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1:
To 金山伯: A-Level pure mathematics
matrix
2010-05-09 7:36 am
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.
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.
回答 (1)
2010-05-10 1:59 am
✔ 最佳答案
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.
收錄日期: 2021-04-24 09:28:44
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20100508000051KK01760