[線代] det

[mqazz1 (無法顯示)]

(f) If det(A)=0, then det(adj(A))=0
        true

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Assume the entries of a matrix A and its inverse A^(-1) are integers

show that det(A) = 正負1


請問這兩題要怎麼證呢?

謝謝

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推 air11       :AA^(-1)=I =>det(AA^(-1))=det(A)*det(A^(-1))=1
→ air11       :and detA=det(A^(-1)),=> (detA)^2=1
→ air11       :so detA=1 or -1
推 air11       :sorry, det(A^(-1))=1/det(A)
→ Sfly        :                           is also an integer
→ Sfly        :so |a|=1 or -1.
→ bineapple   :If det(adj(A)) =/= 0, then adj(A) is invertiblb.
推 bineapple   :Since A*adj(A)=det(A)I, we have adj(A)adj(adj(A))
→ bineapple   :=adj(det(A)I)=det(A)I=0*I=0. But adj(adj(A)) is
→ bineapple   :invertible by that adj(A) is invertible, we get a
→ bineapple   :countradiction.