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
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![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](/uploads/image/z/1096897-49-7.jpg?t=%E8%AE%BEn%E7%BB%B4%E5%90%91%E9%87%8F%E7%BB%84A1+%2CA2+%2CA3%2CA4%2CA5%2C%E7%BA%BF%E6%80%A7%E6%97%A0%E5%85%B3%2CB1%3DA1%2BA2%2CB2%3DA2%2BA3%2CB3%3DA3%2BA4%2CB4%3DA4%2BA5%2CB5%3DA5%2BA1%2C%E8%AF%81%E6%98%8EB1B2B3B4B5%E7%BA%BF%E6%80%A7%E6%97%A0%E5%85%B3%EF%BC%882%EF%BC%89%E8%AE%BEN%E9%98%B6%E7%9F%A9%E9%98%B5A%E6%BB%A1%E8%B6%B3A%5E2-3A-2E%3D0%2C%E8%AF%81%E6%98%8E%E7%9F%A9%E9%98%B5A%E5%8F%AF%E9%80%86%E5%B9%B6%E6%B1%82%E5%87%BA%E5%85%B6%E9%80%86%E7%9F%A9%E9%98%B5A%5E-1)
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
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
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.
满意请采纳^_^