Let A be a 33 real matrix such that A^2(A-2I)-4(A-I)=O, where I,O are the identity and nul…

Question

Let $A$ be a $3	imes3$ real matrix such that $A^2(A-2I)-4(A-I)=O$, where $I,O$ are the identity and null matrices respectively. If $A^5=\alpha A^2+\beta A+\gamma I$ where $\alpha,\beta,\gamma$ are real constants, then $\alpha+\beta+\gamma$ is equal to:

Accepted Answer

$A^2(A-2I)-4(A-I)=0 \Rightarrow A^3-2A^2-4A+4I=0 \Rightarrow A^3=2A^2+4A-4I$.

Multiplying by $A$: $A^4=2A^3+4A^2-4A=2(2A^2+4A-4I)+4A^2-4A=8A^2+4A-8I$.

Multiplying by $A$ again: $A^5=8A^3+4A^2-8A=8(2A^2+4A-4I)+4A^2-8A=20A^2+24A-32I$.

So $\alpha=20,\ \beta=24,\ \gamma=-32$, and $\alpha+\beta+\gamma=20+24-32=12$.

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