# let $A$ be an $n\times n$ matrix. Show that $\det(A^{-1}) = \frac{1}{\det(A)}$

Let $A$ be a $n \times n$ matrix , and then show that $$\det(A^{-1}) = \frac{1}{\det(A)}.$$

Any tips on this one? basically I don't have a clue.

• If you have proved that for square matrices $A$ and $B$ of the same size, we have $\text{det}(AB)=\text{det}(A)\text{det}(B)$, it will be easy. If you have to prove the above multiplication law, not easy. – André Nicolas Apr 8 '14 at 7:16
• I guess I can just assume that det(AB) = det(A)det(B)(aka dont have to prove it) :) but im not sure how this helps me? – yyzzer1234 Apr 8 '14 at 7:20
• $A\cdot A^{-1} = I$. – 5xum Apr 8 '14 at 7:20

Hint: We know that $AA^{-1} = I$. We also have the fact that, in general $det(AB) = det(A)det(B)$. Can you see where to go from here?

• nope, not really and idk what you mean with AA^-1 = I? just matrix A * inverse matrix A = I ? :/ – yyzzer1234 Apr 8 '14 at 7:26
• @yyzzer1234, Try to combine the two facts that Kaj_H wrote down. In other words, try to calculate $\det(AA^{-1})$ – 5xum Apr 8 '14 at 7:38
• Yep, you're on the right track. – Kaj Hansen Apr 8 '14 at 7:39
• its just 1 isnt it? – yyzzer1234 Apr 8 '14 at 7:44
• Certainly. Now bring home the proof: Why is $det(A) = 1/det(A^{-1})$? – Kaj Hansen Apr 8 '14 at 7:46

From properties of the determinant, for square matrices $A$ and $B$ of equal size we have $$|AB|=|A||B|,$$ which means determinants are distributive. This means that the determinant of a matrix inverse can be found as follows: \begin{align} |I|&=\left|AA^{-1}\right|\\ 1&=|A|\left|A^{-1}\right|\\ \left|A^{-1}\right|&=\frac{1}{|A|}, \end{align} where $I$ is the identity matrix.

$$\\$$

$$\Large\color{blue}{\text{# }\mathbb{Q.E.D.}\text{ #}}$$

If $A$ is not defective, there exists an invertible matrix $P$ such that $D=P^{−1}AP$ that diagonalizes $A$.

The diagonal entries of $D^{-1}$ are the reciprocals of the entries of $D$ and since the determinante of a diagonal matrix is the product of all diagonal entries it follows that: $$\det(A^{-1}) = \frac{1}{\det(A)}.$$