Q. 66 - 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.  

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

Step-by-Step Solution

Answer

The given system is inconsistent. 

Step 1 : Given information

The given system of equation is

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

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:    

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

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

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

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

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

Perform the row operations $R_2=-\frac{r_2}{5}$ :

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

Perform the row operations $R_1=r_1-2r_2$:

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

Perform the row operations $R_3=r_3+5r_2$:

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

This matrix is in row echelon form.

Use the obtained matrix to write the system of equations.

$x+\frac{3}{5}z=3 ...\left(1\right)$

$y-\frac{4}{5}z=0 ...\left(2\right)$

Step 4 : Solve the equation

From the above obtained matrix, the bottom row is equivalent to the equation

$0x+0y+0z=1$

which has no solution.

Therefore, the original system is inconsistent.

Step 5 : Conclusion

The given system is inconsistent.