Consider the Circuit Shown in the Figure Below $I _ { 1 } , I _ { 2 } , \text { and } I _ { 3 }$

Consider the circuit shown in the figure below. The currents $I _ { 1 } , I _ { 2 } , \text { and } I _ { 3 }$ ( in amperes ) are the solutions to the system of linear equations $\left\{ \begin{aligned}3 I _ { 1 } + 6 I _ { 3 } & = E _ { 1 } \\2 I _ { 2 } + 6 I _ { 3 } & = E _ { 2 } \\I)_ { 1 } + I _ { 2 } - I _ { 3 } & = 0\end{aligned} \right.$ , where $E _ { 1 } \text { and } E _ { 2 }$ are voltages of 6 and 9 volts, respectively. Use the inverse of the coefficient matrix of this system to find the unknown currents. Round answers to nearest tenth. [Hint: If a current value turns out to be negative, it simply means that the current flow is in the opposite direction from that indicated in the figure below.]
A) $I _ { 1 } = 0.5 \mathrm { amps } ; I _ { 2 } = 0.8 \mathrm { amps } ; I _ { 3 } = 1.3 \mathrm { amps }$
B) $I _ { 1 } = - 0.4 \mathrm { amps } ; I _ { 2 } = 1.6 \mathrm { amps } ; I _ { 3 } = 1.2 \mathrm { amps }$
C) $I _ { 1 } = 1.3 \mathrm { amps } ; I _ { 2 } = 6.4 \mathrm { amps } ; I _ { 3 } = - 1.3 \mathrm { amps }$
D) $I _ { 1 } = - 0.2 \mathrm { amps } ; I _ { 2 } = 1.3 \mathrm { amps } ; I _ { 3 } = 1.1 \mathrm { amps }$
E) $I _ { 1 } = - 4.9 \mathrm { amps } ; I _ { 2 } = 15.0 \mathrm { amps } ; I _ { 3 } = 8.2 \mathrm { amps }$