Q. 68 - 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.   

$\left\{\begin{array}{l}2x+y-z=4\\ -x+y+3z=1\end{array}\right.$

Step-by-Step Solution

Answer

The solution of the system of equations is

$x=1+\frac{4}{3}z , y=2-\frac{5}{3}z$ ; where $z$ is any real number.

Step 1: Given information

The given system of equations is 

$\left\{\begin{array}{l}2x+y-z=4\\ -x+y+3z=1\end{array}\right.$ 

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:     

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

Perform the row operations $R_1=\frac{r_1}{2}$:

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

Perform the row operations $R_1=r_2+r_1$:

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

Perform the row operations $R_2=\frac{2}{3}{r}_2$:

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

Perform the row operations $R_1=r_1-\frac{r_2}{2}$:

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

This matrix is in row echelon form.

Use the obtained matrix to write the system of equations.

$x-\frac{4}{3}z=1 ...\left(1\right)$

$y+\frac{5}{3}z=2 ...\left(2\right)$

Step 4 : Solve the equation

Now, write down some of the solutions, we express both $x$ and $y$ in terms of $z$.

From equation (1),

$x=1+\frac{4}{3}z$

From equation (2),

$y=2-\frac{5}{3}z$

So, the original system of equations is equivalent to the system,

$\left\{\begin{array}{l}x=1+\frac{4}{3}z\\ y=2-\frac{5}{3}z\end{array}\right.$, where $z$ is any real number.

Step 5: Conclusion

The solution of the system of equations is

$x=1+\frac{4}{3}z , y=2-\frac{5}{3}z$ ; where $z$ is any real number.