how to prove that det(adj(A))=(det(A))^n-1?

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We have square matrix A of order n x n. How can we prove that det(adj(A))=(det(A))^n-1 where det(A) is determinant of A and adj(A) is adjoint of matrix A.
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Very clear thank you, this helped me with my final exam in linear algebra 7 years after this video was posted !

meliemelie
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how about when the matrix A is singular?

mohammadobeidat
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finally someone capable of explaining how do you transform det(detA*I) in det(a)n thank you man my bro

biancabadescu
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The best explanation online.
thank you!

rakabadi
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This is great and really helped me figure out a problem, thanks!  Very clearly done!

MyJosef
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thanks a lotsss, you help me gain a different insight in matrix!!!!

wengjunfoong
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Bro ur video helped me even after 8 years ❤️

kaneki
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Thank you bro I used this proof for my linear alg quiz, it was so helpful. continue

mohammedhajomar
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I have to say thank you so much very very so so so so much I make me understand a lot because the book that I read now is not explain how it’s come and you make me understand, thank you. Ps I from Thailand.🙏🏻

diaryck
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thx mate can i contact you somewhere for questions?

jerrerock
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Can u prove that |detA| = 1 with A invertible?

thuanh
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Thank U so much for have been made this video!

muu
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Thanku if you are still reading the comments after 7 years

sanketgautam
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a step by step explanation, thank you! deserved the 0 dislikes xD

mysterynadi
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You are God, thanks for share this video (Y)

daw
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thank you very much, kralsın kardeşiiim

eminedemir