JEE Main 2025 Apr 2 Shift 2, Mathematics Q20: Algebra Of Matrices

JEE Main2025Apr 2, Shift 2Mathematics

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:

A
$12$
B
$20$
C
$76$
D
$4$

Solution

Expanding the given relation: $A^{3} - 2 A^{2} - 4 A + 4 I = O$ , hence $A^{3} = 2 A^{2} + 4 A - 4 I$ .

Multiply by $A$ :

$A^{4} = 2 A^{3} + 4 A^{2} - 4 A = 2 (2 A^{2} + 4 A - 4 I) + 4 A^{2} - 4 A = 8 A^{2} + 4 A - 8 I$ .

Multiply again by $A$ :

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

So $α = 20, β = 24, γ = - 32$ , and $α + β + γ = 20 + 24 - 32 = 12$ .

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JEE Main 2025 Apr 2 Shift 2, Mathematics Q20: Algebra Of Matrices

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