Let the matrix A=[0, 1, 0; 0, 0, 1; 1, 0, 0] and the matrix B₀=A⁴⁹+2 A⁹⁸. If Bₙ=A d j(Bₙ₋₁) for all n ≥ 1, then det(B₄) is equal to :

Solution

Maths sir · 4 steps

Since A³ = I, cut the huge powers down, find det B₀, then square the determinant at every adjoint.

1. Cut down powers. A moves the rows round in a cycle, and multiplying out gives A³ = I. Since 49 = 48 + 1 and 98 = 96 + 2, we get A⁴⁹ = A and A⁹⁸ = A², so B₀ = A + 2A².
2. Find det of B0. A² = (0, 0, 1; 1, 0, 0; 0, 1, 0), so B₀ = (0, 1, 2; 2, 0, 1; 1, 2, 0). Expanding along the first row gives det B₀ = 0 - 1(0 - 1) + 2(4 - 0) = 9 = 3².
3. Adjoint rule. For an n × n matrix, det(adj M) = (det M)ⁿ⁻¹. Here n = 3, so each adjoint squares the determinant.

   det B₄ = (det B₀)^(2⁴) = (3²)¹⁶ = 3³²

4. So. The determinant goes 3² → 3⁴ → 3⁸ → 3¹⁶ → 3³², so det B₄ = 3³², option (c).

The idea to keep. For a permutation matrix, find the power that gives I first. For repeated adjoints of a 3 × 3 matrix, the determinant gets squared each time, so k adjoints give (det M)^(2ᵏ).

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