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:
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:
Show Hint
- 12
- 20
- 76
- 4
The Correct Option is A
Approach Solution - 1
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
Approach Solution -2
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