Let $A$ be a $3 imes 3$ real matrix such that $A^2(A-2 I)-$ $4(\mathrm{~A}-\mathrm{I})=\mathrm{O}$, 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:

Question

Let $A$ be a $3 	imes 3$ real matrix such that $A^2(A-2 I)-$ $4(\mathrm{~A}-\mathrm{I})=\mathrm{O}$, 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:

Aicharya · Accepted Answer

$\begin{aligned} & A^3-2 A^2-4 A+4 I=0 \ & A^3=2 A^2+4 A-4 I \ & A^4=2 A^3+4 A^2-4 A \ & =2\left(2 A^2+4 A-4 Iight)+4 A^2-4 A \ & A^4=8 A^2+4 A-8 I \ & A^5=8 A^3+4 A^2-8 A \ & =8\left(2 A^2+4 A-4 Iight)+4 A^2-8 A \ & A^5=20 A^2+24 A-32 I \ & 	herefore \alpha=20, \beta=24, \gamma=-32 \ & 	herefore \alpha+\beta+\gamma=12\end{aligned}$

Let $A$ be a $3 \times 3$ real matrix such that $A^2(A-2 I)-$ $4(\mathrm{~A}-\mathrm{I})=\mathrm{O}$, where…

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