Q. 69 - 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.    

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

Step-by-Step Solution

Answer

The given system of equations is inconsistent.

Step 1 >: Given information

The given system of equations is

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

Step 2 : Concept used

Row operation :

1.  Interchange any two rows.
2.  Replace a row by a nonzero multiple of that row.
3.  Replace a row by the sum of that row and a constant nonzero multiple of some other row.

Step 3 : Calculation

The augmented matrix of the system is:      

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

Perform the row operations $R_1=\frac{1}{2}{r}_1$:

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

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

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

Perform the row operations $R_3=r_3+r_1$:

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

Perform the row operations $R_4=r_4-r_1$:

$\left[\begin{array}{cc}\begin{array}{c}\begin{array}{ccc}1& \frac{3}{2}& -\frac{1}{2}\\ 1& -\frac{1}{2}& -\frac{1}{2}\end{array}\\ \begin{array}{ccc}0 & \frac{5}{2} & \frac{1}{2}\\ 0 & -\frac{1}{2} & \frac{7}{2}\end{array}\end{array}& \begin{array}{c}\begin{array}{c}\begin{array}{c}\frac{3}{2}\\ -\frac{3}{2}\end{array}\\ \begin{array}{c}\frac{3}{2}\\ \frac{7}{2}\end{array}\end{array}\end{array}\end{array}\right]$

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

$\left[\begin{array}{cc}\begin{array}{c}\begin{array}{ccc}1& \frac{3}{2}& -\frac{1}{2}\\ 0& 1& 1\end{array}\\ \begin{array}{ccc}0& \frac{5}{2}& \frac{1}{2}\\ 0& -\frac{1}{2} & \frac{7}{2}\end{array}\end{array}& \begin{array}{c}\begin{array}{c}\begin{array}{c}\frac{3}{2}\\ 3\end{array}\\ \begin{array}{c}\frac{3}{2}\\ \frac{7}{2}\end{array}\end{array}\end{array}\end{array}\right]$

Perform the row operations $R_1=r_1-\frac{3}{2}{r}_2$:

$\left[\begin{array}{cc}\begin{array}{c}\begin{array}{ccc}1& 0& -2\\ 0& 1& 1\end{array}\\ \begin{array}{ccc}0& \frac{5}{2}& \frac{1}{2}\\ 0& -\frac{1}{2} & \frac{7}{2}\end{array}\end{array}& \begin{array}{c}\begin{array}{c}\begin{array}{c}-3\\ 3\end{array}\\ \begin{array}{c}\frac{3}{2}\\ \frac{7}{2}\end{array}\end{array}\end{array}\end{array}\right]$

Perform the row operations $R_3=r_3-\frac{5}{2}{r}_2$:

$\left[\begin{array}{cc}\begin{array}{c}\begin{array}{ccc}1& 0& -2\\ 0& 1& 1\end{array}\\ \begin{array}{ccc}0& 0& -2\\ 0& -\frac{1}{2} & \frac{7}{2}\end{array}\end{array}& \begin{array}{c}\begin{array}{c}\begin{array}{c}-3\\ 3\end{array}\\ \begin{array}{c}-6\\ \frac{7}{2}\end{array}\end{array}\end{array}\end{array}\right]$

Perform the row operations $R_4=r_4+\frac{1}{2}{r}_2$:

$\left[\begin{array}{cc}\begin{array}{c}\begin{array}{ccc}1& 0& -2\\ 0& 1& 1\end{array}\\ \begin{array}{ccc}0& 0& -2\\ 0& 0 & 4\end{array}\end{array}& \begin{array}{c}\begin{array}{c}\begin{array}{c}-3\\ 3\end{array}\\ \begin{array}{c}-6\\ 5\end{array}\end{array}\end{array}\end{array}\right]$

Perform the row operations $R_3=-\frac{1}{2}{r}_3$:

$\left[\begin{array}{cc}\begin{array}{c}\begin{array}{ccc}1& 0& -2\\ 0& 1& 1\end{array}\\ \begin{array}{ccc}0& 0& 1\\ 0& 0 & 4\end{array}\end{array}& \begin{array}{c}\begin{array}{c}\begin{array}{c}-3\\ 3\end{array}\\ \begin{array}{c}3\\ 5\end{array}\end{array}\end{array}\end{array}\right]$

Perform the row operations $R_1=r_1+2r_3$:

$\left[\begin{array}{cc}\begin{array}{c}\begin{array}{ccc}1& 0& 0\\ 0& 1& 1\end{array}\\ \begin{array}{ccc}0& 0& 1\\ 0& 0 & 4\end{array}\end{array}& \begin{array}{c}\begin{array}{c}\begin{array}{c}3\\ 3\end{array}\\ \begin{array}{c}3\\ 5\end{array}\end{array}\end{array}\end{array}\right]$

Perform the row operations $R_2=r_2-r_3$

$\left[\begin{array}{cc}\begin{array}{c}\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\end{array}\\ \begin{array}{ccc}0& 0& 1\\ 0& 0 & 4\end{array}\end{array}& \begin{array}{c}\begin{array}{c}\begin{array}{c}3\\ 0\end{array}\\ \begin{array}{c}3\\ 5\end{array}\end{array}\end{array}\end{array}\right]$

Perform the row operations $R_4=r_4-4r_3$:

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

This matrix is in row echelon form.

Use the obtained matrix to write the system of equations.

$$\begin{gathered} x=3 ...\left(1\right) \\ y=0 ... \left(2\right) \\ z=3 ...\left(3\right) \end{gathered}$$

Step 4 : Solve the equation

The bottom row of the obtained matrix is equivalent to the equation

$0·x+0·y+0·z=-7$

which has no solution.

Thus, the original system is inconsistent.

Step 5: .Conclusion

The given system of equations is inconsistent.