Translate the Matrix Equations into a System of Linear Equations

Translate the matrix equations into a system of linear equations. $\left[ \begin{array} { c c c } \frac { 1 } { 5 } & 0 & - 1 \\2 & \frac { 1 } { 2 } & 5 \\- 3 & 4 & 8\end{array} \right] \cdot \left[ \begin{array} { l } x \\y \\z\end{array} \right] = \left[ \begin{array} { c } - 9 \\3 \\2\end{array} \right]$

A) $\left\{ \begin{aligned}\frac { 1 } { 5 } x + y - 3 z & = - 9 \\\frac { 1 } { 2 } y + 4 z & = 3 \\- x + 5 y + 8 z & = 2\end{aligned} \right\}$
B) $\left\{ \begin{aligned}\frac { 1 } { 5 } x + y - 3 z + 9 & = 0 \\\frac { 1 } { 2 } y + 4 z + 3 & = 0 \\- x + 5 y + 8 z + 2 & = 0\end{aligned} \right\}$
C) $\left\{ \begin{array} { r } \frac { 1 } { 5 } x - z + 9 = x \\2 x + \frac { 1 } { 2 } y + 5 z - 3 = y \\- 3 x + 4 y + 8 z - 2 = z\end{array} \right\}$
D) $\left\{ \begin{aligned}\frac { 1 } { 5 } x - z & = - 9 \\2 x + \frac { 1 } { 2 } y + 5 z & = 3 \\- 3 x + 4 y + 8 z & = 2\end{aligned} \right\}$
E) $\left\{ \begin{array} { r } \frac { 1 } { 5 } x - z - 9 = 0 \\2 x + \frac { 1 } { 2 } y + 5 z + 3 = 0 \\- 3 x + 4 y + 8 z + 2 = 0\end{array} \right\}$