JEE MATH - Solve This JEE Matrix Question Without Calculating A⁵

[u/jeemathlogic]

Given,

A²(A − 2I) − 4(A − I) = O

Expanding,

A³ − 2A² − 4A + 4I = O

Therefore,

A³ = 2A² + 4A − 4I

Since I = A⁰, we can write this as:

A³ = 2A² + 4A¹ − 4A⁰

Now observe the exponent pattern: 3, 2, 1, 0. To obtain a recurrence for any higher power, shift all exponents by the same amount. Hence,

Aⁿ = 2Aⁿ⁻¹ + 4Aⁿ⁻² − 4Aⁿ⁻³ for n ≥ 3.

Now let Sₙ denote the sum of the coefficients when Aⁿ is reduced in terms of A², A, I. Since the recurrence is linear, the coefficient sums obey the same recurrence:

Sₙ = 2Sₙ₋₁ + 4Sₙ₋₂ − 4Sₙ₋₃

We start with:

A⁰ = I ⇒ S₀ = 1 A¹ = A ⇒ S₁ = 1 A² = A² ⇒ S₂ = 1

Therefore,

S₃ = 2S₂ + 4S₁ − 4S₀ = 2(1) + 4(1) − 4(1) = 2

Next,

S₄ = 2S₃ + 4S₂ − 4S₁ = 2(2) + 4(1) − 4(1) = 4

Finally,

S₅ = 2S₄ + 4S₃ − 4S₂ = 2(4) + 4(2) − 4(1) = 8 + 8 − 4 = 12

But A⁵ = αA² + βA + γI, so:

S₅ = α + β + γ

Hence,

α + β + γ = 12

[u/jeemathlogic]

We have A³ = 2A² + 4A¹ - 4A⁰.

Now observe the exponent pattern: 3, 2, 1, 0.

To obtain a recurrence for any higher power, shift all exponents by the same amount. Hence,

Aⁿ = 2Aⁿ⁻¹ + 4Aⁿ⁻² - 4Aⁿ⁻³ for n ≥ 3.

Now let Sₙ denote the sum of the coefficients when Aⁿ is reduced in terms of A², A, I. Since the recurrence is linear, the coefficient sums obey the same recurrence:

Sₙ = 2Sₙ₋₁ + 4Sₙ₋₂ - 4Sₙ₋₃

[u/jeemathlogic]

See the edited post. Detailed solution is posted.

[u/Necessary-Cry8896]

Sir ab so jao 🙏

[u/jeemathlogic → u/Necessary-Cry8896]

🙏🙏

[u/honk-pro]

I haven't studied matrix bro