JEE MAIN - Mathematics (2025 - 2nd April Evening Shift - No. 6)
Let $A$ be a $3 \times 3$ real matrix such that $A^2(A-2 I)-4(A-I)=O$, 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 :
76
12
4
20
Explanation
$$\begin{aligned} & A^2(A-2 I)-4(A-I)=0 \\ & A^3-2 A^2-4 A+4 I=0 \end{aligned}$$
Multiply by $A$
$$\begin{aligned} & A^4=2 A^3+4 A^2-4 A \\ & A^4=2\left(2 A^2+4 A-4 I\right)+4 A^2-4 A \\ & A^4=8 A^2+4 A-8 I \end{aligned}$$
Multiply again by $A$
$$\begin{aligned} & \Rightarrow A^5=8 A^3+4 A^2-8 A \\ & \Rightarrow A^5=8\left(2 A^2+4 A-4 I\right)+4 A^2-8 A \\ & \Rightarrow A^5=20 A^2+24 A-32 I \end{aligned}$$
Comparing with $A^5=\alpha A^2+\beta A+\gamma I$
$$\begin{aligned} & \alpha=20, \beta=24, \gamma=-32 \\ & \therefore \alpha+\beta+\gamma=20+24-32 \\ & =44-32 \\ & =12 \end{aligned}$$
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