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(A) Given the matrix equation: $A^2(A-2I) - 4(A-I) = O$ Expanding this,we get: $A^3 - 2A^2 - 4A + 4I = O$ Thus,$A^3 = 2A^2 + 4A - 4I$ Now,multiply by

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$12$

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(A) Given the matrix equation: $A^2(A-2I) - 4(A-I) = O$ Expanding this,we get: $A^3 - 2A^2 - 4A + 4I = O$ Thus,$A^3 = 2A^2 + 4A - 4I$ Now,multiply by $A$ to find $A^4$: $A^4 = 2A^3 + 4A^2 - 4A$ Substitute $A^3 = 2A^2 + 4A - 4I$: $A^4 = 2(2A^2 + 4A - 4I) + 4A^2 - 4A$ $A^4 = 4A^2 + 8A - 8I + 4A^2 - 4A = 8A^2 + 4A - 8I$ Now,multiply by $A$ to find $A^5$: $A^5 = 8A^3 + 4A^2 - 8A$ Substitute $A^3 = 2A^2 + 4A - 4I$ again: $A^5 = 8(2A^2 + 4A - 4I) + 4A^2 - 8A$ $A^5 = 16A^2 + 32A - 32I + 4A^2 - 8A$ $A^5 = 20A^2 + 24A - 32I$ Comparing this with $A^5 = \alpha A^2 + \beta A + \gamma I$,we get $\alpha = 20, \beta = 24, \gamma = -32$ Therefore,$\alpha + \beta + \gamma = 20 + 24 - 32 = 12$

Let $A$ be a $3 \times 3$ 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:

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