Q. 63 - 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.

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

Step-by-Step Solution

Answer

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

Step 1. Given information.

The given system of equation are

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

Step 2. Calculation.

The augmented matrix of the system is:  

$\left[\begin{array}{cc}\begin{array}{cccc}1& 1& 1& 1\\ 2& -1& 1& 0\\ 3& 2& 1& -1\\ 1& -2& -2& 2\end{array}& \begin{array}{c}4\\ 16\\ 4\\ -1\end{array}\end{array}\right]$

Perform the row operations  $R_2=r_2-2r, R_3=r_3-3r_1, R_4=r_4-r_1$:

$\left[\begin{array}{cc}\begin{array}{cccc}1& 1& 1& 1\\ 0& -3& -1& -2\\ 0& -1& -2& -4\\ 0& -3& -3& 1\end{array}& \begin{array}{c}4\\ -8\\ -6\\ -5\end{array}\end{array}\right]$

Perform the row operations $R_3=r_3-\frac{1}{3}{r}_2, R_4=r_4-r_2$:

$\left[\begin{array}{cc}\begin{array}{cccc}1& 1& 1& 1\\ 0& -3& -1& -2\\ 0& 0& -\frac{5}{3}& -\frac{10}{3}\\ 0& 0& -2& 3\end{array}& \begin{array}{c}4\\ -8\\ -\frac{10}{3}\\ 3\end{array}\end{array}\right]$

Perform the row operations $R_3=-\frac{5}{3}{r}_3$:

$\left[\begin{array}{cc}\begin{array}{cccc}1& 1& 1& 1\\ 0& -3& -1& -2\\ 0& 0& 1& 2\\ 0& 0& -2& 3\end{array}& \begin{array}{c}4\\ -8\\ 2\\ 3\end{array}\end{array}\right]$

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

$\left[\begin{array}{cc}\begin{array}{cccc}1& 1& 1& 1\\ 0& -3& -1& -2\\ 0& 0& 1& 2\\ 0& 0& 0& 7\end{array}& \begin{array}{c}4\\ -8\\ 2\\ 7\end{array}\end{array}\right]$

Use the obtained matrix to write the system of equations.

$$\begin{gathered} x+y+z+w=4 \\ -3y-z-2w=-8 \\ z+2w=2 \\ 7w=7 \end{gathered}$$

Step 3. Solve the equation.

Solve the equations to find the solution set.

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

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.

$$\begin{gathered} -3y-z-2w=-8 \\ -3y-0-2\left(1\right)=-8 \\ -3y-2=-8 \\ -3y=-8+2 \\ -3y=-6 \\ y=\frac{6}{3} \\ y=2 \end{gathered}$$

Substitute  $y=2, z=0, w=1$ into first equation.

$$\begin{gathered} x+y+z+w=4 \\ x+2+0+1=4 \\ x+3=4 \\ x=4-3 \\ x=1 \end{gathered}$$

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

Step 4. Conclusion.

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