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.

$\left\{\begin{array}{l}2x+y-3z=0\\ -2x+2y+z=-7\\ 3x-4y-3z=7\end{array}\right.$

Short answer

The solution of the system is$x=\frac{56}{13},y=-\frac{7}{13},z=\frac{35}{13}$ ,or, using ordered triplets,$\left(\frac{56}{13},-\frac{7}{13},\frac{35}{13}\right)$ .

Step-by-step solution

Step 1. Given information. 

The given system of equation are

$\left\{\begin{array}{l}2x+y-3z=0\\ -2x+2y+z=-7\\ 3x-4y-3z=7\end{array}\right.$

Step 2. Calculation. 

The augmented matrix of the system is:
$\left[\begin{array}{cc}\begin{array}{ccc}2& 1& -3\\ -2& 2& 1\\ 3& -4& -3\end{array}& \begin{array}{c}0\\ -7\\ 7\end{array}\end{array}\right]$

Perform the row operations $R_3=2r_3+3r_2$:

$\left[\begin{array}{cc}\begin{array}{ccc}2& 1& -3\\ -2& 2& 1\\ 0& -2& -3\end{array}& \begin{array}{c}0\\ -7\\ -7\end{array}\end{array}\right]$

Perform the row operations $R_2=r_2+r_1$:

$\left[\begin{array}{cc}\begin{array}{ccc}2& 1& -3\\ 0& 3& -2\\ 0& -2& -3\end{array}& \begin{array}{c}0\\ -7\\ -7\end{array}\end{array}\right]$

Perform the row operations $R_3=3r_3+2r_2$:

$\left[\begin{array}{cc}\begin{array}{ccc}2& 1& -3\\ 0& 3& -2\\ 0& 0& -13\end{array}& \begin{array}{c}0\\ -7\\ -35\end{array}\end{array}\right]$

Perform the row operations $R_3=-\frac{1}{13}{r}_3$:

$\left[\begin{array}{cc}\begin{array}{ccc}2& 1& -3\\ 0& 3& -2\\ 0& 0& 1\end{array}& \begin{array}{c}0\\ -7\\ \frac{35}{13}\end{array}\end{array}\right]$

Use the obtained matrix to write the system of equations.

$$\begin{gathered} 2x+y-3z=0 \\ 3y-2z=-7 \\ z=\frac{35}{13} \end{gathered}$$

Step 3. Solve the equation.  

Solve the equations to find the solution set.

Substitute $z=\frac{35}{13}$into the second equation.

$$\begin{gathered} 3y-2z=-7 \\ 3y-2\left(\frac{35}{13}\right)=-7 \\ 3y-\frac{70}{13}=-7 \\ 3y=-7+\frac{70}{13} \\ 3y=-\frac{21}{13} \\ y=-\frac{7}{13} \end{gathered}$$

Substitute $y=-\frac{7}{13}\mathrm{and}\mathrm{z}=\frac{35}{13}$into the first equation.

$$\begin{gathered} 2x+y-3z=0 \\ 2x+\left(-\frac{7}{13}\right)-3\left(\frac{35}{13}\right)=0 \\ 2x-\frac{7}{13}-\frac{105}{13}=0 \\ 26x-7-105=0 \\ 26x=112 \\ x=\frac{112}{26}=\frac{56}{13} \end{gathered}$$

Therefore, the solution of the system is$x=\frac{56}{13},y=-\frac{7}{13},z=\frac{35}{13}$or, using ordered triplets,$\left(\frac{56}{13},-\frac{7}{13},\frac{35}{13}\right)$.

Step 4. Conclusion. 

The solution of the system is$x=\frac{56}{13},y=-\frac{7}{13},z=\frac{35}{13}$or, using ordered triplets,$\left(\frac{56}{13},-\frac{7}{13},\frac{35}{13}\right)$.