Product of Matrix with Adjugate equals Determinant by Unit Matrix

Asked Nov 24 '20 at 17:21

Active Nov 24 '20 at 17:57

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Let $ A $ be a square matrix of order $ n $, $ \operatorname {adj} ( A ) $ be the adjugate of $ A $ and $ \det ( A ) $ be the determinant of $ A $. Then $$ \big( \operatorname {adj} ( A ) \big) A = \det ( A ) I = A \big( \operatorname {adj} ( A ) \big) \text , $$ where $ I $ is the unit mtrix of order $ n $.

Proofwiki has no proof, my textbook lacks it as well. Can anyone help?

edited Nov 24 '20 at 17:57

Ben Grossmann

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asked Nov 24 '20 at 17:21

Moran Reznik

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The only facts you need to prove this result are the laplace expansion of the determinant and the fact that a matrix with a repeated row/column has determinant $0$. – Ben Grossmann Nov 24 '20 at 17:57
How does the fact that a matrix with a repeated row/column has determinant 0 helps with this proof? – Moran Reznik Nov 25 '20 at 15:41
With this fact, we can see that the off-diagonal entries of $\operatorname{adj}(A)A$ will be zero – Ben Grossmann Nov 25 '20 at 16:28

Product of Matrix with Adjugate equals Determinant by Unit Matrix

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