Matrix Equation Simplification

The given matrix equation is $A^2(A-2I) - 4(A-I) = O$. Expanding this, we get:

$A^3 - 2A^2 - 4A + 4I = O$

This implies that the matrix $A$ satisfies the polynomial $P(x) = x^3 - 2x^2 - 4x + 4$. We can rearrange this to express $A^3$ in terms of lower powers:

$A^3 = 2A^2 + 4A - 4I$

Calculating Higher Powers of A

We need to find $A^5$. We can do this step-by-step using the relation for $A^3$.

1. Calculate $A^4$:

   $A^4 = A \cdot A^3 = A(2A^2 + 4A - 4I)$ $A^4 = 2A^3 + 4A^2 - 4A$ Substitute the expression for $A^3$: $A^4 = 2(2A^2 + 4A - 4I) + 4A^2 - 4A$ $A^4 = 4A^2 + 8A - 8I + 4A^2 - 4A$ $A^4 = 8A^2 + 4A - 8I$

2. Calculate $A^5$:

   $A^5 = A \cdot A^4 = A(8A^2 + 4A - 8I)$ $A^5 = 8A^3 + 4A^2 - 8A$ Substitute the expression for $A^3$: $A^5 = 8(2A^2 + 4A - 4I) + 4A^2 - 8A$ $A^5 = 16A^2 + 32A - 32I + 4A^2 - 8A$ $A^5 = 20A^2 + 24A - 32I$

Determining Coefficients and Sum

We are given the form $A^5 = \alpha A^2 + \beta A + \gamma I$. Comparing this with our calculated $A^5$:

$20A^2 + 24A - 32I = \alpha A^2 + \beta A + \gamma I$

By matching the coefficients of $A^2$, $A$, and $I$, we find:

• $\alpha = 20$
• $\beta = 24$
• $\gamma = -32$

The question asks for the sum $\alpha + \beta + \gamma$.

$\alpha + \beta + \gamma = 20 + 24 + (-32)$ $\alpha + \beta + \gamma = 44 - 32$ $\alpha + \beta + \gamma = 12$