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.

  • 3
    $\begingroup$ 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. $\endgroup$ – André Nicolas Apr 8 '14 at 7:16
  • $\begingroup$ 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? $\endgroup$ – yyzzer1234 Apr 8 '14 at 7:20
  • $\begingroup$ $A\cdot A^{-1} = I$. $\endgroup$ – 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?

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


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