Solve the system of equations using the matrix method: 3x + y = 9 and 2x − y = 1.

The coefficient matrix $A$, the variable matrix $X$, and the constant matrix $B$ are:

$$A = \begin{bmatrix} 3 & 1 \\ 2 & -1 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \end{bmatrix}, \quad B = \begin{bmatrix} 9 \\ 1 \end{bmatrix}$$

So, the matrix equation is:

$$\begin{bmatrix} 3 & 1 \\ 2 & -1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 9 \\ 1 \end{bmatrix}$$

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

$$\det(A) = (3)(-1) - (1)(2) = -3 - 2 = -5$$

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

$$A^{-1} = \frac{1}{-5} \begin{bmatrix} -1 & -1 \\ -2 & 3 \end{bmatrix}$$ $$A^{-1} = \begin{bmatrix} \frac{-1}{-5} & \frac{-1}{-5} \\ \frac{-2}{-5} & \frac{3}{-5} \end{bmatrix} = \begin{bmatrix} \frac{1}{5} & \frac{1}{5} \\ \frac{2}{5} & -\frac{3}{5} \end{bmatrix}$$

$$\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} \frac{1}{5} & \frac{1}{5} \\ \frac{2}{5} & -\frac{3}{5} \end{bmatrix} \begin{bmatrix} 9 \\ 1 \end{bmatrix}$$

Perform the matrix multiplication:

$$x = \left(\frac{1}{5}\right)(9) + \left(\frac{1}{5}\right)(1) = \frac{9}{5} + \frac{1}{5} = \frac{10}{5} = 2$$ $$y = \left(\frac{2}{5}\right)(9) + \left(-\frac{3}{5}\right)(1) = \frac{18}{5} - \frac{3}{5} = \frac{15}{5} = 3$$

Thus, the solution is $x=2$ and $y=3$.

The final answer is $x=2, y=3$.