Let A be an invertible matrix of order 2, then we have:

$A=
\begin{bmatrix}
a_{11} & a_{12} \\
a_{21} & a_{22}
\end{bmatrix}
$

We know that the inverse of a matrix, A is given by:

$A^{-1}=\frac{1}{det(A)}
\begin{bmatrix}
a_{22} & -a_{12} \\
-a_{21} & a_{11}
\end{bmatrix}
$

Therefore,

$det(A^{-1})=det(\frac{1}{det(A)}
\begin{bmatrix}
a_{22} & -a_{12} \\
-a_{21} & a_{11}
\end{bmatrix}
)$

$=\frac{1}{det(A)}
\begin{vmatrix}
a_{22} & -a_{12} \\
-a_{21} & a_{11}
\end{vmatrix}
$

$=\frac{1}{det(A)}[(a_{22}\times a_{11})-(-a_{12}\times -a_{21})]$

$=\frac{1}{det(A)}[(a_{22}\times a_{11})-a_{12}\times a_{21}]$

We know that the determinant of a 2x2 matrix is given by:

$det(A)=(a_{11}\times a_{22})-(a_{12}\times a_{21})$

Therefore,

$det(A^{-1})=\frac{1}{det(A)}[(a_{22}\times a_{11})-a_{12}\times a_{21}]=\frac{1}{det(A)}det(A)$

$=1$

Hence, det(A¯¹) is equal to 1.