Question

In Problems 37–72, solve each system of equations using matrices (row operations). If the system has no solution, say that it is inconsistent.

$\left\{\begin{array}{l}x-y=6\\ 2x-3z=16\\ 2y+z=4\end{array}\right.$

Short answer

The solution of the system is $x=8,y=2,z=0$,or, using ordered triplets,$\left(8,2,0\right)$.

Step-by-step solution

Step 1. Given information. 

The given system of equation are

$\left\{\begin{array}{l}x-y=6\\ 2x-3z=16\\ 2y+z=4\end{array}\right.$

Step 2. Calculation. 

The augmented matrix of the system is:

$\left[\begin{array}{cc}\begin{array}{ccc}1& -1& 0\\ 2& 0& -3\\ 0& 2& 1\end{array}& \begin{array}{c}6\\ 16\\ 4\end{array}\end{array}\right]$

Perform the row operations $R_2=r_2-2r_1$:

$\left[\begin{array}{cc}\begin{array}{ccc}1& -1& 0\\ 0& 2& -3\\ 0& 2& 1\end{array}& \begin{array}{c}6\\ 4\\ 4\end{array}\end{array}\right]$

Perform the row operations $R_3=r_3-r_2$:

$\left[\begin{array}{cc}\begin{array}{ccc}1& -1& 0\\ 0& 2& -3\\ 0& 0& 4\end{array}& \begin{array}{c}6\\ 4\\ 0\end{array}\end{array}\right]$

Use the obtained matrix to write the system of equations.

$$\begin{gathered} x-y=6 \\ 2y-3z=4 \\ 4z=0 \end{gathered}$$

Step 3. Solve the equation.  

Solve the equations to find the solution set.

$$\begin{gathered} 4z=0 \\ z=0 \end{gathered}$$

Substitute $z=0$into the second equation.

$$\begin{gathered} 2y-3z=4 \\ 2y-3\left(0\right)=4 \\ 2y=4 \\ y=2 \end{gathered}$$

Substitute $y=2$into the first equation.

$$\begin{gathered} x-y=6 \\ x-2=6 \\ x=6+2 \\ x=8 \end{gathered}$$

Therefore, the solution of the system is $x=8,y=2,z=0$or, using ordered triplets, $\left(8,2,0\right)$.

Step 4. Conclusion. 

The solution of the system is $x=8,y=2,z=0$or, using ordered triplets, $\left(8,2,0\right)$.