Question

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^3 = \alpha A^2 + \beta A + \gamma I $, where $ \alpha $, $ \beta $, and $ \gamma $ are real constants, then $ \alpha + \beta + \gamma $ is equal to:

• A. 12
• B. 20
• C. 76
• D. 4

Hint:

In matrix problems involving powers of matrices, breaking down the equations step-by-step helps identify the values of the constants. Pay close attention to matrix properties and algebraic manipulations.

Answer 1

The Correct Option is A

\[ A^3 - 2A^2 - 4A + 4I = 0 \] \[ A^3 = 2A^2 + 4A - 4I \] \[ A^4 = 2A^3 + 4A^2 - 4A \] \[ A^4 = 2(2A^2 + 4A - 4I) + 4A^2 - 4A \] \[ A^4 = 8A^2 + 8A - 8I \] \[ A^5 = 8A^3 + 4A^2 - 8A \] \[ A^5 = 8(2A^2 + 4A - 4I) + 4A^2 - 8A \] \[ A^5 = 20A^2 + 24A - 32I \] \[ \Rightarrow \alpha = 20, \, \beta = 24, \, \gamma = -32 \] \[ \Rightarrow \alpha + \beta + \gamma = 12 \]