Let A be a 3 × 3 real matrix such that A²(A-2 I)-4(A-I)=O, where I and O are the identity and null matrices, respectively. If A⁵=α A²+β A+γ I, where α, β, and γ are real constants, then α+β+γ is equal to :

Solution

Maths sir · 4 steps

Turn the condition into a formula for A³, then climb to A⁵ one power at a time.

1. Expand the condition. A²(A-2I)-4(A-I)=O expands to A³-2A²-4A+4I=O. So every A³ can be replaced by a quadratic in A.

   A³ = 2A² + 4A - 4I

2. Climb to fourth power. Multiply by A: A⁴ = 2A³ + 4A² - 4A. Replace A³: A⁴ = 2(2A²+4A-4I) + 4A² - 4A = 8A² + 4A - 8I.
3. Climb to fifth power. Multiply by A again: A⁵ = 8A³ + 4A² - 8A. Replace A³: A⁵ = 8(2A²+4A-4I) + 4A² - 8A = 20A² + 24A - 32I.
4. So. α = 20, β = 24, γ = -32, and α+β+γ = 20 + 24 - 32 = 12, which is option (b).

The idea to keep. When a matrix satisfies a cubic, any higher power reduces to pA²+qA+rI. Multiply by A, swap out A³ using the cubic, and repeat until you reach the power you need.

Answer from the NTA final answer key. Solution by PrepWiser.