Proof

**kaylakutie** · February 11th 2010, 08:10 PM

Not sure how to approach this one.

Prove $(A^{-1})^T = (A^T)^{-1}$ .

**danielomalmsteen** · February 11th 2010, 08:18 PM

that is equivalent to prove that

$({A}^{-1})^t \cdot {A}^t = I$

Owned by the transpose

$(A \cdot {A}^{-1})^t = I$

**Roam** · February 11th 2010, 09:17 PM

Originally Posted by kaylakutie

Not sure how to approach this one.

Prove $(A^{-1})^T = (A^T)^{-1}$ .

Hi there,

$A^T (A^{-1})^T = (A^{-1})^T A^T = I$

Keeping in your mind the fact that $I^T = I$ you get

$A^T(A^{-1})^T = (A^{-1}A)^T = I ^T = I$

$(A^{-1})^T A^T = (AA^{-1})^T = I^T = I$

This completes your proof.

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