Question

设n阶方阵A满足A^3=2E,且B=A^2+2A-2E,证明 B是可逆矩阵 ,并求B^-1

Answer 1

设f(x) = x²+2x-2, g(x) = x³-2.
先求多项式u(x), v(x)使u(x)f(x)+v(x)g(x) = 1.
带余除法g(x) = (x-2)f(x)+6(x-1), f(x) = (x+3)(x-1)+1.
有6 = 6f(x)-(x+3)(g(x)-(x-2)f(x)) = (x²+x)f(x)-(x+3)g(x).
于是(x²+x)/6·f(x)-(x+3)/6·g(x) = 1.
将x = A代入上式, 由f(A) = B, g(A) = 0, 即得(A²+A)/6·B = E.
因此B可逆, 且B^(-1) = (A²+A)/6.