We are given \( \text{adj}(\text{adj}(A)) = A \). Using the property \( \text{adj}(A) = |A|^{n-1} A^{-1} \) for a square matrix \( A \) of order \( n \), we have \( \text{adj}(\text{adj}(A)) = |A|^{n-1} \text{adj}(A)^{-1} \). Substituting the property for \( \text{adj}(A) \), we get \( \text{adj}(\text{adj}(A)) = |A|^{n-1} (|A|^{n-1} A^{-1})^{-1} = |A|^{n-1} |A|^{-(n-1)} A = |A|^{n-1 - (n-1)} A = A \). This derivation appears to have an error.
Let's re-evaluate using the property \( \text{adj}(A) = |A|^{n-1} A^{-1} \).
Applying this property to \( \text{adj}(\text{adj}(A)) \):
\[
\text{adj}(\text{adj}(A)) = |\text{adj}(A)|^{n-1} (\text{adj}(A))^{-1}
\]
We know that \( |\text{adj}(A)| = |A|^{n-1} \). Substituting this:
\[
\text{adj}(\text{adj}(A)) = (|A|^{n-1})^{n-1} (\text{adj}(A))^{-1} = |A|^{(n-1)^2} (\text{adj}(A))^{-1}
\]
Also, \( (\text{adj}(A))^{-1} = (|A|^{n-1} A^{-1})^{-1} = (|A|^{n-1})^{-1} (A^{-1})^{-1} = |A|^{-(n-1)} A \).
Substituting this back:
\[
\text{adj}(\text{adj}(A)) = |A|^{(n-1)^2} |A|^{-(n-1)} A = |A|^{(n-1)^2 - (n-1)} A = |A|^{(n-1)(n-1-1)} A = |A|^{(n-1)(n-2)} A
\]
Given \( \text{adj}(\text{adj}(A)) = A \):
\[
|A|^{(n-1)(n-2)} A = A
\]
This implies \( |A|^{(n-1)(n-2)} = 1 \).
Consider the case where \( n=2 \). Then \( (n-1)(n-2) = 1 \cdot 0 = 0 \). So \( |A|^0 = 1 \), which is always true.
Consider the case where \( n eq 2 \). For \( |A|^{(n-1)(n-2)} = 1 \), we must have \( |A| = 1 \) or \( |A| = -1 \) if \( (n-1)(n-2) \) is even, or \( |A|=1 \) if \( (n-1)(n-2) \) is odd.
Let's use a more direct property: \( \text{adj}(\text{adj}(A)) = |A|^{n-2} A \).
Given \( \text{adj}(\text{adj}(A)) = A \), we have:
\[
|A|^{n-2} A = A
\]
If \( A \) is invertible, we can multiply by \( A^{-1} \):
\[
|A|^{n-2} = 1
\]
For this equation to hold true for any invertible matrix \( A \), we must have \( n-2 = 0 \), which means \( n=2 \).
If \( n=2 \), then \( |A|^0 = 1 \), which is true for any invertible matrix.
If \( A \) is not invertible, then \( |A| = 0 \). In this case, \( \text{adj}(A) \) is singular, and \( \text{adj}(\text{adj}(A)) \) requires careful consideration.
If \( |A|=0 \) and \( n>2 \), then \( \text{adj}(A) \) is the zero matrix, and \( \text{adj}(0) = 0 \). So \( \text{adj}(\text{adj}(A)) = 0 \). If \( \text{adj}(\text{adj}(A)) = A \), then \( A \) must be the zero matrix, and \( |A|=0 \).
If \( |A|=0 \) and \( n=2 \), then \( \text{adj}(A) \) is a scalar multiple of \( A^T \). If \( A \) is not the zero matrix, \( \text{adj}(A) \) is not the zero matrix, but \( |\text{adj}(A)| = 0 \). In this scenario, \( \text{adj}(\text{adj}(A)) = | \text{adj}(A) |^{2-2} \text{adj}(A) = 0 \cdot \text{adj}(A) = 0 \). So again \( A=0 \).
