Q. 71 - 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. 

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

Step-by-Step Solution

Answer

The solution of the given system is

$x=\frac{7}{6}-\frac{1}{2}z-\frac{1}{3}w , y=z+\frac{7}{3}w-\frac{2}{3}$ ; where $z$ and $w$ are any real numbers.

Step 1: Given information

The given system of equations is

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

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:       

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

Perform the row operations $R_1=\frac{r_1}{4}$:

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

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

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

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

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

Perform the row operations $R_1=r_1-\frac{1}{4}{r}_2$:

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

This matrix is in row echelon form.

Use the obtained matrix to write the system of equations.

$$\begin{gathered} x+\frac{1}{2}z+\frac{1}{3}w=\frac{7}{6} ...\left(1\right) \\ y-z-\frac{7}{3}w=-\frac{2}{3} ...\left(2\right) \end{gathered}$$

Step 4 : Solve the equation

From equation (1),

$x=\frac{7}{6}-\frac{1}{2}z-\frac{1}{3}w$

From equation (2),

$y=z+\frac{7}{3}w-\frac{2}{3}$

Thus, the solution of the system of the equations is

$x=\frac{7}{6}-\frac{1}{2}z-\frac{1}{3}w ,$$y=z+\frac{7}{3}w-\frac{2}{3}$ ; where $z$ and $w$ are any real numbers.

Step 5 : Conclusion

The solution of the given system is

$x=\frac{7}{6}-\frac{1}{2}z-\frac{1}{3}w , y=z+\frac{7}{3}w-\frac{2}{3}$ ; where $z$ and $w$ are any real numbers.