Find the inverse of the matrix A = 1 & 1 3 & 1 .

The determinant of $A$ is given by $\det(A) = (1)(1) - (1)(3) = 1 - 3 = -2$.
The inverse matrix $A^{-1}$ is calculated as:

$$A^{-1} = \frac{1}{\det(A)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$$ $$A^{-1} = \frac{1}{-2} \begin{pmatrix} 1 & -1 \\ -3 & 1 \end{pmatrix}$$ $$A^{-1} = \begin{pmatrix} -\frac{1}{2} & \frac{1}{2} \\ \frac{3}{2} & -\frac{1}{2} \end{pmatrix}$$

The inverse of the matrix is $\boxed{\begin{pmatrix} -\frac{1}{2} & \frac{1}{2} \\ \frac{3}{2} & -\frac{1}{2} \end{pmatrix}}$.

The given equations are:
$x+y=7$
$3x+y=15$
This can be written as $AX=B$:

$$\begin{pmatrix} 1 & 1 \\ 3 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 7 \\ 15 \end{pmatrix}$$

$$X = A^{-1}B$$ $$\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} -\frac{1}{2} & \frac{1}{2} \\ \frac{3}{2} & -\frac{1}{2} \end{pmatrix} \begin{pmatrix} 7 \\ 15 \end{pmatrix}$$

$$x = \left(-\frac{1}{2}\right)(7) + \left(\frac{1}{2}\right)(15) = -\frac{7}{2} + \frac{15}{2} = \frac{8}{2} = 4$$ $$y = \left(\frac{3}{2}\right)(7) + \left(-\frac{1}{2}\right)(15) = \frac{21}{2} - \frac{15}{2} = \frac{6}{2} = 3$$

The point of intersection is $(x, y) = \boxed{(4, 3)}$.