设n维向量组A1 ,A2 ,A3,A4,A5,线性无关,B1=A1+A2,B2=A2+A3,B3=A3+A4,B4=A4+A5,B5=A5+A1,证明B1B2B3B4B5线性无关（2）设N阶矩阵A满足A^2-3A-2E=0,证明矩阵A可逆并求出其逆矩阵A^-1

证(1) 设 k1B1+k2B2+k3B3+k4B4+k5B5 = 0
则 k1(A1+A2)+k2(A2+A3)+k3(A3+A4)+k4(A4+A5)+k5(A5+A1)=0
所以 (k1+k5)A1+(k1+k2)A2+(k2+k3)A3+(k3+k4)A4+(k4+k5)A5=0.
由A1,A2,A3,A4,A5线性无关, 所以
k1+k5 = 0
k1+k2 = 0
k2+k3 = 0
k3+k4 = 0
k4+k5 = 0
因为行列式
1 0 0 0 1
1 1 0 0 0
0 1 1 0 0
0 0 1 1 0
0 0 0 1 1
= 2 ≠ 0
所以 k1=k2=k3=k4=k5=0
所以 B1,B2,B3,B4,B5线性无关.
证(2) 因为 A^2-3A-2E=0
所以 A(A-3E)/2 = E
所以 A 可逆, 且 A^(-1) = (A-3E)/2.
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