Final answer:
The matrix power A⁻⁵⁰ for a rotation matrix A when θ = π/12 is equivalent to the original matrix A due to the periodic properties of the trigonometric functions involved.
Step-by-step explanation:
To find the inverse of a matrix raised to a power, we first need to understand the given matrix A = { cosθ - sinθ, sinθ cosθ }. However, the notation seems to be missing some elements for a standard 2x2 matrix. Assuming we are dealing with a rotation matrix A, where θ = π/12, the usual form would be A = { cosθ, -sinθ; sinθ, cosθ }. The question is asking for A⁻⁵⁰, which is the matrix A to the power of -50.
For a rotation matrix, An will correspond to rotating the plane by nθ. Thus, raising A to the power of -50 implies rotating the plane by -50θ. Since this rotation matrix is periodic with period 2π, we can simplify the problem by recognizing that -50θ mod 2π is actually equivalent to rotating by 2π/12, which brings us back to θ for a single rotation, or simply A1 = A.
Therefore, without providing the matrix multiplication and powers directly, we can conclude that A⁻⁵⁰ when θ = π/12 will be equivalent to the original matrix A, given the periodic properties of sine and cosine in the rotation matrix.