2y \amp \amp \amp {}={} \amp 4 \\ 2x \amp {}+{} \amp y \amp {}-{} \amp 3z \amp {}={} \amp 11 \\ -3x \amp {}-{} \amp 2y \amp {}+{} \amp z \amp {}={} \amp -10 \\ \end{alignedat} \end{equation*}

We will first transform the system into a triangular system so we start by eliminating \(x\) from the second and third equations.

We begin with a replacement operation where we multiply the first equation by -2 and add the result to the second equation.

\begin{equation*} \begin{alignedat}{4} x \amp {}+{} \amp 2y \amp \amp \amp {}={} \amp 4 \\ \amp \amp -3y \amp {}-{} \amp 3z \amp {}={} \amp 3 \\ -3x \amp {}-{} \amp 2y \amp {}+{} \amp z \amp {}={} \amp -10 \\ \end{alignedat} \end{equation*}

Scale the second equation by multiplying it by \(-1/3\text{.}\)

\begin{equation*} \begin{alignedat}{4} x \amp {}+{} \amp 2y \amp \amp \amp {}={} \amp 4 \\ \amp \amp y \amp {}+{} \amp z \amp {}={} \amp -1 \\ -3x \amp {}-{} \amp 2y \amp {}+{} \amp z \amp {}={} \amp -10 \\ \end{alignedat} \end{equation*}

Another replacement operation eliminates \(x\) from the third equation. We multiply the first equation by 3 and add to the third.

\begin{equation*} \begin{alignedat}{4} x \amp {}+{} \amp 2y \amp \amp \amp {}={} \amp 4 \\ \amp \amp y \amp {}+{} \amp z \amp {}={} \amp -1 \\ \amp \amp 4y \amp {}+{} \amp z \amp {}={} \amp 2 \\ \end{alignedat} \end{equation*}

Eliminate \(y\) from the third equation by multiplying the second equation by -4 and adding it to the third.

\begin{equation*} \begin{alignedat}{4} x \amp {}+{} \amp 2y \amp \amp \amp {}={} \amp 4 \\ \amp \amp y \amp {}+{} \amp z \amp {}={} \amp -1 \\ \amp \amp \amp \amp -3z \amp {}={} \amp 6 \\ \end{alignedat} \end{equation*}

After scaling the third equation by \(-1/3\text{,}\) we have found the value for \(z\text{.}\)

\begin{equation*} \begin{alignedat}{4} x \amp {}+{} \amp 2y \amp \amp \amp {}={} \amp 4 \\ \amp \amp y \amp {}+{} \amp z \amp {}={} \amp -1 \\ \amp \amp \amp \amp z \amp {}={} \amp -2 \\ \end{alignedat} \end{equation*}

The system now has a triangular form so we will begin the process of back substitution by multiplying the third equation by -1 and adding to the second.

\begin{equation*} \begin{alignedat}{4} x \amp {}+{} \amp 2y \amp \amp \amp {}={} \amp 4 \\ \amp \amp y \amp \amp \amp {}={} \amp 1 \\ \amp \amp \amp \amp z \amp {}={} \amp -2 \\ \end{alignedat} \end{equation*}

Finally, multiply the second equation by -2 and add to the first to obtain:

\begin{equation*} \begin{alignedat}{4} x \amp \amp \amp \amp \amp {}={} \amp 2 \\ \amp \amp y \amp \amp \amp {}={} \amp 1 \\ \amp \amp \amp \amp z \amp {}={} \amp -2 \\ \end{alignedat} \end{equation*}

Now that we have arrived at a decoupled system, we know that there is exactly one solution to our original system of equations, which is \((x,y,z) = (2,1,-2)\text{.}\)