matrix @ rdfvjur的部落格

標題:

matrix

發問:

1,Let A=(aij) and B=(bij) be square matrices of order 4 such that bij=(-1)^(i+j)aij. Prove that det A = det B 2, Let A and B be distinct nxn matrices with real numbers such that A^3=B^3 and A^2B=B^2A. Prove that A^2+B^2 is not invertible. 更新: To 金山伯: A-Level pure mathematics

最佳解答:

1. det(A)=∑(σ∈S_4) sign(σ) ∏(i=1,..,4) a(i,σi) where sign(σ) is the 1 when the permutation is even and -1 when the permutation is -1. det(B)=∑(σ∈S_4) sign(σ) ∏(i=1,..,4) b(i,σi) =∑(σ∈S_4) sign(σ) ∏(i=1,..,4) (-1)^(i+σi ) a(i,σi) =∑(σ∈S_4) sign(σ) (-1)^[ ∑(i=1,..,4) (i+σi) ] ∏(i=1,..,4) a(i,σi) =∑(σ∈S_4) sign(σ) (-1)^ [ 2 ∑(i=1,..,4) i) ] ∏(i=1,..,4) a(i,σi) =∑(σ∈S_4) sign(σ) (-1)^20 ∏(i=1,..,4) a(i,σi) =∑(σ∈S_4) sign(σ) ∏(i=1,..,4) a(i,σi) = det(A) 2. Suppose AA+BB is invertible As (AA + BB)B = AAB + BBB = BBA + AAA = (AA + BB)A Multiply (AA+BB)^(-1) to both sides of eqation, Then A = B contradicting the fact that A and B are distinct matrices. Hence AA+BB is not invertible. 2010-05-09 18:02:15 補充： Are you studying A-Level pure mathematics or junior level undergraduate linear algebra? The approach may be different.