The number of elements in the set $$\{A = \begin{pmatrix} a & b \\ 0 & d \end{pmatrix} : a, b, d \in \{-1, 0, 1\}$$ and $$(I - A)^3 = I - A^3\}$$, where $$I$$ is $$2 \times 2$$ identity matrix, is _________.

We need to find the number of matrices $$A = \begin{pmatrix} a & b \\ 0 & d \end{pmatrix}$$ with $$a, b, d \in \{-1, 0, 1\}$$ satisfying $$(I - A)^3 = I - A^3$$.

Step 1: Simplify the condition.

Since $$I$$ commutes with every matrix, we expand:

$$(I - A)^3 = I - 3A + 3A^2 - A^3$$

Setting this equal to $$I - A^3$$:

$$I - 3A + 3A^2 - A^3 = I - A^3$$

$$-3A + 3A^2 = 0$$

$$A^2 = A$$

So $$A$$ must be idempotent.

Step 2: Compute $$A^2$$ and set up equations.

$$A^2 = \begin{pmatrix} a & b \\ 0 & d \end{pmatrix}\begin{pmatrix} a & b \\ 0 & d \end{pmatrix} = \begin{pmatrix} a^2 & ab + bd \\ 0 & d^2 \end{pmatrix}$$

Setting $$A^2 = A$$ gives three equations:

$$(i)\; a^2 = a, \quad (ii)\; d^2 = d, \quad (iii)\; b(a + d) = b$$

Step 3: Solve for $$a$$ and $$d$$.

From $$a^2 = a$$ with $$a \in \{-1, 0, 1\}$$:

$$a = -1 \implies 1 \neq -1 \quad (\text{fails})$$

$$a = 0 \implies 0 = 0 \quad (\text{works})$$

$$a = 1 \implies 1 = 1 \quad (\text{works})$$

So $$a \in \{0, 1\}$$, and similarly $$d \in \{0, 1\}$$.

Step 4: Solve for $$b$$ using equation (iii): $$b(a + d - 1) = 0$$.

Case 1: $$(a, d) = (0, 0)$$. Then $$a + d - 1 = -1 \neq 0$$, so $$b = 0$$. This gives 1 matrix.

Case 2: $$(a, d) = (1, 0)$$. Then $$a + d - 1 = 0$$, so $$b$$ can be any of $$\{-1, 0, 1\}$$. This gives 3 matrices.

Case 3: $$(a, d) = (0, 1)$$. Then $$a + d - 1 = 0$$, so $$b \in \{-1, 0, 1\}$$. This gives 3 matrices.

Case 4: $$(a, d) = (1, 1)$$. Then $$a + d - 1 = 1 \neq 0$$, so $$b = 0$$. This gives 1 matrix.

Total count: $$1 + 3 + 3 + 1 = 8$$.