Solve the system of equations using the matrix method: 2x + 3y = 13 and −3x + 2y = 0

$$\begin{bmatrix} 2 & 3 \\ -3 & 2 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 13 \\ 0 \end{bmatrix}$$

Here, $A = \begin{bmatrix} 2 & 3 \\ -3 & 2 \end{bmatrix}$, $X = \begin{bmatrix} x \\ y \end{bmatrix}$, and $B = \begin{bmatrix} 13 \\ 0 \end{bmatrix}$.

For a $2 \times 2$ matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$, the determinant is $ad - bc$.

$$\det(A) = (2)(2) - (3)(-3) = 4 - (-9) = 4 + 9 = 13$$

The inverse of a $2 \times 2$ matrix is $A^{-1} = \frac{1}{\det(A)} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$.

$$A^{-1} = \frac{1}{13} \begin{bmatrix} 2 & -3 \\ 3 & 2 \end{bmatrix}$$

$$X = A^{-1}B$$ $$\begin{bmatrix} x \\ y \end{bmatrix} = \frac{1}{13} \begin{bmatrix} 2 & -3 \\ 3 & 2 \end{bmatrix} \begin{bmatrix} 13 \\ 0 \end{bmatrix}$$ $$\begin{bmatrix} x \\ y \end{bmatrix} = \frac{1}{13} \begin{bmatrix} (2)(13) + (-3)(0) \\ (3)(13) + (2)(0) \end{bmatrix}$$ $$\begin{bmatrix} x \\ y \end{bmatrix} = \frac{1}{13} \begin{bmatrix} 26 + 0 \\ 39 + 0 \end{bmatrix}$$ $$\begin{bmatrix} x \\ y \end{bmatrix} = \frac{1}{13} \begin{bmatrix} 26 \\ 39 \end{bmatrix}$$ $$\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} \frac{26}{13} \\ \frac{39}{13} \end{bmatrix}$$ $$\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 2 \\ 3 \end{bmatrix}$$

Therefore, $x = 2$ and $y = 3$.

The solution is:

$$x=2, y=3$$