Let A=[α, -1; 6, β], α>0, such that det(A)=0 and α+β=1. If I denotes 2 × 2 identity matrix, then the matrix (I+A)⁸ is :

Solution

Maths sir · 4 steps

Find α and β first. Then A² = A, so every power of A is just A and (I + A)^8 collapses to I + 255A.

1. Find α and β. det A = αβ + 6 = 0 gives αβ = -6, and α + β = 1. So α and β are the roots of t² - t - 6 = 0, which are 3 and -2. Since α > 0, α = 3 and β = -2.
2. A is idempotent. By Cayley–Hamilton, A² - (tr A)A + (det A)I = O. Here tr A = 1 and det A = 0, so A² = A. Then Aᵏ = A for every k ≥ 1.
3. Binomial expansion. I commutes with A, so expand (I + A)⁸ binomially. Every term with a power of A becomes A, and the coefficients add up to C(8, 1) + … + C(8, 8) = 2⁸ - 1 = 255.

   (I + A)⁸ = I + 255A = (1, 0; 0, 1) + (765, -255; 1530, -510)

4. So. (I + A)⁸ = (766, -255; 1530, -509), which is option (b).

The idea to keep. When a 2×2 matrix has trace 1 and determinant 0, it satisfies A² = A. Then (I + A)ⁿ = I + (2ⁿ - 1)A, with no repeated multiplication needed.

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