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JEE Mathematics 2025 Question with Solution

Let AA be a 3×33 \times 3 real matrix such that A2(A−2I)−4(A−I)=OA^2(A - 2I) - 4(A - I) = O, where II and OO are the identity and null matrices, respectively. If A3=αA2+βA+γIA^3 = \alpha A^2 + \beta A + \gamma I, where α\alpha, β\beta, and γ\gamma are real constants, then α+β+γ\alpha + \beta + \gamma is equal to:

  • A

    1212

  • B

    2020

  • C

    7676

  • D

    44

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: A2(A−2I)−4(A−I)=OA^2(A - 2I) - 4(A - I) = O

Find: α+β+γ\alpha + \beta + \gamma using the relation obtained for the higher power of AA.

From the given equation,

A2(A−2I)−4(A−I)=OA^2(A - 2I) - 4(A - I) = O

So,

A3−2A2−4A+4I=0A^3 - 2A^2 - 4A + 4I = 0

Hence,

A3=2A2+4A−4IA^3 = 2A^2 + 4A - 4I

Now compute the next powers.

A4=A⋅A3=A(2A2+4A−4I)=2A3+4A2−4AA^4 = A \cdot A^3 = A(2A^2 + 4A - 4I) = 2A^3 + 4A^2 - 4A

Substituting A3=2A2+4A−4IA^3 = 2A^2 + 4A - 4I,

A4=2(2A2+4A−4I)+4A2−4AA^4 = 2(2A^2 + 4A - 4I) + 4A^2 - 4A A4=(4A2+8A−8I)+(4A2−4A)A^4 = (4A^2 + 8A - 8I) + (4A^2 - 4A) A4=8A2+4A−8IA^4 = 8A^2 + 4A - 8I

Next,

A5=A⋅A4=A(8A2+4A−8I)=8A3+4A2−8AA^5 = A \cdot A^4 = A(8A^2 + 4A - 8I) = 8A^3 + 4A^2 - 8A

Again substituting A3=2A2+4A−4IA^3 = 2A^2 + 4A - 4I,

A5=8(2A2+4A−4I)+4A2−8AA^5 = 8(2A^2 + 4A - 4I) + 4A^2 - 8A A5=(16A2+32A−32I)+4A2−8AA^5 = (16A^2 + 32A - 32I) + 4A^2 - 8A A5=20A2+24A−32IA^5 = 20A^2 + 24A - 32I

Therefore,

α=20,β=24,γ=−32\alpha = 20, \quad \beta = 24, \quad \gamma = -32

So,

α+β+γ=20+24−32=12\alpha + \beta + \gamma = 20 + 24 - 32 = 12

Therefore, the correct option is A.

The solution contains a notation mismatch: the question states A3=αA2+βA+γIA^3 = \alpha A^2 + \beta A + \gamma I, but the worked solution correctly uses A5=αA2+βA+γIA^5 = \alpha A^2 + \beta A + \gamma I. Using the worked solution gives α+β+γ=12\alpha + \beta + \gamma = 12.

Repeated Reduction Using the Polynomial Relation

Given: A3−2A2−4A+4I=0A^3 - 2A^2 - 4A + 4I = 0

Find: α+β+γ\alpha + \beta + \gamma from the reduced form of A5A^5.

The key idea is to repeatedly replace every occurrence of A3A^3 using

A3=2A2+4A−4IA^3 = 2A^2 + 4A - 4I

Then,

A4=A⋅A3=2A3+4A2−4AA^4 = A \cdot A^3 = 2A^3 + 4A^2 - 4A A4=2(2A2+4A−4I)+4A2−4A=8A2+4A−8IA^4 = 2(2A^2 + 4A - 4I) + 4A^2 - 4A = 8A^2 + 4A - 8I

Now,

A5=A⋅A4=8A3+4A2−8AA^5 = A \cdot A^4 = 8A^3 + 4A^2 - 8A A5=8(2A2+4A−4I)+4A2−8A=20A2+24A−32IA^5 = 8(2A^2 + 4A - 4I) + 4A^2 - 8A = 20A^2 + 24A - 32I

Thus,

α+β+γ=20+24−32=12\alpha + \beta + \gamma = 20 + 24 - 32 = 12

Therefore, the correct option is A.

Common mistakes

  • Using the relation for A3A^3 and stopping there is incorrect because the constants α,β,γ\alpha, \beta, \gamma are obtained from the reduced form of the higher power used in the solution, namely A5A^5. Reduce powers step-by-step until the required power is reached.

  • Expanding A2(A−2I)−4(A−I)A^2(A - 2I) - 4(A - I) incorrectly is a common error. The correct expansion is A3−2A2−4A+4IA^3 - 2A^2 - 4A + 4I, because distributing −4-4 over (A−I)(A - I) gives −4A+4I-4A + 4I.

  • While computing A4A^4 or A5A^5, students often substitute A3A^3 incorrectly and combine like matrix terms wrongly. After substitution, collect the coefficients of A2A^2, AA, and II carefully before comparing with the required form.

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