Solve the system of linear equations and check any solution algebraically. $\begin{array} { l }
3 x + 3 y = 0 \
\begin{aligned}
6 x + 3 y - z & = 0 \
12 x + 3 y + 3 z & = 0
\end{aligned} \
\end{array}$
A) $( - 3,6,12 )$
B) (0, 0, 0)
C) $( - 3,6 , - 12 )$
D) $( 3,6,12 )$
E) $( 3,6 , - 12 )$

Question

Solve the system of linear equations and check any solution algebraically.  $\begin{array} { l } 
3 x + 3 y = 0 \
\begin{aligned}
6 x + 3 y - z & = 0 \
12 x + 3 y + 3 z & = 0
\end{aligned} \
\end{array}$  
A) $( - 3,6,12 )$  
B) (0, 0, 0)
C)  $( - 3,6 , - 12 )$
D)  $( 3,6,12 )$
E)  $( 3,6 , - 12 )$

Quizplus · Accepted Answer

The system of equations is homogeneous (all constants are zero), which means it has at least the trivial solution (0, 0, 0). Since the equations are linear and dependent (the third equation is a multiple of the second), there are no other distinct solutions.

Solve the System of Linear Equations and Check Any Solution 3x+3y=06x+3y−z=012x+3y+3z=0\begin{array} { l } 3 x + 3 y = 0 \\\begin{aligned}6 x + 3 y - z & = 0 \\12 x + 3 y + 3 z & = 0\end{aligned} \\\end{array}3x+3y=06x+3y−z12x+3y+3z​=0=0​​

Solve the System of Linear Equations and Check Any Solution $\begin{array} { l } 3 x + 3 y = 0 \\\begin{aligned}6 x + 3 y - z & = 0 \\12 x + 3 y + 3 z & = 0\end{aligned} \\\end{array}$