Let A = bmatrix 1 & 0 & 0 0 & -1 & 0 0 & 0 & -1 bmatrix and B₀ = A²⁰²⁵ + A²⁰²⁴ , where > 0 . If B_n = adj (B_ n-1 ) for all n 1 and (B₃) = 2³² , then the value of is

Step-by-step solution

We are given A = bmatrix 1 & 0 & 0 0 & -1 & 0 0 & 0 & -1 bmatrix . Observe that A^2 = bmatrix 1 & 0 & 0 0 & 1 & 0 0 & 0 & 1 bmatrix = I . Since A^2 = I , we have A²⁰²⁴ = (A^2)¹⁰¹² = I and A²⁰²⁵ = A²⁰²⁴A = A . Substitute these into the expression for B₀ : B₀ = A + I = bmatrix + 1 & 0 & 0 0 & 1 - & 0 0 & 0 & 1 - bmatrix . The determinant of B₀ is: |B₀| = ( + 1)(1 - )^2 = ( + 1)( - 1)^2 . We are given the sequence B_n = adj (B_ n-1 ) . For a 3 3 matrix X , | adj (X)| = |X|^2 . Applying this repeatedly: |B₁| = |B₀|^2 |