Use the matrix capabilities of a graphing utility to solve the following system of linear equations: $$\left\{ \begin{array} { l l }
- 14 x + 14 y + 21 z & = 10 \
7 x - 7 y & = - 5 \
7 x + 28 z & = 20
\end{array} \right.$$
A) $\left[ \begin{array} { l }
x \
y \
z
\end{array} \right] = \left[ \begin{array} { c }
\frac { 20 } { 7 } \
\frac { 25 } { 7 } \
0
\end{array} \right]$
B) $\left[ \begin{array} { l }
x \
y \
z
\end{array} \right] = \left[ \begin{array} { c }
\frac { 10 } { 7 } \
0 \
\frac { - 15 } { 7 }
\end{array} \right]$
C) $\left[ \begin{array} { l }
x \
y \
z
\end{array} \right] = \left[ \begin{array} { c }
\frac { - 15 } { 7 } \
\frac { 10 } { 7 } \
\frac { 5 } { 7 }
\end{array} \right]$
D) $\left[ \begin{array} { l }
x \
y \
z
\end{array} \right] = \left[ \begin{array} { c }
\frac { 5 } { 7 } \
\frac { - 20 } { 7 } \
\frac { 10 } { 7 }
\end{array} \right]$
E) $\left[ \begin{array} { l }
x \
y \
z
\end{array} \right] = \left[ \begin{array} { c }
\frac { - 15 } { 7 } \
\frac { - 10 } { 7 } \
\frac { 10 } { 7 }
\end{array} \right]$

Question

Use the matrix capabilities of a graphing utility to solve the following system of linear equations:   $$\left\{ \begin{array} { l l } 
- 14 x + 14 y + 21 z & = 10 \
7 x - 7 y & = - 5 \
7 x + 28 z & = 20
\end{array} \right.$$  
A)  $\left[ \begin{array} { l } 
x \
y \
z
\end{array} \right] = \left[ \begin{array} { c } 
\frac { 20 } { 7 } \
\frac { 25 } { 7 } \
0
\end{array} \right]$  
B)  $\left[ \begin{array} { l } 
x \
y \
z
\end{array} \right] = \left[ \begin{array} { c } 
\frac { 10 } { 7 } \
0 \
\frac { - 15 } { 7 }
\end{array} \right]$
C)  $\left[ \begin{array} { l } 
x \
y \
z
\end{array} \right] = \left[ \begin{array} { c } 
\frac { - 15 } { 7 } \
\frac { 10 } { 7 } \
\frac { 5 } { 7 }
\end{array} \right]$
D)  $\left[ \begin{array} { l } 
x \
y \
z
\end{array} \right] = \left[ \begin{array} { c } 
\frac { 5 } { 7 } \
\frac { - 20 } { 7 } \
\frac { 10 } { 7 }
\end{array} \right]$
E) $\left[ \begin{array} { l } 
x \
y \
z
\end{array} \right] = \left[ \begin{array} { c } 
\frac { - 15 } { 7 } \
\frac { - 10 } { 7 } \
\frac { 10 } { 7 }
\end{array} \right]$

Quizplus · Accepted Answer

To solve the system of equations using matrix capabilities, we can set up the augmented matrix and then use row operations or a graphing utility to find the solution. The correct solution is $\left[ \begin{array} { l } x \ y \ z \end{array} \right] = \left[ \begin{array} { c } \frac { 20 } { 7 } \ \frac { 25 } { 7 } \ 0 \end{array} \right]$, which matches option A. This can be verified by substituting the values of $x$, $y$, and $z$ back into the original equations.

Use the Matrix Capabilities of a Graphing Utility to Solve {−14x+14y+21z=107x−7y=−57x+28z=20\left\{ \begin{array} { l l } - 14 x + 14 y + 21 z & = 10 \\7 x - 7 y & = - 5 \\7 x + 28 z & = 20\end{array} \right.⎩⎨⎧​−14x+14y+21z7x−7y7x+28z​=10=−5=20​

Use the Matrix Capabilities of a Graphing Utility to Solve $\left\{ \begin{array} { l l } - 14 x + 14 y + 21 z & = 10 \\7 x - 7 y & = - 5 \\7 x + 28 z & = 20\end{array} \right.$