The solution to the system of equations is \[\left[ {\begin{array}{*{20}{c}} 2&5 \\ { – 4}&3 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {\text{x}} \\ {\text{y}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 2 \\ { – 30} \end{array}} \right]\] A. 6, 2 B. -6, 2 C. -6, -2 D. 6, -2

The correct answer is $\boxed{\text{(B)}}$.

To solve the system of equations, we can use Gaussian Elimination. First, we add $\frac{2}{3}$ times the first equation to the second equation:

$$\left[ {\begin{array}{{20}{c}} 2&5 \ { 0}&\frac{23}{3} \end{array}} \right]\left[ {\begin{array}{{20}{c}} {\text{x}} \ {\text{y}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 2 \ { – 30} \end{array}} \right]$$

Next, we multiply the second equation by $\frac{3}{23}$:

$$\left[ {\begin{array}{{20}{c}} 2&5 \ { 0}&1 \end{array}} \right]\left[ {\begin{array}{{20}{c}} {\text{x}} \ {\text{y}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 2 \ { – 15} \end{array}} \right]$$

Finally, we subtract $15$ times the second equation from the first equation:

$$\left[ {\begin{array}{{20}{c}} 2&0 \ { 0}&1 \end{array}} \right]\left[ {\begin{array}{{20}{c}} {\text{x}} \ {\text{y}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 0 \ { 0} \end{array}} \right]$$

This tells us that $\text{x}=0$ and $\text{y}=0$. Therefore, the solution to the system of equations is $\boxed{\text{(B)}}$.

Here is a step-by-step solution using Gaussian Elimination:

1. Add $\frac{2}{3}$ times the first equation to the second equation:

$$\left[ {\begin{array}{{20}{c}} 2&5 \ { 0}&\frac{23}{3} \end{array}} \right]\left[ {\begin{array}{{20}{c}} {\text{x}} \ {\text{y}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 2 \ { – 30} \end{array}} \right]$$

1. Multiply the second equation by $\frac{3}{23}$:

$$\left[ {\begin{array}{{20}{c}} 2&5 \ { 0}&1 \end{array}} \right]\left[ {\begin{array}{{20}{c}} {\text{x}} \ {\text{y}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 2 \ { – 15} \end{array}} \right]$$

1. Subtract $15$ times the second equation from the first equation:

$$\left[ {\begin{array}{{20}{c}} 2&0 \ { 0}&1 \end{array}} \right]\left[ {\begin{array}{{20}{c}} {\text{x}} \ {\text{y}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 0 \ { 0} \end{array}} \right]$$

This tells us that $\text{x}=0$ and $\text{y}=0$. Therefore, the solution to the system of equations is $\boxed{\text{(B)}}$.