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

Prove $(A^T)^{-1}$ = $(A^{-1})^T$ for any invertible matrix "A".

I actually don't know where to start - I do not think I can just apply index laws.

Any help is cool! Thanks.

Try taking the transpose of the equation $$AA^{-1}=I.$$

• using that, I got $(A^{-1})^T A^T = I$ and $A^T (A^{-1})^T = I$, then $(A^{-1})^T A^T = A^T (A^{-1})^T$, but what next? Sorry if I've been asking too much - been stoning at this for a while – Clinton May 12 '13 at 2:26
• @Clinton Which means that $(A^T)^{-1}=(A^{-1})^T$, by the definition of inverses. $(A^T)^{-1}$ is defined to be the matrix $B$ such that $BA^T=A^TB=I$. – Alex Becker May 12 '13 at 2:27
• @AlexBecker ah. thanks! – Clinton May 12 '13 at 2:28

first we define a "B" matrix like below : $$B = (A^{-1})$$ so : $$B^T=(A^{-1})^T *$$ and we can write : $$AB = I$$ then $$(AB)^T=I^T \textrm{ and } B^T A^T =I$$ From this equation we can say that : $$B^T=(A^T)^{-1}$$ and finally from * we can write :
$$B^T=(A^{-1})^T=(A^T)^{-1}$$

• Here is some MathJax tutorial – ASB Feb 22 '15 at 16:01

As we know $AA^{-1}=I$. Now taking transpose both sides, we get $$(AA^{-1})^T = I^T$$ which implies $$[ (A^{-1})^T ](A^T) = I$$ Now multiply both sides with $[(A^T)^{-1}]$ at right side, $$[(A^{-1})^T]{(A^T)(A^T)^{-1}} = (A^T)^{-1}$$ Here $(A^T)(A^T)^{-1}$ will form identity $I$, Since we know $AA^{-1} = I$, Therefore $$(A^{-1})^T = (A^T)^{-1}$$ Hence Proved!

Alternatively: $$A^{-1}=\frac{\text{adj} (A)}{|A|}$$ Transpose: \begin{align}(A^{-1})^T&=\left(\frac{\text{adj}(A)}{|A|}\right)^T=\\ &=\frac{(\text{adj} (A))^T}{|A|}=\\ &=\frac{\text{adj} (A^T)}{|A|}=\\ &=\frac{\text{adj} (A^T)}{|A^T|}=\\ &=(A^T)^{-1}.\end{align} The following properties of adjoint and transpose were used: \begin{align}(\text{adj} (A))^T&=\text{adj} (A^T);\\ (cA)^T&=c(A)^T;\\ |A|&=|A^T|.\end{align}