Q. 54 - Precalculus Enhanced with Graphing Utilities

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. 

$\left\{\begin{array}{l}2x-3y-z=0\\ 3x+2y+2z=2\\ x+5y+3z=2\end{array}\right.$

Step-by-Step Solution

Answer

The solution of the system is $x=\frac{6}{13}-\frac{4}{13}z, y=\frac{4}{13}-\frac{7}{13}z$where, z is any real number. 

Step 1. Given information.

The given system of equation are  

$\left\{\begin{array}{l}2x-3y-z=0\\ 3x+2y+2z=2\\ x+5y+3z=2\end{array}\right.$

Step 2. Calculation.

The augmented matrix of the system is:

$\left[\begin{array}{cc}\begin{array}{ccc}2& -3& -1\\ 3& 2& 2\\ 1& 5& 3\end{array}& \begin{array}{c}0\\ 2\\ 2\end{array}\end{array}\right]$

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

$\left[\begin{array}{cc}\begin{array}{ccc}2& -3& -1\\ 1& 5& 3\\ 1& 5& 3\end{array}& \begin{array}{c}0\\ 2\\ 2\end{array}\end{array}\right]$

Perform the row operations $R_2=r_2-\frac{1}{2}{r}_1, R_3=r_3-r_2$:

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

Perform the row operations $R_2=2r_2$:

$\left[\begin{array}{cc}\begin{array}{ccc}2& -3& -1\\ 0& 13& 7\\ 0& 0& 0\end{array}& \begin{array}{c}0\\ 4\\ 0\end{array}\end{array}\right]$

The above matrix is in reduced row echelon form. The corresponding system of equations is 

$$\begin{gathered} 2x-3y-z=0 \left(1\right) \\ 13y+7z=4 \left(2\right) \end{gathered}$$

Step 3. Solve the equation.

From equation (2):

$$\begin{gathered} 13y+7z=4 \\ 13y=4-7z \\ y=\frac{4}{13}-\frac{7}{13}z \end{gathered}$$

Substitute $y=\frac{4}{13}-\frac{7}{13}z$ into equation (1).

$$\begin{gathered} 2x-3y-z=0 \\ 2x-3\left(\frac{4}{13}-\frac{7}{13}z\right)-z=0 \\ 2x-\frac{12}{13}+\frac{21}{13}z-z=0 \\ 2x+\frac{8}{13}z=\frac{12}{13} \\ 2x=\frac{12}{13}-\frac{8}{13}z \\ x=\frac{6}{13}-\frac{4}{13}z \end{gathered}$$

Therefore, the solution of the system is $x=\frac{6}{13}-\frac{4}{13}z, y=\frac{4}{13}-\frac{7}{13}z$, where, z is any real number. 

Step 4. Conclusion.

The solution of the system is $x=\frac{6}{13}-\frac{4}{13}z, y=\frac{4}{13}-\frac{7}{13}z$ where, z is any real number.