Q. 72 - 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. 

$\left\{\begin{array}{c}-4x+y=5\\ 2x-y+z-w=5\\ z+w=4\end{array}\right.$

Step-by-Step Solution

Answer

The solution of the given system is

$x=3-w,y=-7-4w, z=4-w$ ; where $w$ is any real numbers.

Step 1: Given information

The given system of equations is

$\left\{\begin{array}{c}-4x+y=5\\ 2x-y+z-w=5\\ z+w=4\end{array}\right.$

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:       

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

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

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

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

$\left[\begin{array}{cc}\begin{array}{cccc}1& -\frac{1}{4}& 0& 0\\ 0& -\frac{1}{2}& 1& -1\\ 0& 0& 1& 1\end{array}& \begin{array}{c}\begin{array}{c}-\frac{5}{4}\\ \frac{15}{2}\\ 4\end{array}\end{array}\end{array}\right]$

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

$\left[\begin{array}{cc}\begin{array}{cccc}1& -\frac{1}{4}& 0& 0\\ 0& 1& -2& 2\\ 0& 0& 1& 1\end{array}& \begin{array}{c}\begin{array}{c}-\frac{5}{4}\\ -15\\ 4\end{array}\end{array}\end{array}\right]$

Perform the row operations $R_1=r_1+\frac{r_2}{4}$:

$\left[\begin{array}{cc}\begin{array}{cccc}1& 0& -\frac{1}{2}& \frac{1}{2}\\ 0& 1& -2& 2\\ 0& 0& 1& 1\end{array}& \begin{array}{c}\begin{array}{c}-5\\ -15\\ 4\end{array}\end{array}\end{array}\right]$

Perform the row operations $R_1=r_1+\frac{r_3}{2}$:

$\left[\begin{array}{cc}\begin{array}{cccc}1& 0& 0& 1\\ 0& 1& -2& 2\\ 0& 0& 1& 1\end{array}& \begin{array}{c}\begin{array}{c}-3\\ -15\\ 4\end{array}\end{array}\end{array}\right]$

Perform the row operations $R_2=r_2+2r_3$:

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

This matrix is in row echelon form.

Use the obtained matrix to write the system of equations.

$$\begin{gathered} x+w=-3 ...\left(1\right) \\ y+4w=-7 ...\left(2\right) \\ z+w=4 ...\left(3\right) \end{gathered}$$

Step 4 : Solve the equation

Present all equation in terms of $w$.

From equation (1),

$x=-3-w$

From equation (2),

$y=-7-4w$

From equation (3),

$z=4-w$

Thus, the solution is

$x=3-w,y=-7-4w, z=4-w$ ; where $w$ is any real number.

Step 5 : Conclusion

The solution of the system of the equation is

$x=3-w,y=-7-4w, z=4-w$ ; where $w$ is any real numbers.