Q. 65 - 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. 

$\left\{\begin{array}{l}x+2y+z=1\\ 2x-y+2z=2\\ 3x+y+3z=3\end{array}\right.$

Step-by-Step Solution

Answer

The solution of the system is $x=1-z, y=0$; where $z$ is any real number.

Step 1: Given information

The given system of equation are 

$\left\{\begin{array}{l}x+2y+z=1\\ 2x-y+2z=2\\ 3x+y+3z=3\end{array}\right.$

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:   

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

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

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

Perform the row operations  $R_3=r_3-3r_1$

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

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

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

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

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

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

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

Use the obtained matrix to write the system of equations.

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

Step 4: Solve the equation.

From equation (1),

$x=1-z$

So, the solution of the system is $x=1-z,y=0$; where $z$ is any real number.

  

Step 5: Conclusion

The solution of the system is $x=1-z , y=0$; where $z$ is any real number.