Solve the system of linear equations $$\left\{ \begin{array} { l l }
- 6 x + 18 y + 6 z & = 1 \
12 x + 30 y & = 2 \
18 x + 6 y - 12 z & = - 1
\end{array} \right.$$ using an inverse matrix.
A) $$\left[ \begin{array} { l }
x \
y \
z
\end{array} \right] = \left[ \begin{array} { c }
\frac { 1 } { 6 } \\
\frac { - 1 } { 2 } \\
\frac { 1 } { 3 }
\end{array} \right]$$
B) $$\left[ \begin{array} { l }
x \
y \
z
\end{array} \right] = \left[ \begin{array} { c }
\frac { - 1 } { 6 } \\
\frac { - 1 } { 2 } \\
\frac { - 1 } { 3 }
\end{array} \right]$$
C) $$\left[ \begin{array} { l }
x \
y \
z
\end{array} \right] = \left[ \begin{array} { c }
\frac { - 1 } { 6 } \
0 \
\frac { 1 } { 2 }
\end{array} \right]$$
D) $$\left[ \begin{array} { l }
x \
y \
z
\end{array} \right] = \left[ \begin{array} { c }
0 \
\frac { 1 } { 3 } \
\frac { - 1 } { 6 }
\end{array} \right]$$
E) $$\left[ \begin{array} { l }
x \
y \
z
\end{array} \right] = \left[ \begin{array} { l }
\frac { 1 } { 6 } \
0 \
\frac { 1 } { 3 }
\end{array} \right]$$

Question

​Solve the system of linear equations $$\left\{ \begin{array} { l l } 
- 6 x + 18 y + 6 z & = 1 \
12 x + 30 y & = 2 \
18 x + 6 y - 12 z & = - 1
\end{array} \right.$$ using an inverse matrix. ​
A)​ $$\left[ \begin{array} { l } 
x \
y \
z
\end{array} \right] = \left[ \begin{array} { c } 
\frac { 1 } { 6 } \\
\frac { - 1 } { 2 } \\
\frac { 1 } { 3 }
\end{array} \right]$$ ​
B)​ $$\left[ \begin{array} { l } 
x \
y \
z
\end{array} \right] = \left[ \begin{array} { c } 
\frac { - 1 } { 6 } \\
\frac { - 1 } { 2 } \\
\frac { - 1 } { 3 }
\end{array} \right]$$
C)​ $$\left[ \begin{array} { l } 
x \
y \
z
\end{array} \right] = \left[ \begin{array} { c } 
\frac { - 1 } { 6 } \
0 \
\frac { 1 } { 2 }
\end{array} \right]$$
D)​ $$\left[ \begin{array} { l } 
x \
y \
z
\end{array} \right] = \left[ \begin{array} { c } 
0 \
\frac { 1 } { 3 } \
\frac { - 1 } { 6 }
\end{array} \right]$$
E)​ $$\left[ \begin{array} { l } 
x \
y \
z
\end{array} \right] = \left[ \begin{array} { l } 
\frac { 1 } { 6 } \
0 \
\frac { 1 } { 3 }
\end{array} \right]$$

Quizplus · Accepted Answer

The Answer of ​Solve the system of linear equations $$\left\{ \begin{array} { l l } 
- 6 x + 18 y + 6 z & = 1 \
12 x + 30 y & = 2 \
18 x + 6 y - 12 z & = - 1
\end{array} \right.$$ using an inverse matrix. ​
A)​ $$\left[ \begin{array} { l } 
x \
y \
z
\end{array} \right] = \left[ \begin{array} { c } 
\frac { 1 } { 6 } \\
\frac { - 1 } { 2 } \\
\frac { 1 } { 3 }
\end{array} \right]$$ ​
B)​ $$\left[ \begin{array} { l } 
x \
y \
z
\end{array} \right] = \left[ \begin{array} { c } 
\frac { - 1 } { 6 } \\
\frac { - 1 } { 2 } \\
\frac { - 1 } { 3 }
\end{array} \right]$$
C)​ $$\left[ \begin{array} { l } 
x \
y \
z
\end{array} \right] = \left[ \begin{array} { c } 
\frac { - 1 } { 6 } \
0 \
\frac { 1 } { 2 }
\end{array} \right]$$
D)​ $$\left[ \begin{array} { l } 
x \
y \
z
\end{array} \right] = \left[ \begin{array} { c } 
0 \
\frac { 1 } { 3 } \
\frac { - 1 } { 6 }
\end{array} \right]$$
E)​ $$\left[ \begin{array} { l } 
x \
y \
z
\end{array} \right] = \left[ \begin{array} { l } 
\frac { 1 } { 6 } \
0 \
\frac { 1 } { 3 }
\end{array} \right]$$

​Solve the System of Linear Equations {−6x+18y+6z=112x+30y=218x+6y−12z=−1\left\{ \begin{array} { l l } - 6 x + 18 y + 6 z & = 1 \\12 x + 30 y & = 2 \\18 x + 6 y - 12 z & = - 1\end{array} \right.⎩⎨⎧​−6x+18y+6z12x+30y18x+6y−12z​=1=2=−1​

Solve the System of Linear Equations $\left\{ \begin{array} { l l } - 6 x + 18 y + 6 z & = 1 \\12 x + 30 y & = 2 \\18 x + 6 y - 12 z & = - 1\end{array} \right.$