Let $$A = \begin{pmatrix} 0 & 1 & 2 \\ a & 0 & 3 \\ 1 & c & 0 \end{pmatrix}$$, where $$a, c \in \mathbb{R}$$. If $$A^3 = A$$ and the positive value of $$a$$ belongs to the interval $$(n-1, n]$$, where $$n \in \mathbb{N}$$, then $$n$$ is equal to _______.

Let $$A = \begin{pmatrix} 0 & 1 & 2 \\ a & 0 & 3 \\ 1 & c & 0 \end{pmatrix}$$ with $$A^3 = A$$. We need to find the positive value of $$a$$ and determine $$n$$ such that $$a \in (n-1, n]$$.

The condition $$A^3 = A$$ implies $$A^3 - A = 0$$, so $$A(A^2 - I) = 0$$. Hence the minimal polynomial of $$A$$ divides $$x^3 - x = x(x-1)(x+1)$$ and the eigenvalues of $$A$$ can only be $$0,1,-1$$.

Computing $$A^2$$ gives $$A^2 = \begin{pmatrix} 0 & 1 & 2 \\ a & 0 & 3 \\ 1 & c & 0 \end{pmatrix}\begin{pmatrix} 0 & 1 & 2 \\ a & 0 & 3 \\ 1 & c & 0 \end{pmatrix},$$ and the row computations yield Row 1: $$(0+a+2, 0+0+2c, 0+3+0) = (a+2,2c,3),$$ Row 2: $$(0+0+3, a+0+3c, 2a+0+0) = (3,a+3c,2a),$$ Row 3: $$(0+ac+0, 1+0+0, 2+3c+0) = (ac,1,2+3c).$$ Therefore $$A^2 = \begin{pmatrix} a+2 & 2c & 3 \\ 3 & a+3c & 2a \\ ac & 1 & 2+3c \end{pmatrix}.$$

Next $$A^3 = A^2\cdot A,$$ and multiplying out gives the entries Row 1: $$(a+2)(0)+2c(a)+3(1),\;(a+2)(1)+2c(0)+3(c),\;(a+2)(2)+2c(3)+3(0) = (2ac+3,\;a+2+3c,\;2a+4+6c),$$ Row 2: $$3(0)+(a+3c)(a)+2a(1),\;3(1)+(a+3c)(0)+2a(c),\;3(2)+(a+3c)(3)+2a(0) = (a^2+3ac+2a,\;3+2ac,\;6+3a+9c),$$ Row 3: $$ac(0)+1(a)+(2+3c)(1),\;ac(1)+1(0)+(2+3c)(c),\;ac(2)+1(3)+(2+3c)(0) = (a+2+3c,\;ac+2c+3c^2,\;2ac+3).$$

Setting $$A^3 = A$$ and comparing with $$A = \begin{pmatrix} 0 & 1 & 2 \\ a & 0 & 3 \\ 1 & c & 0 \end{pmatrix}$$ gives from the (1,1) entry $$2ac+3=0\implies ac=-\dfrac{3}{2},$$ from (1,2) $$a+2+3c=1\implies a+3c=-1,$$ from (1,3) $$2a+4+6c=2\implies a+3c=-1$$ (same), and from (2,1) $$a^2+3ac+2a=a\implies a^2+a-\dfrac{9}{2}=0.$$

Rewriting $$a^2+a-\dfrac{9}{2}=0$$ as $$2a^2+2a-9=0$$ leads to $$a=\dfrac{-2\pm\sqrt{4+72}}{4}=\dfrac{-2\pm\sqrt{76}}{4}=\dfrac{-2\pm2\sqrt{19}}{4}=\dfrac{-1\pm\sqrt{19}}{2},$$ so the positive value is $$a=\dfrac{-1+\sqrt{19}}{2}.$$ Since $$\sqrt{19}\approx4.359$$, we have $$a\approx1.68.$$

Because $$a\approx1.68\in(1,2]$$, it follows that $$n-1=1$$ and hence $$n=2$$. Therefore the answer is $$2$$.