Question ID: #1031
Let $A$ be a $3\times3$ real matrix such that $A^{2}(A-2I)-4(A-I)=O$ 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:
- (1) $12$
- (2) $20$
- (3) $76$
- (4) $4$
Solution:
Given the matrix polynomial equation[cite: 389]:
$$A^{2}(A-2I) – 4(A-I) = O$$
Expand the equation:
$$A^{3} – 2A^{2} – 4A + 4I = O$$
Express $A^3$ in terms of lower powers of $A$
$$A^{3} = 2A^{2} + 4A – 4I$$
Multiply the entire equation by $A$ to find $A^4$
$$A^{4} = 2A^{3} + 4A^{2} – 4A$$
Substitute the expression for $A^3$ back into this equation
$$A^{4} = 2(2A^{2} + 4A – 4I) + 4A^{2} – 4A$$
$$A^{4} = 4A^{2} + 8A – 8I + 4A^{2} – 4A$$
$$A^{4} = 8A^{2} + 4A – 8I$$
Multiply the equation by $A$ again to find $A^5$
$$A^{5} = 8A^{3} + 4A^{2} – 8A$$
Substitute the expression for $A^3$ one more time
$$A^{5} = 8(2A^{2} + 4A – 4I) + 4A^{2} – 8A$$
$$A^{5} = 16A^{2} + 32A – 32I + 4A^{2} – 8A$$
$$A^{5} = 20A^{2} + 24A – 32I$$
Compare this result with the given equation $A^{5} = \alpha A^{2} + \beta A + \gamma I$:
$$\alpha = 20$$
$$\beta = 24$$
$$\gamma = -32$$
Calculate the sum $\alpha+\beta+\gamma$
$$\alpha+\beta+\gamma = 20 + 24 – 32 = 12$$
Ans. (1)
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