Question

急急急，线性代数，已知矩阵A=[1，2，4，3]，求A100

Answer 1

det(A-sE)=(1-s)(3-s)-2*4 =s^2 -4s -5 = (s-5)(s+1)=0得到特征值为s=5/s=-1
取f(x)=x^100, g(x)=a+bx
f(5)=g(5)得到5^100 = a+5b
f(-1) =g(-1)得到 1 = a-b
得到b = (5^100-1)/4, a=b+1 = (5^100 +3/4)
A^100=aE+bA 带入即可