Let $$A$$ be a $$n \times n$$ matrix such that $$|A| = 2$$. If the determinant of the matrix $$\text{Adj}\left(2 \cdot \text{Adj}(2A^{-1})\right)$$ is $$2^{84}$$, then $$n$$ is equal to ______.

1. Simplify the given determinant expression

The given condition is:

$$\left\vert{} \text{Adj}\left(2 \cdot \text{Adj}(2A^{-1})\right) \right\vert{} = 2^{84}$$

Using the property of the adjoint determinant $$\left\vert{} \text{Adj}(B) \right\vert{} = \vert{}B\vert{}^{n-1}$$ for a matrix of order $$n \times n$$, we can rewrite the equation by setting $$B = 2 \cdot \text{Adj}(2A^{-1})$$:

$$\left\vert{} 2 \cdot \text{Adj}(2A^{-1}) \right\vert{}^{n-1} = 2^{84}$$

2. Evaluate the determinant inside the brackets

Using the scalar multiplication property $$\left\vert{} k \cdot B \right\vert{} = k^n \vert{}B\vert{}$$ for an $$n \times n$$ matrix, we expand the inner determinant:

$$\left\vert{} 2 \cdot \text{Adj}(2A^{-1}) \right\vert{} = 2^n \left\vert{} \text{Adj}(2A^{-1}) \right\vert{}$$

Applying the adjoint property again to $$\left\vert{} \text{Adj}(2A^{-1}) \right\vert{}$$ gives:

$$\left\vert{} \text{Adj}(2A^{-1}) \right\vert{} = \left\vert{} 2A^{-1} \right\vert{}^{n-1}$$

3. Simplify the inverse matrix determinant

Using the properties $$\left\vert{} k \cdot A^{-1} \right\vert{} = k^n \left\vert{} A^{-1} \right\vert{}$$ and $$\left\vert{} A^{-1} \right\vert{} = \frac{1}{\vert{}A\vert{}}$$, we evaluate $$\left\vert{} 2A^{-1} \right\vert{}$$:

$$\left\vert{} 2A^{-1} \right\vert{} = 2^n \cdot \frac{1}{\vert{}A\vert{}}$$

Since we are given that $$\vert{}A\vert{} = 2$$, this simplifies to:

$$\left\vert{} 2A^{-1} \right\vert{} = 2^n \cdot \frac{1}{2} = 2^{n-1}$$

4. Substitute back to find the final equation

Now substitute $$\left\vert{} 2A^{-1} \right\vert{} = 2^{n-1}$$ back into the adjoint expression:

$$\left\vert{} \text{Adj}(2A^{-1}) \right\vert{} = (2^{n-1})^{n-1} = 2^{(n-1)^2}$$

Next, substitute this into the expression for the entire inner determinant:

$$\left\vert{} 2 \cdot \text{Adj}(2A^{-1}) \right\vert{} = 2^n \cdot 2^{(n-1)^2} = 2^{n + n^2 - 2n + 1} = 2^{n^2 - n + 1}$$

Finally, substitute this back into our original equation:

$$(2^{n^2 - n + 1})^{n-1} = 2^{84}$$

$$2^{(n^2 - n + 1)(n - 1)} = 2^{84}$$

5. Solve for n

Equating the exponents gives:

$$(n^2 - n + 1)(n - 1) = 84$$

Let us test integer values for n:

If n = 5:

$$(5^2 - 5 + 1)(5 - 1) = (25 - 5 + 1)(4) = 21 \cdot 4 = 84$$

Since this satisfies the equation, the value of n is 5.