Let $$A = [a_{ij}]$$ be a real matrix of order $$3 \times 3$$, such that $$a_{i1} + a_{i2} + a_{i3} = 1$$, for $$i = 1, 2, 3$$. Then, the sum of all entries of the matrix $$A^3$$ is equal to:

We are given a $$3 \times 3$$ real matrix $$A = [a_{ij}]$$ where each row sums to 1: $$a_{i1} + a_{i2} + a_{i3} = 1$$ for $$i = 1, 2, 3$$.

Let $$\mathbf{e} = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$$. The row-sum condition means $$A\mathbf{e} = \mathbf{e}$$.

Then $$A^2 \mathbf{e} = A(A\mathbf{e}) = A\mathbf{e} = \mathbf{e}$$, and similarly $$A^3 \mathbf{e} = \mathbf{e}$$.

This means each row of $$A^3$$ also sums to 1. The sum of ALL entries of $$A^3$$ is the sum of the three row sums = $$1 + 1 + 1 = 3$$.

The answer is $$3$$, which is Option C.