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Matrices and Determinants question

2025 · 2 Apr · Shift 2 · Q31
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Matrices and Determinants question

2025 · 2 Apr · Shift 2 · Q31

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1
Let AAA be a 3×33 \times 33×3 real matrix such that A2(A−2I)−4(A−I)=OA^2(A-2 I)-4(A-I)=OA2(A−2I)−4(A−I)=O, where III and OOO are the identity and null matrices, respectively. If A5=αA2+βA+γIA^5=\alpha A^2+\beta A+\gamma IA5=αA2+βA+γI, where α,β\alpha, \betaα,β, and γ\gammaγ are real constants, then α+β+γ\alpha+\beta+\gammaα+β+γ is equal to :
  1. A
    76
  2. B
    12
  3. C
    4
  4. D
    20
View written solutionFree

Correct answer: B

  1. Given matrix equation

We are given A2(A−2I)−4(A−I)=O.A^2(A-2I)-4(A-I)=O.A2(A−2I)−4(A−I)=O.

Expand it: A3−2A2−4A+4I=O.A^3-2A^2-4A+4I=O.A3−2A2−4A+4I=O.

So, A3=2A2+4A−4I.(1)A^3=2A^2+4A-4I. \qquad (1)A3=2A2+4A−4I.(1)

  1. Find A4A^4A4 in terms of A2,A,IA^2,A,IA2,A,I

Multiply (1) by AAA: A4=2A3+4A2−4A.A^4=2A^3+4A^2-4A.A4=2A3+4A2−4A.

Now substitute A3=2A2+4A−4IA^3=2A^2+4A-4IA3=2A2+4A−4I from (1): A4=2(2A2+4A−4I)+4A2−4A.A^4=2(2A^2+4A-4I)+4A^2-4A.A4=2(2A2+4A−4I)+4A2−4A.

Simplify: A4=4A2+8A−8I+4A2−4AA^4=4A^2+8A-8I+4A^2-4AA4=4A2+8A−8I+4A2−4A A4=8A2+4A−8I.(2)A^4=8A^2+4A-8I. \qquad (2)A4=8A2+4A−8I.(2)

  1. Find A5A^5A5 in terms of A2,A,IA^2,A,IA2,A,I

Multiply (2) by AAA: A5=8A3+4A2−8A.A^5=8A^3+4A^2-8A.A5=8A3+4A2−8A.

Again use (1): A5=8(2A2+4A−4I)+4A2−8A.A^5=8(2A^2+4A-4I)+4A^2-8A.A5=8(2A2+4A−4I)+4A2−8A.

Simplify: A5=16A2+32A−32I+4A2−8AA^5=16A^2+32A-32I+4A^2-8AA5=16A2+32A−32I+4A2−8A A5=20A2+24A−32I.A^5=20A^2+24A-32I.A5=20A2+24A−32I.

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

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

  1. Check with options

The correct option is: 12\boxed{12}12​ which is Option B.

  1. Comparison with stored answer

Stored correct answer: B

Our derived answer is also B. Hence they agree.

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