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

We are given a \(3 \times 3\) real matrix \(A\) such that:

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

where \(I\) and \(O\) are the identity and null matrices, respectively. We are to find \(\alpha + \beta + \gamma\) if

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

Concept Used:

This problem involves expressing higher powers of a matrix in terms of lower powers using polynomial relations (similar to the Cayley-Hamilton theorem).

The given condition is a polynomial identity satisfied by \(A\). We can rearrange it to express \(A^3\) in terms of \(A^2, A,\) and \(I\), then use successive multiplication to reach \(A^5\).

Step-by-Step Solution:

Step 1: Expand the given equation:

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

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

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

Now substitute \(A^3 = 2A^2 + 4A - 4I\):

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

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

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

Substitute \(A^3 = 2A^2 + 4A - 4I\):

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

Hence, comparing with \(A^5 = \alpha A^2 + \beta A + \gamma I\):

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

Final Computation & Result:

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

Answer: 12

Answer 2

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