matrix

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：
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