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

To solve the system of equations using the inverse coefficient matrix, we multiply the inverse matrix by the constants on the right side of the equations. The given inverse matrix is:$$\left[ \begin{array} { c c c } \frac { 3 } { 3 } & \frac { 3 } { 2 } & - \frac { 11 } { 6 } \- \frac { 2 } { 3 } & - \frac { 1 } { 2 } & \frac { 3 } { 6 } \- \frac { 2 } { 3 } & 0 & \frac { 1 } { 3 }\end{array} \right]$$And the constants from the equations are:$$\left[ \begin{array} { c } 1 \4 \2\end{array} \right]$$Multiplying the inverse matrix by the constants gives the solution for $x$, $y$, and $z$. The correct solution, given the provided inverse matrix, is $(-8, 3, 0)$, which corresponds to choice C.

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