The property \( \text{adj}(\text{adj}(A)) = |A|^{n-2} A \) implies that if \( \text{adj}(\text{adj}(A)) = A \), then \( |A|^{n-2} = 1 \).
This condition is satisfied if \( n=2 \) or \( |A|=1 \).
The initial derivation which led to \( |A|=1 \) had an error in the manipulation of the inverse of the adjugate.
Using \( \text{adj}(\text{adj}(A)) = |A|^{n-2} A \), and given \( \text{adj}(\text{adj}(A)) = A \), we get \( |A|^{n-2} A = A \).
This implies \( |A|^{n-2} = 1 \).
Therefore, either \( n=2 \) or \( |A|=1 \).
The problem statement does not specify the order of the matrix \( n \). If \( n=2 \), then the determinant can be any value. If \( n eq 2 \), then the determinant must be 1.
Let's re-examine the initial steps:
We are given that \( \text{adj}(\text{adj}(A)) = A \).
Using the property \( \text{adj}(A) = |A|^{n-1} A^{-1} \).
Applying this property for the adjugate of the adjugate:
\[
\text{adj}(\text{adj}(A)) = |\text{adj}(A)|^{n-1} (\text{adj}(A))^{-1}
\]
We know \( |\text{adj}(A)| = |A|^{n-1} \).
\[
\text{adj}(\text{adj}(A)) = (|A|^{n-1})^{n-1} (\text{adj}(A))^{-1} = |A|^{(n-1)^2} (\text{adj}(A))^{-1}
\]
And \( (\text{adj}(A))^{-1} = (|A|^{n-1} A^{-1})^{-1} = (|A|^{n-1})^{-1} (A^{-1})^{-1} = |A|^{-(n-1)} A \).
Substituting this:
\[
\text{adj}(\text{adj}(A)) = |A|^{(n-1)^2} |A|^{-(n-1)} A = |A|^{(n-1)^2 - (n-1)} A = |A|^{(n-1)(n-2)} A
\]
Given \( \text{adj}(\text{adj}(A)) = A \):
\[
|A|^{(n-1)(n-2)} A = A
\]
For this to hold for a non-zero matrix \( A \), we must have \( |A|^{(n-1)(n-2)} = 1 \).
This implies that either \( n=2 \) (in which case the exponent is 0, and \( |A|^0 = 1 \) is always true for invertible A) or \( |A|=1 \) (if \( (n-1)(n-2) eq 0 \)).
Therefore, if \( n eq 2 \), then \( |A|=1 \). If \( n=2 \), the determinant can be any value.
The original derivation that concluded \( |A|=1 \) without considering the order \( n \) is incomplete.
The initial text attempts to use the property \( \text{adj}(A) = |A|^{n-1} A^{-1} \) in a way that incorrectly substitutes \( \text{adj}(A) \) for \( A \) within the formula for \( \text{adj}(\text{adj}(A)) \). The correct application of the property \( \text{adj}(X) = |X|^{n-1} X^{-1} \) for \( X = \text{adj}(A) \) leads to \( \text{adj}(\text{adj}(A)) = |\text{adj}(A)|^{n-1} (\text{adj}(A))^{-1} \). Using \( |\text{adj}(A)| = |A|^{n-1} \) and \( (\text{adj}(A))^{-1} = |A|^{-(n-1)} A \), we derive \( \text{adj}(\text{adj}(A)) = |A|^{(n-1)(n-2)} A \). Given \( \text{adj}(\text{adj}(A)) = A \), this implies \( |A|^{(n-1)(n-2)} = 1 \). Thus, if \( n eq 2 \), it must be that \( |A| = 1 \). If \( n = 2 \), this equation is always satisfied for any invertible matrix \( A \). The conclusion that \( |A|=1 \) is therefore only universally true when \( n eq 2 \).