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

Question

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.

Answer 1

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?

Answer 2

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: $$ \det(A^{-1}) = \frac{1}{\det(A)}. $$

Answer 3

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.

$$\\$$

---

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