Q. 64 - 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.

$\left\{\begin{array}{l}x+y+z+w=4\\ -x+2y+z=0\\ 2x+3y+z-w=6\\ -2x+y-2z+2w=-1\end{array}\right.$

Step-by-Step Solution

Answer

The solution of the system is $x=2,y=1,z=0,w=1$ or, using ordered quads, $\left(2,1,0,1\right)$.

Step 1. Given information.

The given system of equation are  

$\left\{\begin{array}{l}x+y+z+w=4\\ -x+2y+z=0\\ 2x+3y+z-w=6\\ -2x+y-2z+2w=-1\end{array}\right.$

Step 2. Calculation.

The augmented matrix of the system is:  

$\left[\begin{array}{cc}\begin{array}{cccc}1& 1& 1& 1\\ -1& 2& 1& 0\\ 2& 3& 1& -1\\ -2& 1& -2& 2\end{array}& \begin{array}{c}4\\ 0\\ 6\\ -1\end{array}\end{array}\right]$

Perform the row operations $R_4=r_4+2r_1$:

$\left[\begin{array}{cc}\begin{array}{cccc}1& 1& 1& 1\\ -1& 2& 1& 0\\ 2& 3& 1& -1\\ 0& 3& 0& 4\end{array}& \begin{array}{c}4\\ 0\\ 6\\ 7\end{array}\end{array}\right]$

Perform the row operations $R_2=r_2+r_1, R_3=r_3-2r_1$:

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

Substitute $R_3 \mathrm{and} R_2$:

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

Perform the row operations $R_3=r_3-3r_2$:

$\left[\begin{array}{cc}\begin{array}{cccc}1& 1& 1& 1\\ 0& 1& -1& -3\\ 0& 0& 5& 10\\ 0& 0& -2& 3\end{array}& \begin{array}{c}4\\ -2\\ 10\\ 3\end{array}\end{array}\right]$

Perform the row operations $R_3=\frac{1}{5}{r}_3$:

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

Perform the row operations $R_4=r_4+2r_3$:

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

Use the obtained matrix to write the system of equations.

$$\begin{gathered} x+y+z+w=4 \\ y-z-3w=-2 \\ z+2w=2 \\ 7w=7 \end{gathered}$$

Step 3. Solve the equation.

Solve the equations to find the solution set.

$$\begin{gathered} 7w=w \\ w=1 \end{gathered}$$

Substitute $w=1$into the third equation.

$$\begin{gathered} z+2w=2 \\ z+2\left(1\right)=2 \\ z=2-2 \\ z=0 \end{gathered}$$

Substitute $z=0 \mathrm{and} \mathrm{w}=1$into the second equation.

$$\begin{gathered} y-z-3w=-2 \\ y-0-3\left(1\right)=-2 \\ y-3=-2 \\ y=-2+3 \\ y=1 \end{gathered}$$

Substitute $y=1,z=0,w=1$into first equation.

$$\begin{gathered} x+y+z+w=4 \\ x+1+0+1=4 \\ x+2=4 \\ x=4-2 \\ x=2 \end{gathered}$$

Therefore, the solution of the system is  $x=2,y=1,z=0,w=1$or, using ordered quads $\left(2,1,0,1\right)$.

Step 4. Conclusion.

The solution of the system is $x=2,y=1,z=0,w=1$or, using ordered quads $\left(2,1,0,1\right)$.