Let A be a $3 \times 3$ real matrix such that $A^{2}(A-2 I)- 4(A-I)=O$, 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:

JEE Main · 2025 · 02 Apr 2025 · Shift 2 · Matrices

Let A be a

3 \times 3

real matrix such that

A^{2} (A - 2 I) - 4 (A - I) = O

, where I and O are the identity and null matrices, respectively. If

A^{5} = α A^{2} + β A + γ I

, where

α, β

and

γ

are real constants, then

α + β + γ

is equal to:

Answer: (A)

Expanding the given relation,

A^{3} - 2 A^{2} - 4 A + 4 I = O

, so

A^{3} = 2 A^{2} + 4 A - 4 I

. Multiplying by

A

and reducing with this identity each time,

A^{4} = 2 A^{3} + 4 A^{2} - 4 A = 8 A^{2} + 4 A - 8 I

, and

A^{5} = 8 A^{3} + 4 A^{2} - 8 A = 20 A^{2} + 24 A - 32 I

. Matching with

A^{5} = α A^{2} + β A + γ I

gives

α = 20

β = 24

γ = - 32

, so

α + β + γ = 12

JEE Main · 2025 · 02 Apr 2025 · Shift 2 · Single correct option