Solve the system of equations $\left\{ \begin{array} { r }
x + 3 y - 2 z = 1 \
2 x + 4 y + z = 4 \
2 x + 6 y - z = 2
\end{array} \right.$ by converting to a matrix equation and using its inverse coefficient matrix $$\left[ \begin{array} { c c c }
\frac { 3 } { 3 } & \frac { 3 } { 2 } & - \frac { 11 } { 6 } \
- \frac { 2 } { 3 } & - \frac { 1 } { 2 } & \frac { 3 } { 6 } \
- \frac { 2 } { 3 } & 0 & \frac { 1 } { 3 }
\end{array} \right]$$
A) $( - 14 , - 3,0 )$
B) $( 2,3,0 )$
C) $( - 8,3,0 )$
D) $( - 8,0,1 )$
E) $( - 6 , - 3,0 )$

Question

Solve the system of equations $\left\{ \begin{array} { r } 
x + 3 y - 2 z = 1 \
2 x + 4 y + z = 4 \
2 x + 6 y - z = 2
\end{array} \right.$ by converting to a matrix equation and using its inverse coefficient matrix $$\left[ \begin{array} { c c c } 
\frac { 3 } { 3 } & \frac { 3 } { 2 } & - \frac { 11 } { 6 } \
- \frac { 2 } { 3 } & - \frac { 1 } { 2 } & \frac { 3 } { 6 } \
- \frac { 2 } { 3 } & 0 & \frac { 1 } { 3 }
\end{array} \right]$$
A) $( - 14 , - 3,0 )$
B) $( 2,3,0 )$
C) $( - 8,3,0 )$
D) $( - 8,0,1 )$
E) $( - 6 , - 3,0 )$

Quizplus · Accepted Answer

The correct solution is found by multiplying the inverse coefficient matrix by the constants on the right side of the equations. The given inverse matrix correctly applied to the constants (1, 4, 2) yields the solution $(-8, 3, 0)$.

Solve the System of Equations {x+3y−2z=12x+4y+z=42x+6y−z=2\left\{ \begin{array} { r } x + 3 y - 2 z = 1 \\2 x + 4 y + z = 4 \\2 x + 6 y - z = 2\end{array} \right.⎩⎨⎧​x+3y−2z=12x+4y+z=42x+6y−z=2​

Solve the System of Equations $\left\{ \begin{array} { r } x + 3 y - 2 z = 1 \\2 x + 4 y + z = 4 \\2 x + 6 y - z = 2\end{array} \right.$