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 \[ A^{5} = \alpha A^{2} + \beta A + \gamma I, \] where \( \alpha, \beta, \gamma \) 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

Given a \(3 \times 3\) real matrix \(A\) satisfying the equation:

\[ A^2(A - 2I) - 4(A - I) = O \]

where \(I\) is the identity matrix and \(O\) is the null matrix. Determine the value of \(\alpha + \beta + \gamma\) given the expression:

\[ A^5 = \alpha A^2 + \beta A + \gamma I. \]

Underlying Principle:

This problem leverages matrix polynomial identities, analogous to the Cayley-Hamilton theorem, to express higher matrix powers in terms of lower ones.

The provided matrix equation can be manipulated to express \(A^3\) in terms of \(A^2, A,\) and \(I\). Subsequent multiplications will yield \(A^5\).

Detailed Solution:

Step 1: Simplify the given equation:

\[ A^2(A - 2I) - 4(A - I) = 0 \] \[ A^3 - 2A^2 - 4A + 4I = 0 \] Therefore, \[ A^3 = 2A^2 + 4A - 4I \]

Step 2: Calculate \(A^4\).

\[ A^4 = A \cdot A^3 \] Substitute the expression for \(A^3\):

\[ A^4 = A(2A^2 + 4A - 4I) \] \[ A^4 = 2A^3 + 4A^2 - 4A \]

Now, substitute the expression for \(A^3\) again:

\[ A^4 = 2(2A^2 + 4A - 4I) + 4A^2 - 4A \] \[ A^4 = 4A^2 + 8A - 8I + 4A^2 - 4A \] \[ A^4 = 8A^2 + 4A - 8I \]

Step 3: Calculate \(A^5\).

\[ A^5 = A \cdot A^4 \] Substitute the expression for \(A^4\):

\[ A^5 = A(8A^2 + 4A - 8I) \] \[ A^5 = 8A^3 + 4A^2 - 8A \]

Substitute the expression for \(A^3\):

\[ A^5 = 8(2A^2 + 4A - 4I) + 4A^2 - 8A \] \[ A^5 = 16A^2 + 32A - 32I + 4A^2 - 8A \] \[ A^5 = 20A^2 + 24A - 32I \]

Comparing this with \(A^5 = \alpha A^2 + \beta A + \gamma I\), we identify:

\[ \alpha = 20, \quad \beta = 24, \quad \gamma = -32 \]

Final Calculation & Result:

\[ \alpha + \beta + \gamma = 20 + 24 - 32 \] \[ \alpha + \beta + \gamma = 12 \]

Answer: 12