Invertibilty of the inverse matrix

Question

Suppose I have a matrix $A$ which is invertible. Call $A^{-1}$ the inverse of $A$.

Is $A^{-1}$ invertible just because is the inverse of another matrix?

Suppose also I know the determinant of $A$, $\det(A)$. What can I say about $\det(A^{-1})$?

Answer 1

Suppose that $A$ is a square matrix.

$A$ is defined to be "invertible" if there exists another square matrix $B$ such that $AB=BA=I$.

We can show that this choice of $B$ is unique, that if $A$ is invertible there is one and only one $B$ such that $AB=BA=I$, and we call $B$ the "inverse of $A$" and we can choose to denote it instead as $A^{-1}$.

Now, notice that if we ask about whether or not there exists an inverse for $A^{-1}$, we have that $A^{-1}A=AA^{-1}=I$ so $A$ plays the role of the inverse of $A^{-1}$ and so yes $A^{-1}$ is indeed invertible.

As for the determinant, you have that $\det(A^{-1}) = \det(A)^{-1}$, which follows from the fact that $\det(AB)=\det(A)\det(B)$. Here we have $1=\det(I)=\det(AA^{-1})=\det(A)\det(A^{-1})$ so dividing by $\det(A)$ we get $\frac{1}{\det(A)}=\det(A^{-1})$

Answer 2

For the first question the answer is yes. $A$ is the inverse of $A^{-1}$.

$1=\det(I)=\det (AA^{-1})=\det(A)\det(A^{-1}).$ So $\det(A^{-1})=\dfrac 1 {\det(A)}$.

Answer 3

A matrix $X$ is invertible if there exists some matrix $Y$ such that $XY=YX=I$.

For $X=A^{-1}$, can you find any matrix such that $XY=YX=I$?

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For the determinant part, think about the fact that $\det(XY)=\det(X)\det(Y)$.