Multiply matrix A by matrix B.

From the image, the matrices are:

Matrix A has dimensions $3 \times 2$ (3 rows, 2 columns).
Matrix B has dimensions $2 \times 3$ (2 rows, 3 columns).
Since the number of columns in A (2) equals the number of rows in B (2), the multiplication $AB$ is possible. The resulting matrix will have dimensions $3 \times 3$.

To find each element $(AB)_{ij}$, multiply the elements of the $i$-th row of A by the elements of the $j$-th column of B and sum the products.

For the first row of $AB$:
$(AB)_{11} = (1)(-3) + (3)(1) = -3 + 3 = 0$
$(AB)_{12} = (1)(-2) + (3)(3) = -2 + 9 = 7$
$(AB)_{13} = (1)(1) + (3)(1) = 1 + 3 = 4$

For the second row of $AB$:
$(AB)_{21} = (2)(-3) + (1)(1) = -6 + 1 = -5$
$(AB)_{22} = (2)(-2) + (1)(3) = -4 + 3 = -1$
$(AB)_{23} = (2)(1) + (1)(1) = 2 + 1 = 3$

For the third row of $AB$:
$(AB)_{31} = (3)(-3) + (4)(1) = -9 + 4 = -5$
$(AB)_{32} = (3)(-2) + (4)(3) = -6 + 12 = 6$
$(AB)_{33} = (3)(1) + (4)(1) = 3 + 4 = 7$

The final answer is $\boxed{\begin{bmatrix} 0 & 7 & 4 \\ -5 & -1 & 3 \\ -5 & 6 & 7 \end{bmatrix}}$.