This problem requires us to use a given polynomial matrix equation to reduce higher powers of a matrix into a linear combination of its lower powers. This is a common technique in matrix algebra, especially when dealing with matrices that satisfy their characteristic polynomial (Cayley-Hamilton Theorem).

1. Key Concepts and Formulas

• Polynomial Matrix Equation: A matrix $A$ can satisfy a polynomial equation, such as $P(A) = c_n A^n + c_{n-1} A^{n-1} + \dots + c_1 A + c_0 I = O$.
• Reduction of Powers: If a matrix $A$ satisfies a polynomial equation, we can express the highest power of $A$ in terms of lower powers. For instance, if $A^3 = pA^2 + qA + rI$, we can use this relation to express $A^4, A^5$, and so on, as linear combinations of $A^2, A,$ and $I$.
• Matrix Algebra Properties: Standard algebraic properties apply: $A \cdot I = I \cdot A = A$, $A \cdot O = O \cdot A = O$, and scalar multiplication commutes with matrix multiplication ($k(A+B) = kA+kB$).

2. Step-by-Step Solution

We are given the matrix equation $A^2(A-2 I)-4(A-I)=O$ and need to express $A^5$ in the form $\alpha A^2+\beta A+\gamma I$ to find $\alpha+\beta+\gamma$.

• Step 1: Simplify the Given Matrix Equation to Find a Reduction Formula for $A^3$

  First, expand the given matrix equation and rearrange it to express the highest power of $A$ (which is $A^3$) in terms of lower powers ($A^2, A, I$). This will be our primary reduction formula.

  $A^2(A-2 I)-4(A-I)=O$ Expand the terms using distributive property: $A^2 \cdot A - A^2 \cdot 2I - 4 \cdot A + 4 \cdot I = O$ $A^3 - 2A^2 - 4A + 4I = O$ Now, isolate $A^3$ to obtain the reduction formula: $A^3 = 2A^2 + 4A - 4I \quad (*)$ Reasoning: This equation is crucial because it allows us to replace any instance of $A^3$ with a simpler expression involving $A^2$, $A$, and $I$. This is the foundation for reducing higher powers of $A$.

• Step 2: Calculate $A^4$ in terms of $A^2, A, I$

  To find $A^4$, multiply both sides of the reduction formula $(*)$ by $A$: $A \cdot A^3 = A \cdot (2A^2 + 4A - 4I)$ $A^4 = 2A^3 + 4A^2 - 4A$ Now, substitute the expression for $A^3$ from $(*)$ back into this equation. This is the application of our reduction principle. $A^4 = 2(2A^2 + 4A - 4I) + 4A^2 - 4A$ Distribute the scalar and combine like terms: $A^4 = 4A^2 + 8A - 8I + 4A^2 - 4A$ $A^4 = (4A^2 + 4A^2) + (8A - 4A) - 8I$ $A^4 = 8A^2 + 4A - 8I \quad (**)$ Reasoning: We systematically multiplied by $A$ to reach the next higher power, then immediately used the reduction formula for $A^3$ to ensure that $A^4$ is also expressed solely in terms of $A^2$, $A$, and $I$.

• Step 3: Calculate $A^5$ in terms of $A^2, A, I$

  Similarly, to find $A^5$, multiply both sides of the equation for $A^4$ from $(**)$ by $A$: $A \cdot A^4 = A \cdot (8A^2 + 4A - 8I)$ $A^5 = 8A^3 + 4A^2 - 8A$ Again, substitute the expression for $A^3$ from $(*)$ into this equation to reduce the power: $A^5 = 8(2A^2 + 4A - 4I) + 4A^2 - 8A$ Distribute the scalar and combine like terms: $A^5 = 16A^2 + 32A - 32I + 4A^2 - 8A$ $A^5 = (16A^2 + 4A^2) + (32A - 8A) - 32I$ $A^5 = 20A^2 + 24A - 32I$ Reasoning: This completes the process of expressing $A^5$ in the desired form, $\alpha A^2+\beta A+\gamma I$, by consistently applying the reduction formula.

• Step 4: Determine $\alpha, \beta, \gamma$ and their Sum

  We are given that $A^5=\alpha A^2+\beta A+\gamma I$. By comparing our derived expression for $A^5$ with this general form: $A^5 = 20A^2 + 24A - 32I$ We can identify the coefficients: $\alpha = 20$ $\beta = 24$ $\gamma = -32$ Finally, calculate the sum $\alpha+\beta+\gamma$: $\alpha+\beta+\gamma = 20 + 24 + (-32)$ $\alpha+\beta+\gamma = 44 - 32$ $\alpha+\beta+\gamma = 12$

3. Common Mistakes & Tips

• Sign Errors: Be meticulous with signs during expansion and combination of terms. A single sign error can propagate and lead to an incorrect final answer.
• Distributive Property: Remember to distribute scalar multiples to all terms inside the parentheses. For example, $k(X+Y) = kX+kY$.
• Matrix Identity: Always include the identity matrix $I$ when a scalar term appears in a matrix polynomial equation (e.g., $4$ becomes $4I$).
• Iterative Substitution: The key is to repeatedly substitute the lowest power reduction formula (in this case, for $A^3$) whenever a higher power appears, ensuring the expression is always in terms of $A^2, A,$ and $I$.

4. Summary

This problem effectively demonstrates how to use a given polynomial relationship for a matrix to simplify expressions involving higher powers of that matrix. By deriving a reduction formula for $A^3$ from the initial equation, we systematically calculated $A^4$ and then $A^5$ as linear combinations of $A^2$, $A$, and $I$. The coefficients $\alpha, \beta, \gamma$ were found to be $20, 24,$ and $-32$ respectively, leading to a sum of $12$.

5. Final Answer

The calculated value of $\alpha+\beta+\gamma$ is $12$. The final answer is $\boxed{12}$, which corresponds to option (B).