Let A be a $$3 \times 3$$ real matrix such that $$A^2(A - 2I) - 4(A - I) = O$$, where I and O are the identity and null matrices, respectively. If $$A^5 = \alpha A^2 + \beta A + \gamma I$$, where $$\alpha, \beta$$ and $$\gamma$$ are real constants, then $$\alpha + \beta + \gamma$$ is equal to :

The given matrix equation is $$A^2(A-2I)-4(A-I)=O$$.

Expand the left-hand side:
$$A^2(A-2I)-4(A-I)=A^3-2A^2-4A+4I=O.$$

Hence $$A^3-2A^2-4A+4I=O \quad\Longrightarrow\quad A^3=2A^2+4A-4I \; -(1).$$

Step 1: Find $$A^4$$.
Multiply both sides of $$(1)$$ by $$A$$ on the left:
$$A^4=A\bigl(2A^2+4A-4I\bigr)=2A^3+4A^2-4A.$$

Replace $$A^3$$ using $$(1)$$ again:
$$A^4=2(2A^2+4A-4I)+4A^2-4A\\ \phantom{A^4}=4A^2+8A-8I+4A^2-4A\\ \phantom{A^4}=8A^2+4A-8I \; -(2).$$

Step 2: Find $$A^5$$.
Multiply $$(2)$$ by $$A$$ on the left:
$$A^5=A\bigl(8A^2+4A-8I\bigr)=8A^3+4A^2-8A.$$

Again substitute $$A^3$$ from $$(1)$$:
$$A^5=8(2A^2+4A-4I)+4A^2-8A\\ \phantom{A^5}=16A^2+32A-32I+4A^2-8A\\ \phantom{A^5}=20A^2+24A-32I.$$

Thus $$A^5=\alpha A^2+\beta A+\gamma I$$ with
$$\alpha=20,\qquad \beta=24,\qquad \gamma=-32.$$

Therefore $$\alpha+\beta+\gamma=20+24-32=12.$$

Hence the required value is $$12$$, which corresponds to Option A.